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The Electric Potential (Cont.) Dr Miguel Cavero July 29, 2014 Electric Potential (Cont.) July 29, 2014 1 / 18 V and E The electric potential V is related by the electric field E as follows: Z B ∆VAB = VB − VA = − E dr A where dr = |dr| and E = |E|. Alternatively, the electric field is the gradient of the potential: E=− dV dr The electric field at some point is the negative gradient of the electric potential at that point. Electric Potential (Cont.) Electric Potential And The Electric Field July 29, 2014 3 / 18 Electric Potential Consider the graph of a constant electric potential vs distance. For a positive charge positioned at some point, in which direction will the charge accelerate? What is the direction of the electric field? Electric Potential (Cont.) Electric Potential And The Electric Field July 29, 2014 4 / 18 Uniform Electric Field For a uniform (constant) electric field in one dimension, a positive charge experiences a force as shown: The field is directed in the +x-direction, so the positive charge will experience a force in that direction. The work done in moving the charge in the +x-direction is given by W = qE∆x Electric Potential (Cont.) Electric Potential And The Electric Field July 29, 2014 5 / 18 Uniform Electric Field For a positive work done W , the charge must have moved from a region of higher potential to a region of lower potential. The electric potential V must then decrease in the direction of the electric field. Electric Potential (Cont.) Electric Potential And The Electric Field July 29, 2014 6 / 18 Electric Potential The electric field is related to the electric potential by E=− dV dx From the graph, the potential is described by ∆V V (x) = x + V0 ∆x = V0 − Ex Electric Potential (Cont.) Electric Potential And The Electric Field July 29, 2014 7 / 18 Electric Potential V (x) = V0 − Ex The potential V decreases in the direction of the electric field E (in the +x-direction). Positive charges must accelerate down the gradient of the potential, since the force they experience is in the direction of the electric field. Negative charges, therefore, must accelerate up the gradient of the electric potential. Electric Potential (Cont.) Electric Potential And The Electric Field July 29, 2014 8 / 18 Field Between Two Uniformly Charged Plates Consider a region of space where the electric field is uniform. Let the distance between the parallel plates be d. Since the potential V (x) varied with distance by V = V0 − Ex, setting the potential V to zero gives −V0 = −Ed ⇒ E = Electric Potential (Cont.) Consequence Of E = −∇V V0 d July 29, 2014 10 / 18 Field Between Two Uniformly Charged Plates For parallel plates maintained at a constant potential difference V0 , the electric field E within the plates is constant: E= V0 d Note that the unit of the electric field (N C−1 ) also has units of V m−1 . Electric Potential (Cont.) Consequence Of E = −∇V July 29, 2014 11 / 18 The Potential Due To A Stationary Point Charge How does the potential vary with distance for a point charge? Z B ∆V = VB − VA = − E dr ZAr V (r) = − ke ∞ q dr r2 q = ke r The electric field of a point charge extends through all space, therefore the potential does as well. The electric potential (work per unit charge) required to move a positive test charge q0 from infinity to a distance r from a positive point charge q, increases as the test charge gets closer to q. Electric Potential (Cont.) Consequence Of E = −∇V July 29, 2014 12 / 18 The Potential Due To A Stationary Point Charge The electric field generated by a positive charge, and the electric potential due to the charge, are shown below: The dashed lines are equipotential surfaces, all points at the same distance from the positive charge that are at the same potential. Note that the electric field is everywhere perpendicular to the equipotential surface. Electric Potential (Cont.) Consequence Of E = −∇V July 29, 2014 13 / 18 Superposition And The Electric Potential The principle of superposition also applies to the electric potential. The electric potential at some point P depends on the electric potential due to all charges, and their distances to P . The electric potential at some point P due to three charges q1 , q2 and q3 is given by q1 q2 q3 V = ke + ke + ke r1 r2 r3 Note that the electric potential is a scalar quantity, but can be positive or negative depending on the sign of the charges that are generating the electric fields. Electric Potential (Cont.) Consequence Of E = −∇V July 29, 2014 14 / 18 Superposition Example Two charges lie on a line as shown in the diagram: Calculate the electric field at P . The electric field at P points in the same direction due to both charges (to the left). The magnitude of the electric field is EP = ke Electric Potential (Cont.) 3 × 10−9 2 × 10−9 + ke 2 (0.02) (0.01)2 Consequence Of E = −∇V July 29, 2014 15 / 18 Superposition Example Two charges lie on a line as shown in the diagram: Calculate the electric potential at P . The electric potential at P has no direction associated to it. The electric potential is VP = −ke Electric Potential (Cont.) 3 × 10−9 2 × 10−9 + ke 0.02 0.01 Consequence Of E = −∇V July 29, 2014 16 / 18 Electric Potential Energy Let V1 be the electric potential due to a charge q1 at a point P . Work must be done to bring a charge q2 from infinity to P . The work done is given by q2 V1 , and is the potential energy of this two-particle system. The electrical potential energy for the pair of particles is therefore PE = q2 V1 = ke Electric Potential (Cont.) q1 q2 r Electric Potential Energy July 29, 2014 18 / 18