potential due to an electric dipole

possible to calculate the electric potential. For more information:http://www.7activestudio.com info@7activestudio.comhttp://www.7activemedical.com/ info@7activemedical.comhttp. Accordingly we can find the electric field component wise. Calculate the field due to an electric dipole of length 10 cm and consisting of charges of - plus 100 C at appoint 20cm from each charge? Derive the expression for the potential energy of a system of two charges in the absence of the external electric field. Reason Any point on the perpendicular bisector of the dipole axis is equidistant from the the two equal and opposite charges. We defined the electrostatic potential in lecture EML-4, as the work done against the Coulomb electrostatic force experienced by a test charge (of unit magnitude and positive sign) from infinite distance to a given reference point r. In the scenario that no charge distributions are present at infinite dimensions we obtain an expression for the potential of a single charge Q (that produces the field and the potential in the first place). . The electric dipole moment associated with two equal charges of opposite polarity separated by a distance, d is defined as the vector quantity having a magnitude equal to the product of the charge and the distance between the charges and having a direction from the negative to the positive charge along the line between the charges. The potential due to an electric dipole important points falls as 1/r 2 and the potential due to a single point charge falls as 1/ r. The potential due to the dipole r falls is much more than a monopole (point charge). The gradient operator ( a vector operator) which appears on the right hand side (and acts on scalar as well as vector fields, albeit in different ways, but here on the scalar field ) is given as: , in spherical polar coordinate system. potential due to electric dipole is V = p.r / 4(pi)(epsilon)r3 show the tangential component of the electrical field is = psin(theta). We also mentioned the superposition principle where if two source charges (of equal magnitude q) are present in their respective locations, their potential would be simply the sum of the potential due to the individual source charges. Thus the magnitude of the dipole moment in our example here isp = = qdand its direction isi.e. In the limit d << r (i.e. ) (i) The potential due to an electric dipole Important points falls as 1/r2 and the potential due to a single point charge falls as 1/r. So the spherical coordinate form of the dipole electric field now becomes: . The amount of work done in moving a unit positive charge from infinity to a given point is known as: NEET Repeater 2023 - Aakrosh 1 Year Course, Relation Between Electric Field and Electric Potential, Dielectric Polarization and Electric Dipole Moment, Potential Energy of Charges in an Electric Field, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. The net force on an electric dipole present in a uniform electric field is zero as forces act in opposite directions for oppositely signed charges. An electric dipole: a system of 2 equal and oppositely signed charge. The electric potential on the axis of the electric dipole: Let's logically derive an expression for electric potential due to an electric dipole, at a point far away from it. on the electric fence for themselves." Take the origin at the centre of the dipole. Electric potential due to a Dipole (V) Suppose there are two charges, "-q", placed at A, and "+q" placed at B, separated by a distance "d", forming a dipole. 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Let us assume that the potential at infinity will be zero. (i) Equipotential surfaces due to single point charge are concentric sphere having charge at the centre. (The electric field must be uniform for this assertion). Question . Get Most Demanded Physics Notes, Chemistry notes and Biology notesfor 11th, 12th, NEET & IIT Subscribe E-Books (on Our App https://dhxel.courses.store. But makes angle +(/2) with (i.e. ) Potential due to an electric dipole. We will add some content here if time permits in the near future. This leads to the final expression for the coordinate independent form of the electric field of the electric dipole moment: . Electric potential: The amount of effort required to transmit a unit positive charge from one location inside an electric field to another without causing acceleration. (ii) potential energy of the dipole, if the dipole has charges of 8nC. The rest of them have to pee Cosine law gives magnitude of a resultant vector from its components, by using the dot product of the resultant vector with itself as the square of its magnitude. Electric potential isthe amount of work needed to move a unit charge from a reference point to a specific point against anelectricfield Let an electric dipole consist of two equal and opposite point chargesqat Aandqat B separated by a small distance AB2a with centre at O The dipole momentpq2a We will calculate potential at any point P . its useful only where the spherical polar coordinate system is purported to be implemented. All articles in this series will be foundhere. With a surge in distance from electric dipole, the effects of positive and negative charges will nullify each other. is the angle between vector and radial unit vector so . (ii) In constant electric field along z-direction, the perpendicular distance between equipotential surfaces remains same. Suppose the midpoint of AB is "O". >> Potential due to Electric Dipole >> The electric potential at a point on the. Will Rogers, Dr. C. L. Davis The Electric potential due to a dipole at any point P, such that OP = r will be: V = 1 4 p c o s r 2 Case 1: If = 90 Electric potential = V = 0 It can also be shown that the potential energy of an electric dipole in a uniform electric field is given by . 7 Active driving force \"The Joy of Happy Learning\" -- is what makes difference from other digital content providers. The following diagram shows the spherical polar coordinate system of r, and juxtaposed on the Cartesian coordinate system of x, y, z for any vector r. The dipole moment vector p is aligned along the positive z-axis making an angle with the reference vector r. From the given expression above for the electrostatic potential of an electric dipole moment we can calculate the electrostatic or electric field created by the dipole. For one charge, the -ve one the separation vector is: and for the other the positive the separation vector is: . Welcome to- #OpenYourMindwithMurugaMPJoin Our Membership:https://www.youtube.com/channel/UCVJc7bS5lP8OrZGd7vs_yHw/join Remember to SUBSCRIBE my channel a. To use this online calculator for Electric Potential of Dipole, enter Electric Dipole Moment (p), Angle between any two vectors () & Magnitude of position vector (r) and hit the calculate button. We consider Student needs, Lecturer needs and College needs in designing the 3D \u0026 2D Animated Video Lectures. For more information:http://www.7activestudio.com info@7activestudio.comhttp://www.7activemedical.com/ info@7activemedical.comhttp://www.sciencetuts.com/Contact: +91- 9700061777, 040-64501777 / 65864777 7 Active Technology Solutions Pvt.Ltd. Calculate the (i) magnitude of the electric field. As you can see in the diagram above, we have taken the z-axis to be the direction of the dipole moment vector p. The electric field is the negative potential gradient given by the expression, which we have discussed in detail here. Sort by: Questions Tips & Thanks Video transcript Therefore, If we choose the origin at mid point of the dipole then the positive charge +q lies at and the negative charge -q lies at . To log in and use all the features of Khan Academy, please enable JavaScript in your browser. "There are three kinds of men. You can see the explanation how this is derived in the following show/hide box. Enter your email address to follow this blog and receive updates by email. The point is r distance . Physics Department An electric dipole of length 4 cm, when placed with its axis making an angle of 60 with a uniform electric field experiences a torque of 4 3 Nm. Just as we calculated the electric field due to a dipole it is a dipole) can easily be written; . Just as we calculated the electric field due to a dipole it is possible to calculate the electric potential. + Click to see more about the cosine law. Click to share on Pocket (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on WhatsApp (Opens in new window), Click to share on Reddit (Opens in new window), Click to email a link to a friend (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on scoopit (Opens in new window), Click to share on Skype (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Twitter (Opens in new window). The one that learns by reading. It means field due to electric dipole will be same at every point on the surface of a right circular cylinder imagined with the electric dipole as the axis. In the meanwhile if you cant wait and you need some of these concepts at the earliest, here is a slide-share presentation I had made roughly 5 years ago that consists of some of the things an undergrad needs:Electricity and Magnetism slides. However, in this case it is convenient to use polar co-ordinates. | Physics | Khan Academy - YouTube Let's logically derive an expression for electric potential due to an. Where should a third charge -q be placed between them, so that the electrical potential energy of the system is minimum? this case it is convenient to use polar co-ordinates. is an educational 3D digital content provider for K-12. The few who learn by observation. Click on image to go to Gravatar profile of founder. 1) In electric potential due to point charge we are calculating potential of single point at a point in the space which is separated by small distance. If you're seeing this message, it means we're having trouble loading external resources on our website. 2) Electric potential due to point charge is depend on the distance r. 3) Electric potential due to point charge is given by: V = Q/4 0 r. We would like to determine electrostatic potential and electrostatic field at a point referenced by position vector r. + Click if you wanna see the explanation how this is done. Potential due to point charge. Creative Commons Attribution/Non-Commercial/Share-Alike. At what points on the line joining of the two charges will be the electric potential zero? For that we first off all need the following transformation rules (for unit vectors) in Spherical Polar Coordinate and Rectangular Cartesian Coordinate system. As you recall, dipole is a point charge system which consists of two point charges with equal magnitudes and opposite signs, separated from one another by a very small distance. An electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point inside an electric field without producing acceleration. The cosine law gives magnitude of a resultant vector from its components, by using the dot product of the resultant vector with itself as the square of its magnitude. r 2 = P R R C = O R + O C = r + a c o s . Charge m is mass, charge v is speed, and charge m is mass. E = k 2qcos r2 ^i (2) (2) E = k 2 q cos r 2 i ^. In this Physics video for Class 12 in Hindi we derived the expression for electric potential at a point due to a an electric dipole. We can cast this expression into its coordinate independent form so that its useful in any arbitrary coordinate system for vector analysis. The radial component is , the polar angle component is and the azimuthal component is . Science Class 12 Physics (India) Electrostatic potential and capacitance Potential due to an electric dipole and other systems of charges. We also customise the content as per your requirement for companies platform providers colleges etc . We are carrying a huge 3D Digital Library ready to use. The electric potential due to an electric dipole at a point on the perpendicular bisector of the dipole axis is $$0$$. Calculating torque on an electric dipole in a uniform electric field: . Note, unit vector transformation rules between Spherical polar and Cartesian system: . But there is a net amount of torque: , which does not vanish in a uniform electric field. Potential due to dipole (logical derivation) | Electric potential & cap. Electric Potential Electric Potential of Dipole Dielectric Constant Capacitor & Capacitance Parallel Plate Capacitor Spherical Capacitor Numerical Problems Current Electricity Electric Current Ohm's Law Specific Resistance Thermocouple Electric Circuit Electromotive Force Combination of Cells Kirchoff's Law Potentiometer Moving Coil Galvanometer Lectures on Electricity and Magnetism new series of lectures EML 9. The electrostatic potential for such an arrangement (i.e. So, the above equation becomes, Where r is the unit vector along the vector O R. From the above equation, we can see that the electric potential due to electric dipole does not only depend on r but also depends . An electric dipple has dipple moment 4 1 0 1 1 and the two charges have magnitudes 4nC each. An electric dipole is a system of two equal and oppositely signed charges +q and -q separated by a distance d. Atomic phenomena can often be modelled in terms of dipoles, so its important to study the dipole in detail. Electric potential due to an electric dipoleConsider an electric dipole as shown in Fig. p== qd. Assuming no azimuthal variation we may write, When r >> a, then r 1, r 2 and r are approximately equal and r 2 - r 1 = 2acos. Since potential is related to the work done by the field, electrostatic potential also follows thesuperposition principle. Click on link to left or search for menu E AND M BASICS on top. eg the current lecture will be namedEML 9 . half yearly exam | electric potential due to dipole| vidyut dvidhruv ke karn vibhavyour queris:1.vidyut dvidhruv ke karn vibhav2.vidyut dvidhruv ke karn kisi. email: c.l.davis@louisville.edu. So, , this leads to: . Assume the drops to be spherical. In vector form if the unit vector towards x-direction is ^i i ^, the above equation is. Positive and negative point charges of equal magnitude are kept at $\left(0,0,\dfrac{a}{2}\right)$ and $\left(0,0,\dfrac{-a}{2}\right)$, respectively. An electric dipole is a system of two equal and oppositely signed charges +q and -q separated by a distance d. Atomic phenomena can often be modelled in terms of dipoles, so its important to study the dipole in detail. If we choose the origin at mid point of the dipole then the positive charge +q lies at and the negative charge -q lies at . Thus, the potential due to the dipole is the sum of potentials due to the charges q and -q. where r 1 and r 2 are the distances of the point P from q and -q, respectively. The electric potential energy is determined by the distance between charges and the strength of the electric field. Here is how the Electric Potential of Dipole calculation can be explained with given input values -> 0.128003 = [Coulomb]*12*cos(1.5707963267946)/(0 . The resultant electric field then due to the dipole oriented along the z-axis ( = 0) . Lectures on Electricity and Magnetism new series of lectures EML 9, All articles in this series will be found, Click on link to left or search for menu E AND M BASICS on top. This lecture was delivered to honors students on 15th Feb 2017. ($90V$). There are other slides on different topics at that account of mine onslideshare.net (such as; Introduction to Quantum Mechanics , and these are quite well received by the community for their usefulness). Let us evaluate the potential V at a point P at a distance r from th. The dipole moment vector p is aligned along the positive z-axis making an angle with the reference vector r in the spherical polar coordinate system of r, and juxtaposed on the Cartesian coordinate system of x, y, z for any vector r. Note, unit vector transformation rules between Spherical polar and Cartesian system, Calculating torque on an electric dipole in a uniform electric field. Lecture-10, 11. We define theelectric dipole momentvectororpas a vector with the magnitude of product of charge of one pole (qhere) into the distance between the poles (dhere), whose direction is given by a unit vector running from the negative pole towards the positive pole. Laplace And Poisson Equation, Lecture 5, 6, 7 And 8. The vector r makes an angle of with the vector d (see Diagram 1 above) so we can use the simple results of vector algebra called cosine law (see diagram and expression below), and write; . You know the electric field magnitude E E from the above equation and therefore, the total electric field is. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. i.e. Now the potential due to the dipole takes the form: . This series is on Electricity and Magnetism and bears the name sakeElectricity and Magnetism Lecturesand the number of the lecture will be appended to the end to reflect the same. This potential (r) is given as . The electric potential due to an electric dipole can be measured at different points: The electric potential on the axis of the electric dipole; The electric potential on the equatorial line of the electric dipole; The electric potential at any point of the electric dipole; 1. Its based on the expression for potential of a single charge and superposition principle. This article belongs to a group of lectures I intend to prepare for their online dissemination these were delivered in a physical format, beginning with hand written notes that were delivered in a classroom full of students. See Diagram 1 above for this assertion. Potential of an electric dipole. Equipotential surface is a surface which has equal potential at every Point on it. Two charges $5\times {{10}^{-8}}C$ and $-3\times {{10}^{-8}}C$ are placed $16cm$ apart. Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App.Download the App from Google. The torque N is calculated as follows. As the distance increases from electric dipole, the effects of positive and negative charges nullify each other. . University of Louisville Potential due to dipole. Since the charges are no more present at origin we need to take the relative vector called separation vector as the distance at which we are calculating field and potential (this vector is usually expressed with the curly r symbol, but we are technically not equipped to do so here). What will be the potential if all the charged drops are made to combine to form one large drop? #ssp #cbse2021#sachinsirphysicsin this video i have explained about concept of potential and potential differencepotential due to point charge potential due . We then replaced Q by q, where we understand that q is no more the test charge (reference charge) but the source charge that produces the field and the potential. Potential due to dipole (logical derivation) About Transcript Let's logically derive an expression for electric potential due to an electric dipole, at a point far away from it. A dipole is a pair of opposite charges with equal magnitudes separated by a distance, d. Electrostatic Potential Energy Of Different Charge Configurations. Thus the potential due to the dipoler falls faster than that due to a monopole (point charge). (b) If P is far from the dipole, the lines of lengths r (+) and r (-) are approximately parallel to the line of length r, and the E = k2qcos r2 (1) (1) E = k 2 q cos r 2. This is a coordinate dependent form. Two point charges 2q and 8q are placed at a distance r apart. as we mentioned in the note above. However, in The electric potential energy of a dipole can be described in three steps. But there is a net amount of torque. The E symbol is determined by the number - (1/2)mv2 and thus the equation - (1/2). Calculate the electric field due to the dipole i) At a point on the axis at a distance 0.01m from one of the charges and ii) At a point on the perpendicular bisector at a distance 0.02m from the middle point. Calculate the field due to an electric dipole of length . Created by Mahesh Shenoy. Lets redraw a more fullscape diagram to show the situation. where p is the dipole moment, r is the distance at which the potential V of the dipole is calculated and is the angle between the distance vector and the dipole. The line OP makes an angle with the dipole axis. Hint: We knew that there are two types of dipole in nature: electric dipole and the magnetic dipole. M Dash Foundation: C Cube Learning, Conductors in Electrostatic Field M Dash Foundation: C Cube Learning, Creative Commons Attribution-NoDerivs 3.0 Unported License. 24-4 Potential due to a Electric Dipole (a) Point P is a distance r from the midpoint O of a dipole. The net force on an electric dipole present in a uniform electric field is zero as forces act in opposite directions for oppositely signed charges. This is shown in the following diagram. Now we can use the definition of electric dipole moment that we mentioned above, with the angle between the reference vector r (from the mid point as shown in Diagram 1 above) and the dipole vector p: . we can use the binomial expression, after retaining only first order term (d/r). Magnetic dipole is a closed system while electric dipole is a system made up of two opposite charges separated by small distances. Electric potential experienced by the dipole at any point is the algebraic sum of potential created by both the positive and negative charges. Also so angle between and is (/2)+. 2IU If the electric potential on the axis of an electric dipole at a distance 'r' from it is V, then the potential at a point on its equatorial line at the same distance away from it will be (1) 2 V sev (3) O (4) - V TI etioloo bouing charges a 20 . Lecture-10, 11. V= 4 01 r 2pcos. From equation (1), we get, (Consider r >> a) The magnitude of dipole is, | p | = 2 q a. The work done by the electric field when another positive point charge is moved from $(-a,0,0)$ to $(0,a,0)$ is: Three point charges q, -2q and -2q are placed at the vertices of an equilateral triangle of side a find the work done by external force to increase their separation to 2a(in joules). Let's say we're interested with the potential of such a system at a certain distance, r, away from the center of a dipole. imagine we have an electric dipole basically a negative charge and a positive charge of same value separated by some distance let's say we call it 2a traditionally we call that 2a our goal is to figure out what the potential due to this dipole is going to be at some point p let me write that show that at some point p far away from that dipole at some distance are far away these dotted lines are showing far away and just to show you the real picture um this is what it would look like if i were to actually show you this ins to scale imagine it this way that this distance r is way bigger way bigger compared to a that's what i mean when i say i want to calculate the distance potential very far away all right so let me just keep this picture somewhere over here now the question is how do i do that well i already know how to calculate potential due to a point charge we've seen the expression for that it's k q one by four epsilon naught which is called k so k q divided by r so all i have to do is figure out what's the potential due to this charge over here what's the potential due to this charge over there and then just add them up but you could say hey the problem is i don't know what's the distance of that point p from this charge and from this charge that r is the distance from the center so how do i what is this distance what is this distance that's not given to me so what i'll do is like like in most cases in physics when something is not given well you draw that and you assume something so let's do that so i'm going to take i'm going to draw some lines from here let's call this distance from here to here let's call that r1 so this is my r1 and from here to here let's call that distance to be r2 and now in terms of r1 and r2 we can write down what the electric potential is going to be it's due to it's due to this plus due to this so can you pause the video and write down the what the expression for the electric potential p turns out to be in terms of r1 r2 q and minus q pause the video and give it a shot all right let's do this so the electric potential at point p is going to be the potential due to a positive q at point p plus the potential due to negative q you might be thinking here there's a negative sign right where the negative sign comes here itself when i substitute all right so if i do that if i substitute due to plus q it's going to be k q divided by r2 so r2 and you will have due to the negative q you'll have minus k q the minus sign comes from the negative charge here divided by r1 divided by r1 and we're done now comes the question what do i do further this is not done because we have to find out the equation in terms expression in terms of r that's given to us not r1 and r2 so what do i do is the question well i know the condition is going far away and what that means is that this distance r is way bigger than that distance to a so let me write that down so that's our condition so we are calculating this far away and we're doing that what does that mean um that means that this distance r sorry this distance r is way bigger than this distance 2a or hey whatever you want to call that okay so how do i bake this in over here how can i use this to somehow simplify this well when r is very far away something very interesting happens when you compare r1 r2 and r let me show you let me draw r1 and r2 here also and you will see something really interesting what you now find is that the length of the length of all the three lines are almost equal to each other can you see that because p is so so far away this looks like almost a point these three lines will almost have the same value which means because it is far away i can now say this means this means r1 is almost equal to r2 which is almost equal to r okay they're almost equal so now comes the question what do i do can i somehow use this over here and my first my first thought would be since we want an approximate value let me just go ahead and substitute and see what happens and why don't you give it a shot you don't have to actually you can do it in your head what will happen if i substitute r1 is r and r2 also as r what will i get substituency okay if i put both of them the new denominators become same numerators are also same when i subtract i will get 0 so let me just write that down if i were to directly substitute so direct substitution what does that give me we get potential at point p to be zero and that is wrong why is that wrong why can't that be an approximate value because although they are approximately equal to each other i know from the diagram i can see that r1 is slightly bigger than r2 slightly bigger than r2q plus q is slightly closer compared to minus q therefore the potential over here must be slightly positive so there is some slight small value which is not zero and my goal is to figure out the expression for that small value are you getting that okay so how do i figure that out how do i figure out that that expression for that small value that you're gonna get so what i learned whenever i'm doing this is if direct substitution doesn't work you simplify that one more step let's say and then try substituting and see what happens so let's go ahead and simplify this now this is no longer physics we're just doing algebra simplifying i can take common denominator so let me go ahead and do that so if i take the common denominator i get vp to be equal to let's see i can take kq out and when i take common denominator i get r1 minus r2 so i get r1 minus r2 divided by r1 times r2 so that's going to be r1 times r2 and now let's see if we can substitute this over here and see what we get if i were to substitute in the denominator r1 is r and r2 also as r i will get r squared r into r is r squared and i don't get 0 so yeah i can substitute that in the denominator so what i'll do is in the denominator i'll call this r squared and what about in the numerator if i call this r and if i call that r now r minus r is 0 and again i'll get this as 0 and so i can't substitute so in the numerator i will not substitute and at this point i'm sure you may be thinking what is this business of somewhere you will substitute and sometimes you will not substitute what is this what is going on so let me take let me walk you through this it took me some time to realize but now finally i understand what's going on so let me take some numbers it will make sense so imagine r1 was say 11 and let's say r2 was a little smaller so let's call it nine and let's say r somewhere in between the two numbers is 10. okay let's write that down over here as to um what happens when you approximate so what is the actual value what is the real value well that is 11 times 9 that's 99 and what's our approximation in this case it's 10 square which is 100. and so what's the error we can say right when you approximate you get an error so what's the error over here the error is one there's a difference of one positive one but it's okay you get one and compare the error with the real value it's 1 out of 99 so it's almost like saying 1 out of 100 so it's one percent error so we're saying this equation this this is one percent wrong with these values and i can say i'm okay with one percent wrong no problem for me but now let's do the same thing for the numerator and see what happens so what is the real value if i have to substitute the values over here 11 minus 9 is 2. so this is 2 in the numerator what if i substitute 10 here and here i get 0 that's the approximate value what's the error the error is also 2. now 2 out of 2 is 2 divided by 2 is 1 that's 100 error and that's not acceptable so if i were to approximate over here i'm getting 100 wrong answer and i don't know about you but i don't want to get a hundred percent wrong answer so when you get zero you end up getting a hundred percent wrong answer so hundred percent error and that's why that's why we can't substitute in the numerator but we can substitute in the denominator and pause this if it didn't make a lot of sense it didn't make a lot of sense to me initially as well it takes some time to soak in take something digest that so pause and then we can continue all right let's continue so we are now reaching an interesting point very very interesting point and important point in this derivation as well where we need to find what the difference between r1 and r2 is we need to find that under the condition that they are almost equal to each other okay how do you do that how do you say that they're almost equal to each other and somehow find find the difference between the two now there are multiple ways to do that but the way i like to do it is zoom in on this this real picture zoom in zoom in zoom in go close to my dipole and you notice something you notice that if the point p is very far away then all these three lines look parallel to each other and that's the secret we're going to use look at them they look so parallel to each other so i'm going to go back i'm going to redraw this and this time i'm going to draw r1 and r2 parallel to each other and the point p is really really far away so how does that help us in figuring out what this is remember what i need to figure out is the difference between these two lines and one of the ways to calculate the difference between two lines which are parallel to each other in this case would be let me show you to drop a perpendicular from the shorter line onto the bigger line so let me try that one more time all right there you go so this is a perpendicular and this is a perpendicular you may ask how does that help well it helps because now if i call this point or something else the point p is at the same distance from here to here from here to p and here to p distance is the same which means that extra distance r1 minus r2 is this let me just quickly write that down this is that extra r1 minus r2 so that is our delta r and if you're wondering why why why does it work out that way well look at the look at this picture when i draw when i draw that perpendicular which you cannot see properly let me zoom in a little bit okay and i draw that perpendicular notice we end up with a giant triangle and in that triangle this angle is almost zero and these two angles are almost 90 degrees okay you may wonder how can you have a triangle like that well think of it as 89.99 89.99 and 0.0001 or something like that okay so so look at that triangle these triangles are equal that means this is an isosceles triangle and as a result this side is exactly equal to this side and therefore that extra distance should be r1 minus r2 so this should be r r2 so this should be r1 minus r2 does that make sense so this is that extra distance okay the final question we have is what does that extra distance really depend on and if you look carefully you can see that that really depends upon the angle between this r and the dipole let me show you that i have an animation over here let me show you so here you go this is our delta r and notice what happens as this angle decreases and this point p comes on the equator oh sorry on the axis notice what happens look at that look at that look at that can you see when the point comes on the axis hopefully you can see that r2 minus r1 minus r2 delta r automatically becomes 2a maximum all right notice what happens as this now comes off this particular axis and comes back what happens this will become smaller and smaller and smaller and if it now comes somewhere over here notice eventually becomes even smaller and finally when that point p is right in between these two that difference becomes zero which makes sense right if that point piece right in between these two distance from here to p and here to there would be zero would be equal and so there will be no difference so hopefully we can see that let me get rid of this that this delta r depends on this angle and so let's break that angle in so let's call that angle as theta okay and so if that angle is theta we're assuming that all lines are parallel this angle also becomes theta and so now comes the question using theta and using this right angle triangle can you figure out what delta r is going to be from this i'm going to give you a clue it says trigonometry so pause the video and see final pause this is the final part all right so in this triangle notice 2 is the hypotenuse and this but what we need is the adjacent side so i'm going to use cos theta so from from here i can say cos theta equals delta r divided by 2a and so what is delta r so i can substitute that directly over here this is our delta r okay so what is delta r delta r is just 2a cos theta let me write that delta r equals 2a cos theta and so if i substitute that i now have my final equation so vp is going to be k q into 2a cos theta to a cos theta divided by r square and that is our expression for the electric potential due to a dipole and so what you notice is that the potential at up due to a dipole not only depends on the charge but also depends upon the distance between the two charges not so surprising because we've seen that before when it comes to electric field as well and so this is what we call the product is what we call the dipole moment but notice here we also see that it's dependence on this angle theta and we saw why that is the case because this r1 minus r2 the difference between the two that depends on that angle and so the most important step over here was to figure this out in the limit that r1 is almost equal to r2 and the trick was to imagine that they are assume that they are parallel to each other and use trigonometry, Middle school Earth and space science - NGSS, World History Project - Origins to the Present, World History Project - 1750 to the Present, Potential due to an electric dipole and other systems of charges. 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