Handout 3: Electric potential and electric potential energy. Electric potential

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1 Handout 3: Electric potential and electric potential energy Electric potential Consider a charge + fixed in space as in Figure. Electric potential V at any point in space is defined as the work done by electric force in bringing a unit positive charge from infinity to that point. V r = r F dr = r 4πε 0 r dr = 4πε 0 r. The above is the expression for electric potential of charge + at distance r from the charge. It should be noted that if the charge is negative, the potential is negative too. Electric potential is a scalar quantity and its unit is volt (V). Figure : Electric potential from a point charge Example Four electric charges are situated at the corners of a square. Determine the electric potential at point P at the center of the square in terms of q and d. *Example A straight rod of length L has charge uniformly distributed on it and lies on the x-axis. One end of the rod is at distance x from the origin. x L Show that the electric potential at the origin is given by V = 4πε 0 L ln + L x

2 Electric potential energy Charge q at point with potential V has electric potential U = qv. Note that U can be positive or negative depending on signs of q and V. Sometimes the unit of energy is given as electronvolts (ev) where ev = J. When a charged particle moves from point with kinetic energy K and electric potential energy U to point with kinetic energy K and electric potential energy U, conservation of energy requires that K + U = K + U. Example 3 An electron is at distance r = 0.5 m from a fixed charge =.7 nc. What is value of the electric potential energy of the electron? Example 4 A proton is held at rest at distance r =.0 m from a charge =.4 μc. After the proton is released, calculate the speed of the proton when it reaches a) r =.0 m, b) infinity. Example 5 Calculate the total electric potential of the following cofiguration of electric charges.

3 Example 6 An alpha particle (charge +e, mass 4u) of kinetic energy 4.6 MeV is projected towards a gold nuclues (charge +79e). What is the minimum separation between the alpha particle and the gold nuclues? 3 Relationship between electric field and electric potential Consider the positive charge q moving from point to point in uniform electric filed E as shown in Figure. Work done by electric force is W = F dr = qe dr. Figure : Work done by electric force At point, the potential V is greater than the potentialv at point. The change in the potential energy of the charge q is Since W = U, U = q V V, V > V. V V = E dr dv = E dr. The above equation can be expressed as The direction of E is along r. E = dv dr. Example 7 From Example, the expression for V(x) is given by V x = 4πε 0 L ln + L x. Find the expression for electric field E x at the origin. The answer must be the consistent with the result in Handout, Example 4.

4 4 Equipotential surface Adjacent points that have the same electric potential form an equipotential surface. All points on this surface have equal electric potential. For a point (spherical symmetric) charge, the equipotential surfaces are a series of concentric spheres around the charge (Figure 3). In a uniform field, Figure 4, the surfaces of equal potential are a series of planes perpendicular to the field line. Figure 5 shows equipotential surfaces of two point charges with opposite signs. Figure 3: Equipotential surfaces around a point charge Since all points on a equipotential surfaces have equal electric potential, there is no work done by electric force along the path on the equipotential surface. The electric field lines must be perpendicular to the equipotential surfaces. If the electric field passed through an equipotential surface at angle, there would be a component of electric field on the equipotential surface and there would be work done on the equipotential surface by electric force. Figure 4: Equipotential surfaces in uniform field Parallel charged plates In Figure 6, two parallel plates of opposite charges are separated by distance d. The potential difference between the plates is V. The electric field E between the plates is uniform. From equation, one can integrate V dv V+ d = E dr 0 Since V + V = V, we have V + V = Ed. Figure 5: Equipotential surfaces in a system of two opposite charges V = Ed or E = V d. The above equation relates the electric field strength between the parallel plates to the potential difference and the separation between the plates. Figure 6: Electric field between parallel plates

5 Example 8 Two horizontal plate X and Y are 0.75 cm apart. A positively charged particle of mass kg is situated in vacuum between the plates. The potential difference between the plates is 630 V. 5 a) State, with reason, which plate, X or Y, is positively charged. b) Calculate the charge on the particle. Example 9 Two parallel charged square plates, separated by distance d = 0.5 cm, are connected a source with potential difference V = 500 V. The length of each side of the square is a = 0 cm. Calculate the magnitude of the charge on each plate.

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