Electric Potential

 Physics requires  practice to develop your solving problems skills and your logical thinking . I strongly recommend to study carefully the

examples, and solve the following problems in this webpage


Electric potential: 1,4,7,8,9,11,12,14,15,17

Capacitors: 24,25,26,27,28,30,31,35,36,37.



1. (3G). How much kinetic energy will an electron gain (in joules and eV) if it accelerates through a potential difference of 23,000 V in a TV picture tube?


2.(10G). The work done by an external force to move a –8.50 mC charge from point a to point b is 15.0 x 10-4 J If the charge was started from rest and had 4.82 x 10-4J of kinetic energy when it reached point b, what must be the potential difference between a and b?


3.(1S).   How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro’s number of electrons from an initial point where the electric potential is 9.00 V to a point where the potential is –5.00 V? (The potential in each case is measured relative to a common reference point.)


4.(4S)    (a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V. (b) Calculate the speed of an electron that is accelerated through the same potential difference.


5. (5S). A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A +12.0-μC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm). (a) What is the change in the potential energy of the charge–field system? (b) Through what potential difference does the charge move?


6.(7S)    An electron moving parallel to the x axis has an initial speed of 3.70 × 106 m/s at the origin. Its speed is reduced to 1.40 × 105 m/s at the point x = 2.00 cm. Calculate the potential difference between the origin and that point. Which point is at the higher potential?


7. (15S).            (a) Find the potential at a distance of 1.00 cm from a proton. (b) What is the potential difference between two points that are 1.00 cm and 2.00 cm from a proton? (c) What If? Repeat parts (a) and (b) for an electron.


 8. (14G) What is the electric potential 15.0 cm from a 4.0 m C point charge?


9. (15G).            A point charge Q creates an electric potential of +125 V at a distance of 15 cm. What is Q?


10. (16G) A +35 m C point charge is placed 32 cm from an identical  charge +35 m C . How much work would be required to move a – 0.50 m C test charge from a point midway between them to a point 12 cm closer to either of the charges?


11. (18G). (a) What is the electric potential a distance of 2.5 x 10-15 away from a proton? (b) What is the electric potential energy of a system that consists of two protons

         2.5 x 10-15  apart — as might occur inside a typical nucleus?

12. (19G).  Three point charges are arranged at the corners of a square of side L as shown in Fig. 17–25. What is the potential at the fourth corner (point A), taking V= 0 at a great distance?


13. (23G). How much work must be done to bring three electrons from a great distance apart to 1.0 x 10-10 m from one another (at the corners of an equilateral triangle)?


14.(16S)  Given two 2.00-μC charges, as shown in Figure P25.16, and a positive test charge q = 1.28 × 10–18 C at the origin, (a) what is the net force exerted by the two 2.00-μC charges on the test charge q? (b) What is the electric field at the origin due to the two 2.00-μC charges? (c) What is the electrical potential at the origin due to the two 2.00-μC charges?


15.(17S).   At a certain distance from a point charge, the magnitude of the electric field is 500 V/m and the electric potential is –3.00 kV. (a) What is the distance to the charge? (b) What is the magnitude of the charge?


16.(19S) The three charges in Figure P25.19 are at the vertices of an isosceles triangle. Calculate the electric potential at the midpoint of the base, taking q = 7.00 μC.


17. (20S). Two point charges, Q1 = +5.00 nC and Q2 = –3.00 nC, are separated by 35.0 cm. (a) What is the potential energy of the pair? What is the significance of the algebraic sign of your answer? (b) What is the electric potential at a point midway between the charges?


18. (23S) Show that the amount of work required to assemble four identical point charges of magnitude Q at the corners of a square of side s is 5.41keQ2/s.


19.(28S)   Two particles, with charges of 20.0 nC and –20.0 nC, are placed at the points with coordinates (0, 4.00 cm) and (0, –4.00 cm), as shown in Figure P25.28. A particle with charge 10.0 nC is located at the origin. (a) Find the electric potential energy of the configuration of the three fixed charges. (b) A fourth particle, with a mass of 2.00 × 10–13 kg and a charge of 40.0 nC, is released from rest at the point (3.00 cm, 0). Find its speed after it has moved freely to a very large distance away.


20.(34S)                Calculate the energy required to assemble the array of charges shown in Figure P25.34, where a = 0.200 m, b = 0.400 m, and q = 6.00 μC.


21.(37S)            The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.0 V and b = –7.00 V/m. Determine (a) the potential at x = 0, 3.00 m, and 6.00 m, and (b) the magnitude and direction of the electric field at x = 0, 3.00 m, and 6.00 m.


22.(38S)            The electric potential inside a charged spherical conductor of radius R is given by V = keQ/R, and the potential outside is given by V = keQ/r. Using Er = ‑dV/dr, derive the electric field (a) inside and (b) outside this charge distribution.


23.(39S) Over a certain region of space, the electric potential is V = 5x – 3x2y + 2yz2. Find the expressions for the x, y, and z components of the electric field over this region. What is the magnitude of the field at the point P that has coordinates (1, 0, –2) m?





24. (31G) The two plates of a capacitor hold +2500 mC and - 2500 mC of charge, respectively, when the potential difference is 850 V. What is the capacitance?


25. (34G). How much charge flows from each terminal of a 12.0-V battery when it is connected to a 7.0 mF capacitor?


26. (35G).  A 0.20-F capacitor is desired. What area must the plates have if they are to be separated by a 2.2-mm air gap?


27.( 37G). An electric field of 8.50 x 105 V/m is desired between two parallel plates, each of area 35.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?


28.( 39G). How strong is the electric field between the plates of a 0.80 mF air-gap capacitor if they are 2.0 mm apart and each has a charge of 72 mC


29.(41G). A 2.50 mF capacitor is charged to 857 V and a 6.80 mF  capacitor is charged to 652 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: charge is conserved.]


30. (43G). What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 3.2 mm of mica?


31. (47G). A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in this device is charged to 5.0 kV and stores 1200 J of energy. What is its capacitance?


32.(1S).  (a) How much charge is on each plate of a 4.00-μF capacitor when it is connected to a 12.0-V battery? (b) If this same capacitor is connected to a 1.50-V battery, what charge is stored?


33.(2S).  Two conductors having net charges of +10.0 μC and –10.0 μC have a potential difference of 10.0 V between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to +100 μC and –100 μC?


34.(3S). An isolated charged conducting sphere of radius 12.0 cm creates an electric field of 4.90 × 104 N/C at a distance 21.0 cm from its center. (a) What is its surface charge density? (b) What is its 5S.          Two conducting spheres with diameters of 0.400 m and 1.00 m are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to 7.00 μC. (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be V = 0 at r = ∞?


35.(16S)  Two capacitors, C1 = 5.00 μF and C2 = 12.0 μF, are connected in parallel, and the resulting combination is connected to a 9.00-V battery. (a) What is the equivalent capacitance of the combination? What are (b) the potential difference across each capacitor and (c) the charge stored on each capacitor?


36.(17S) What If? The two capacitors of Problem 16 are now connected in series and to a 9.00-V battery. Find (a) the equivalent capacitance of the combination, (b) the potential difference across each capacitor, and (c) the charge on each capacitor.


37.(33S)  Two capacitors, C1 = 25.0 μF and C2 = 5.00 μF, are connected in parallel and charged with a 100-V power supply. (a) Draw a circuit diagram and calculate the total energy stored in the two capacitors. (b) What If? What potential difference would be required across the same two capacitors connected in series in order that the combination stores the same amount of energy as in (a)? Draw a circuit diagram of this circuit.



Electric Potential. Answers


1.(3E). Vba= 3.7 x 10-15J

              = 2.3 x 104 eV


2.(10G).  ( Vb – Va ) = -1.20 x 102 V


3.(1S). W = 1.35 MJ


4.(4S). W = -(-1.60 x 10-19C) V   From which, ∆V = -0.502 V


5.(5S). a)  ∆U = -6.00 x 10-4 J

          b) ∆V = -50.0 V


6.(7S). ∆V = -38.9 V. The origin is at highest potential.


7.(15S). a) V1 = 1.44 x 10-7 V

          b) V2 = 0.719 x 10-7 V

          c) ∆V = 7.19 x 10-8  V


8.(14G). V = 2.41 x 105 V.


9.(15G). V = 2.1 x 10-9 C


10.(16G). W = 2.5 J


11.(18G).  a)  V = 5.8 x 105V

              b) PE = 9.2 x 10-14 J


12.(19G). V =  (¸ KQ / 2 L) (¸ + 1 )         


13.(23G). W = 6.9 x 10-18 J


14.(16S).  a) F = 0

             b) E = 0

             c) V = 45.0 kV


15.(17S). a) r = 6.00 m

            b) Q = -2.00 mC


16.(19S). V = -11.0 MV


17.(20S). a) U = -3.86 x 10-7 J

             b) V =  103 V


18.(23S). U = 5.41 keQ2/s

 19.(28S).    a) V1 = 4.50 kV

                      U12 + U23 + U13 = -4.50 x 10-5 J         

b)      v = 3.46 x 104 m/s  


20.(34S). U = -3.96 J   


21.(37S). a) At  x = 0              V = 10.0 V

                 At x = 3.00 m      V = -11.0 V

                 At x = 6.00 m      V = - 32.0 V

            b) E = 7.00 N/C  in the +  x  direction


22.(38S).  a) For  r< R       Er = 0

              b) For  r≥ R      Er = keQ / r2


23.(39S).   E = 7.07 N/C




Capacitors. Answers


24.(31G). C = 2.9 x 10-6 F


25.(34G).  Q = 8.4 x 10-5 C


26.(35G).  A = 5.0 x 107 m2


27.(37G). Q = 2.63 x 10-8 C


28.(39G). E = 4.5 x 104 V/m


29.(41G). V1  =  V2 = 711.95 V


              Q1final = 1.78 x 10-3 C

              Q2final = 4.84 x 10-3 C


30.( 43G). C = 1.5 x 10-10 F


31.(47G).   C =  9.6 x 10-5 F


32.(1S). a)   C = 48.0 mC

b)      C = 6.00 mC


33.(2S). a)  F = 1.00 mF

b)      DV = 100 V


34.(3S).  a)  s = 1.33 mC / m2


             b) C = 13.3 pF


35.(16S). a)  Ceq = 17.0 mF

b)      DV = 9.00 V

c)      Q5 = 45.0 mC             Q12 = 108 mC


36.(17S).  a)   Ceq = 3.53 mF

             b)   DV1 = 6.35 V               DV2 = 2.65 V

d)      Q1 = Q2 = 31.8 mC


37.(33S). a)  U = 0.150 J

             b)  DV = 268 V