\[r_{i}\] is the distance of the point P from the ith charge \[Q_{i}\] and \[r_{i}\] is a unit vector directed from \[\widehat{Q_{i}}\] to the point P. ri is a unit vector directed from Qi to the point P. Lets say charge Q1, Q2Qn are placed in vacuum at positions r, r,.,r respectively. Your physics teacher turns through your test paper, a slight frown upon their face. If we have a uniform electric field, we know that the electric field lines will be parallel to each other and point in the same direction. Coulomb's Law for calculating the electric field due to a given distribution of charges. Three point charges are placed on the y axis as shown. This is impossible. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). What is the voltage 5.00 cm away from the center of a 1-cm diameter metal sphere that has a \(-3.00 \mathrm{nC}\) static charge? \[\overrightarrow{E}({r}) = \frac { \overrightarrow{F}(r)} {q_o}\], \[\overrightarrow{E} = \frac{1}{4\pi \epsilon_{0}} / r^2 (r)\]. Thus we can find the voltage using Equation \ref{eq1}. Note that the isolines are always perpendicular to the field lines. Electric Field due to a System of Charges. The electric field of a negative point charge points towards the point charge as a result of the definition of the electric field of a point charge. We can represent the strength and direction of an electric field at a point using electric field lines. And I'll call that blue E x because it was the horizontal component created by the blue, positive charge. The next day, your physics teacher brings in a Van der Graaf generator, which causes your classmate's hair to stand on end when they touch it. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. For the field of a point charge, field lines are radial and isolines form concentric circles centered on the charge. The isolines or lines of equipotential are also parallel to each other but are perpendicular to the field lines at all times. (\overrightarrow{r_{2}} - \overrightarrow{r_{1}})}}\], Here, \[AB = \overrightarrow{r_{12}} = \overrightarrow{r_{2}} - \overrightarrow{r_{1}}\], As, \[\overrightarrow{E} = \frac{\overrightarrow{F}}{q_{2}}\], \[\overrightarrow{F} = \frac{1} {4 \pi \epsilon_{0}}{\frac{ q_{1}}{|\overrightarrow{r_{2}} - \overrightarrow{r_{1}}|^{3} . To find the voltage due to a combination of point charges, you add the individual voltages as numbers. For a point charge, how is the electric potential \(V\) related to the distance \(r\) from the charge? If the electric field is known, then the electrostatic force on any charge q is simply obtained by multiplying charge times electric field, or F = q E. Consider the electric field due to a point charge Q. 3. In other words, the electric field due to a point charge obeys an inverse square law, which means, that the electric field due to a point charge is proportional to the reciprocal of the square of the distance that the point in space, at which we wish to know the electric field, is from the point charge that is causing the electric field to exist. Question: Electric Field due to a point charge.The electric field in the xy-plane due to a point charge at (0,0) is a gradient field with a potential V(x,y) = k/((sqrt(x^2 + y^2)) where k > 0 is a physical constant.a. \(\left|\vec{E}\right| =4\,000\,\mathrm{V\,m^{-1}}\). Electric potential of a point charge is [latex]\boldsymbol{V = kQ/r}[/latex]. No sweeter sound could have entered your ears and you return to your seat satisfied that you have atoned for any physics-related sins. The electric field intensity at any point is the strength of the electric field at that point. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. More about Electric Potential due to a Point Charge, Charged Particle in Uniform Electric Field, Electric Field Between Two Parallel Plates, Magnetic Field of a Current-Carrying Wire, Mechanical Energy in Simple Harmonic Motion, Galileo's Leaning Tower of Pisa Experiment, Electromagnetic Radiation and Quantum Phenomena, Centripetal Acceleration and Centripetal Force, Total Internal Reflection in Optical Fibre, For a point charge, the potential \(V\) is related to the distance \(r\) from the charge \(q\), \[V=\frac{1}{4\pi \varepsilon_0}\frac{q}{r}.\], The SI unit of measurement of potential is the \(\text{volt, V.}\), The average magnitude of the electric field \(\left|\vec{E}\right|\) between two points is equal to the magnitude of change in electric potential \(\Delta V\) divided by the change in position between those points \(\Delta r\) in the field, \[\left|\vec{E}\right|=\left|\frac{\Delta V}{\Delta r}\right|.\]. 4 below shows the field lines and isolines due to a positive point charge. \label{eq1}\], where \(k\) is a constant equal to \(9.0 \times 10^{9}\, \mathrm{N}\cdot \mathrm{m^{2}/C^{2}}.\). In this article, we will discuss the electric potential due to a point charge, so that you may never make this mistake again. In fact, \[V=\frac{1}{4\pi \varepsilon_0}\frac{q}{r},\] where \(\varepsilon_0\) is a constant known as the permittivity of free space and has the value \(\varepsilon_0 = 8.85\times10^{-12}\,\mathrm{F\,m^{-1}}.\) The SI unit of measurement of potential is the \(\text{volt, V,}\) which is equivalent to the \(\text{joule-per-coulomb, } \mathrm{J\,C^{-1}}.\) A graph of due potential against distance due to a positive charge and due to a negative charge is shown in Fig. When a glass rod is rubbed with a piece of silk, the rod acquires the property of attracting objects like bits of paper, etc towards it. Electric Field due to point charge Formula Electric Field = [Coulomb]*Charge/ (Separation between Charges^2) E = [Coulomb]*q/ (r^2) What is Electric Field? by Michel van Biezen. Entering known values into the expression for the potential of a point charge, we obtain, \[ \begin{align*} V&=k\dfrac{Q}{r} \\[5pt] &=(8.99 \times 10^{9} \, \mathrm{N}\cdot \mathrm{m^{2}/C^{2}}) \left(\dfrac{-3.00\times 10^{-9}\,\mathrm{C}}{5.00\times 10^{-2}\,\mathrm{m}}\right) \\[5pt] &= -539\, \mathrm{V}. Electric Field Due to a Point Charge Formula The concept of the field was firstly introduced by Faraday. The circular isolines mean that the potential is constant along a circular path of radius \(r\) surrounding the point charge. What is the charge in the inner and outer surface of the enclosing sphere? 3 below. And similarly, for the electric field this negative charge creates, it has a horizontal component that points to the right. by Patrick Ford. The voltage of a battery is an example of the electric potential difference between its ends. The isolines are always perpendicular to the field lines and so form concentric circles centered on the charge. However, along a line that is parallel to the surface, the potential will be constant, as all points on that line are equidistant from the surface. Click Start Quiz to begin! Electric potential is a scalar, and electric field is a vector. from Q to q. You wipe the sweat from your brow as you eagerly await even a few words of encouragement. The change in potential \(\Delta V\) between two points is also called the potential difference between those points. Addition of voltages as numbers gives the voltage due to a combination of point charges, whereas addition of individual fields as vectors gives the total electric field. Electric charge is a scalar quantity. The outside field is often written in terms of charge per unit length of the cylindrical charge. Calculate the electric potential \(V\) of a \(2.0\,\mathrm{\mu C}\) point charge at a distance of \(0.50\,\mathrm{cm}\) from the charge. Question: The electric potential energy between an electron and proton is \(9.6\times 10^{-17}\,\mathrm{J}.\) Calculate the electric potential of the electron at the position of the proton assuming that both can be treated as point charges. If a charge q is brought around at any point near Q, Q itself experiences an electrical force due to q and will gradually move away. The test charge would be repelled as q1 is positive, therefore E1 originates from q1. \(\left|\vec{E}\right|=\left|\frac{\Delta V}{\Delta r}\right|\). \[\overrightarrow{F} = \frac{1}{4\pi \epsilon_{0}} q_{0} \sum_{i=1}^{i=n} \frac{\overrightarrow Q_{i}}{|\overrightarrow{r} - \overrightarrow{r_{i}}|^{3} . Here, if force acting on this unit positive charge +q at a point r, then electric field intensity is given by: \[\overrightarrow{E}({r}) = \frac {\overrightarrow{F}{(r)}}{q_o}\]. Every charge in the universe exerts a force on every other charge in the universe is a bold yet true statement of physics. Now we examine an arbitrary location on the line connecting the charges. This is analogous to taking sea level as \(h=0\) when considering gravitational potential energy, \(\mathrm{PE_{g}}=mgh\). 4 - The field lines for the electric field of a positive point charge point radially outward. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. To reach a point in the electric field where the unit positively charges from infinity to the point, you must do a lot of work. ELECTRIC FIELD: The region around a charged body within which its influence can be realized by other charges is called electric field. The electrostatic force produced by a point charge on a test charge at a distance r is proportional to the charge of both charges and the distance between them. Lastly, we can take a look at how a potential difference between two points affects the magnitude of the electric field in that region. Electric potential is a scalar, and electric field is a vector. We can thus determine the excess charge using Equation \ref{eq1}. We can verify it graphically by adding the vectorsE1 and E2 together with the help of the parallelogram law as seen in the next figure. Therefore, we can say that the electric field of charge Q as space by virtue of which the presence of charge Q modifies the space around itself leading to the generation of force F on any charge q held in this space, given by: Here, from the above figure, we have the following parameters, r = The separation between source charge and test charge, \[k = \frac{1}{4\pi \epsilon_{0}} = 9\times 10^{9} N m^{2} C^{-1}\]. The test charge has to be small enough to have no effect on the field. (5.12.2) V 21 = r 1 r 2 E d l. Suppose we have to calculate the electric field intensity or strength at any point P due to a point charge Q at O. In other words we can define the electric field as the force per unit charge. In Sections 5.8 and 5.9, it was determined that the potential difference measured from position r 1 to position r 2 is. This happens due to the discharge of electric charges by rubbing of insulating surfaces. Thus, F = (k|q 1 q 2 |)/r 2, where q 2 is defined as the test charge that is being used to "feel" the electric field. It is flipped about the distance axis for a negative charge. The concept of electric field was introduced by Faraday during the middle of the 19th century. Electric potential of a point charge is \(V=kQ/r\). V As we have discussed in Electric Charge and Electric Field, charge on a metal sphere spreads out uniformly and produces a field like that of a point charge located at its center. Coulomb's law can be used to express the field strength due to a point charge Q. |\overrightarrow{r} - \overrightarrow{r_{i}}|}\]], Putting \[\frac {1}{4 \pi \epsilon_{0}}\] = k, \[\overrightarrow{E} = k \frac {Q_{1}} {r_{1^2}} + k \frac {Q_{2}}{r_{2^2}} + . 111 views. Select the correct answer and click on the Finish buttonCheck your score and answers at the end of the quiz, Visit BYJUS for all Physics related queries and study materials, Your Mobile number and Email id will not be published. electric field due to finite line charge at equatorial point electric field due to a line of charge on axis We would be doing all the derivations without Gauss's Law. The electric field due to point charge +Q at a distant point P is V volt/meter. \end{align}\] The electric potential due to the electron at the position of the proton is \(600\,\mathrm{V}.\). Thus is directed along the axis of the ring. \(V=\frac{1}{4\pi \varepsilon_0}\frac{q}{r}\). Strategy We can find the electric field created by a point charge by using the equation E=\frac {kQ} {r^2}\\ E = r2kQ . \end{align}\] The electric potential of this charge is \(3\,600\,\mathrm{V}\), at a distance of \(0.50\,\mathrm{cm}\) from the charge. The electric field at a point due to the presence of a charge q 1 is simply given by the relation We can now move on to slightly more complex examples. Assume a point between the charges where the electric field due to each charge points to the left, so the net electric force cannot be zero. Fig. People who viewed this item also viewed. We need to define a new quantity, the electric potential, and we will do so for a point charge specifically. Identify your study strength and weaknesses. Earn points, unlock badges and level up while studying. Apart from having a magnitude and direction, a quantity to be termed a vector should also obey the laws of vector addition, such as triangle law of vector addition and parallelogram law of vector addition; only then the quantity is said to be a vector quantity. . If E is the electric field created by Q at P, then by definition: E = lim q 0 0 F q 0 = lim q 0 0 k e Q q 0 r 2 r . Your Mobile number and Email id will not be published. To see this, recall that the electric field of a point charge q is defined as. Find the electric field at point P on the x axis. From electric field due to multiple point charges we find that the resultant field produced by one portion is given by. The electric field for +q is directed radially outwards from the charge while for - q, it will be radially directed inwards. Source Charge: The point charge which produces the electric field is known as source charge. This page titled 19.3: Electrical Potential Due to a Point Charge is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Knowledge is free, but servers are not. Its 100% free. Two point charges q1 = q2 = 10-6 C are respectively located at the points of coordinates (-1, 0) y (1, 0) (the coordinates are expressed in meters). \(\begin{array}{l}\overrightarrow {F} = \frac{1}{4~\pi~_0} \frac {q q_0}{r^2} \hat r \end{array} \). They are generated by electric charges, and charge configurations such as capacitors or by varying magnetic fields. The potential on the surface will be the same as that of a point charge at the center of the sphere, 12.5 cm away. Rod length =. Consider two points A and B separated by a small distance dx in an electric field. To help visualize how a charge, or a collection of charges, influences the region around it, the concept of an electric field is used. A +0.05 C charge is placed in a uniform electric field pointing downward with a strength of 100 . Consider a system of charges q1, q2,q3 , qn placed at a distance r1,r2,r3 , rn relative to some origin O.Let q be the test charge at point P where the total electric field due to n charges is to be determined..Let test charge q placed in a medium at a distance r from origin (i.e. Solving for \(Q\) and entering known values gives, \[ \begin{align*} Q &=\dfrac{rV}{k} \\[5pt] &= \dfrac{(0.125 \,\mathrm{m})(100\times 10^{3}\, \mathrm{V})}{8.99\times 10^{9}\, \mathrm{N\cdot m^{2}/C^{2}}} \\[5pt] &= 1.39\times 10^{-6} \,\mathrm{C} \\[5pt] &= 1.39\, \mathrm{\mu C}.\end{align*}\]. 3 - The field lines for a uniform electric field are parallel to each other. What is the SI unit of measurement of electric potential? Test your knowledge with gamified quizzes. Required fields are marked *. find the compnents of the electric field in the x- and y- directions, where E(x,y) = - ? Electric field lines always point in one direction. We can deduce from the previous figure that the charge q3 has to be negative, because the field E3 has to be oriented toward the point charge (recall that negative charges are sinks of field lines). Even if we pass electricity through it as a stream of charged particles, there wont be a spark. Subatomic particles carry electric charges. We know that, in reality, charged particles like protons and ions have a definite size and occupy some volume in space. The isolines are always perpendicular to the field lines and so form concentric circles centered on the charge. We will find the electric field E 1 caused by charge q 1, the electric field E 2 caused by charge q 2, and the electric field E 3 caused by charge q 3. 3. In the present post, I will consider the problem of calculating the electric field due to a point charge surrounded by a conductor which has the form of a thick spherical shell. Explain point charges and express the equation for electric potential of a point charge. If two points lie on the same isoline, no work is done in moving a charged particle between those points. \(\begin{array}{l} \overrightarrow{E} 5 N downward 5 N upward 2000 N downward 2000 N upward Stop procrastinating with our smart planner features. We can also relate the electric potential to the average magnitude of the electric field \(\left|\vec{E}\right|\) as follows, \[\left|\vec{E}\right|=\left|\frac{\Delta V}{\Delta r}\right|.\] The average magnitude of the electric field between two points is equal to the magnitude of change in electric potential \(\Delta V\) divided by the change in position between those points \(\Delta r\) in the field. Where r is a unit vector directed from Q towards q. Test Charge: The small charge which experiences the electrostatic force due to the source charge is known as test charge. (a) Arrows representing the electric field's magnitude and direction. The force that a charge q 0 = - 2 10 -9 C situated at the point P would experience. Everything you need for your studies in one place. It is defined as the force experienced by a unit positive charge placed at a particular point. The force \(F_{qQ}\) that charge \(q\) exerts on \(Q\) is equal and opposite to the force \(F_{Qq}\) that charge \(Q\) exerts on \(q.\) We can call the magnitude of this force \(F.\) From Coulomb's law, \[F=\frac{1}{4\pi \varepsilon_0}\frac{qQ}{r^2},\] and the electric potential energy \(E_\mathrm{P}\) is the same as the work done \(W\) to bring two charges to points at which their separation is \(r,\) \[E_\mathrm{P}=W=\frac{1}{4\pi \varepsilon_0}\frac{qQ}{r}.\] The definition of electric potential tells us that the work done per unit charge in bringing charge \(Q\) from infinity to a distance \(r\) from charge \(q\) is given by \[\begin{align}V&=\frac{W}{Q}\\&=\frac{1}{\cancel{Q}} \cdot \frac{1}{4\pi \varepsilon_0} \frac{q\cancel{Q}}{r}\\&=\frac{1}{4\pi \varepsilon_0}\frac{q}{r}, \end{align}\] which is the same as the first equation stated above. Fig. In other words, check this out. If an electron orbits the nucleus on a circular path, what work is done on the electron? Case (i) Electric field due to an electric dipole at points on the axial line. To calculate the unit vector ur1 we will divide the vector A that goes from the location of q1 to the point P by its magnitude. Suppose the point charge +Q is located at A, where OA = r1. The electric field is nonuniform. Legal. And this electric field is gonna have a vertical component, that's gonna point upward. We know how to find the electric field caused by a single point charge. Calculate the strength and direction of the electric field E due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge. Thanks! Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. For a uniform field, field lines are parallel to each other and isolines are parallel to each other but perpendicular to the field lines. The electric potential energy between an electron and proton is \(1.92\times 10^{-16}\,\mathrm{J}.\) Calculate the electric potential \(V\) of the electron at the position of the proton assuming that both can be treated as point charges. Since dx is small, the electric field E is assumed to be uniform along AB. Electric Field Due To An Infinitely Long Straight Uniformly Charged Wire Let us learn how to calculate the electric field due to infinite line charges. Free shipping. Please consider supporting us by disabling your ad blocker on YouPhysics. Charge Q producing electric field at a given point r is: E (r) = r = r. Where r = , which is a vector unit from the origin to the point r, F = r. Assuming the SI unit of the electric field as N/C. Simply we can write this mathematically as \[V=\frac{W}{q}.\] Adjacent points that have equal electric potential form lines of equipotential, also called isolines. \[\begin{align}V&=\frac{W}{Q}\\[4 pt]&=\frac{9.6\times 10^{-17}\,\mathrm{J}}{1.60\times 10^{-19}\,\mathrm{C}}\\[4 pt] &=600\,\mathrm{J\,C^{-1}}\\[4 pt]&=600\,\mathrm{V}. Sponsored. Multiplying 0 0 by R2 R 2 will give charge per unit length of the cylinder. Let the linear charge density of this wire be . P is the point that is located at a perpendicular distance from the wire. An electric field at a distance d from a straight charged conductor is known as the electric field. No, electric fields and electric charges cannot exist in a vacuum because all the charge carriers are massive particles or holes in a crystal of massive particles (atoms). Charges in static electricity are typically in the nanocoulomb \((\mathrm{nC})\) to microcoulomb \((\mu \mathrm{C})\) range. Everything we learned about gravity, and how masses respond to . You have demonstrated potential." by Patrick Ford. The electric field at point {4,4,0} is the vector sum of two fields: (a) that from the infinite plate and (b) that from the point charge at {4,0,0}. Therefore, the test charge used to measured the electric field must be too small. Hence, we obtained a formula for the electric field due to a system of point charges. Set individual study goals and earn points reaching them. Addition of voltages as numbers gives the voltage due to a combination of point charges, whereas addition of individual fields as vectors gives the total electric field. The electric field intensity due to a point charge q at the origin is (see Section 5.1 or 5.5) (5.12.1) E = r ^ q 4 r 2. 2 below. The field lines would be radial but we would require that the isolines always be perpendicular to them. of the users don't pass the Electric Potential due to a Point Charge quiz! Electric Field Of Point Charge Electric Field Due To Point Electric Charges "Every charge in the universe exerts a force on every other charge in the universe" is a bold yet true statement of physics. The net electric field E net is the _vector_ sum of these three fields, E net = E 1 + E 2 + E 3. (r_{i})\]. The electric field strength due to a dipole, far away, is always proportional to the dipole moment and inversely proportional to the cube of the distance. 4. (Figure \(\PageIndex{1}\)) What excess charge resides on the sphere? Well, if the electric field points to the right and this charge is negative, then the electric force has to point to the left. The scenario is different for a point charge. Now that we have seen how the electric potential of a point charge varies with distance, we can work our way through some examples relating to this concept. If you are using this example for a problem in which you are . Now the point charge +Q is enclosed by the hollow conducting sphere, a. The negative value for voltage means a positive charge would be attracted from a larger distance, since the potential is lower (more negative) than at larger distances. Note that the electric field is a vector quantity that is defined at every pint in space, the value of which is dependent only upon the radial distance from q. The Electric Field due to a Half-Ring of Charge | by Rhett Allain | Geek Physics | Medium 500 Apologies, but something went wrong on our end. To find out an electric field of a charge q, we can establish a test charge q0 and gauge the force exerted on it. In parallel plates, a 1600 n/c electric field is between two plates with a diameter of 2.0 - 10 - 2 m each. A Few Important Points To Remember Which Understanding Electric Field Due To Point Charges: The electrostatic force field that surrounds a charged item stretches outward in all directions. The unit vectors we are going to use to calculate the electric fields E1 and E2 are represented in red in the figure. 5. Electric potential of a point charge is V = k Q / r. Electric potential is a scalar, and electric field is a vector. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. Create beautiful notes faster than ever before. By registering you get free access to our website and app (available on desktop AND mobile) which will help you to super-charge your learning process. The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal ( x )-components of the field cancel, so that the net field points in the z -direction. Here are my starting parameters. Answer: Recall that the charge of a proton is \(1.60\times 10^{-19}\,\mathrm{C}.\) The electric potential \(V\) due to the electron at the position of the proton is the work done per unit charge in bringing the proton to that point in the electric field of the electron. Coulomb's law states that if another point charge q is placed at a position P where OP = r, the charge Q will exert a force on q. The electric field lines for point charges are radial and point into or away from the charge (depending on the sign of the charge). One way to understand the ability of a charge to influence other charges anywhere in space is by imagining the influence of the charge as a field. Consider an electric dipole placed on the x-axis as shown in Figure 1.17. Dipole moment is the product of the charge and distance between the two charges. [7] That direction will be determined by the sign of the charge on the surface of the object generating the potential. Answer: We can use the equation relating potential \(V\) to distance \(r,\) \[\begin{align} V&=\frac{1}{4\pi \varepsilon_0}\frac{q}{r}\\[2 pt]&=\frac{1}{4\pi \left(8.85\times10^{-12}\,\mathrm{F\,m^{-1}}\right)}\left(\frac{2.0\times 10^{-9}\,\mathrm{C}}{0.50 \times 10^{-2}\,\mathrm{m}}\right)\\[4 pt]&=3\,600\,\mathrm{C\,F^{-1}}\\[4 pt]&=3\,600\,\mathrm{V}. \end{align*}\]. Let's check this formally. This is always necessary since any component of the electric field along the direction of an isoline will cause an electric force on a charge along that line. The electric lines of force are shown in figure above. By principle of superposition, the Electric field at a point will be the sum of electric field due to the two charges +8q and -2q This is consistent with the fact that \(V\) is closely associated with energy, a scalar, whereas \(\mathbf{E}\) is closely associated with force, a vector. Be perfectly prepared on time with an individual plan. Fig. The value of a point charge q 3 situated at the origin of the cartesian coordinate system in order for the electric field to be zero at point P. Givens: k = 9 10 9 N m 2 /C 2 The super position principle says that the total electric field at some point is the vector sum of the electric field due to individual point charges. 1 - A graph of electric potential vs distance shows an inverse relationship for a positive charge and the curve is flipped about the distance axis for a negative charge. What is an example of electric potential due to a point charge? Question: Calculate the average magnitude of the electric field between two points which have a potential difference of \(150\,\mathrm{V}\) between them, and are separated by a distance of \(2.5\,\mathrm{cm}.\). 3. To find the direction of the vector E1 we perform a thought experiment that consists in placing a positive test charge at point P and to identify the direction of the force it would experience in presence of charge q1. If we think quite classically and assume that electrons orbit the nucleus of an atom in a circular path, this would be why the nucleus does not work on electrons. Thus the force exerted per unit charge is: \(\begin{array}{l} \overrightarrow{E} = \frac {\overrightarrow F}{q_0} = \frac {1}{4~\pi~_0} \frac{q}{r^2} \hat r \end{array} \). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. This means that at all points on the spherical surface drawn around the point charge, the magnitude of E is same and doesn't depend on the direction of r . Hence, E is a vector quantity and is in the direction of the force and along the direction in which the test charge +q tends to move. The electric field E is a vector quantity whose direction is the same as that of the force F exerted on a positive test charge. 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