Question

A thin spherical shell of radius R = 20.0 cm has total charge Q = 58.0 nC uniformly distributed on its surface.

Part A

A test particle with charge q = 8.00 nC is initially at a position r = 43.0 cm from the center of the shell. The particle moves to the surface of the shell closest to its initial position. What is the change in potential energy of the test particle as a result of this move?

Part B

What is the potential difference between the final position of the test charge in part A and the initial position of the test charge in part A?

Part C

The test particle now moves from the surface of the charged shell to the center of the shell. What is the change in potential energy of the test particle as a result of this 20.0 cm move?

Part D

What is the potential difference between the center of the shell and the surface of the shell?

Answer 1

(a) Write the equation of Gauss law. E-A= E(4T2) _ E= 4TE 4πε,r Write the formula for the potential E dr V=J - dr 4πε, " = 4TEr Write the formula ofpotential difference. AV =V2-V (1) Here, AV is the potential difference, Vis the potential at the surface and V is the potential at r 43.00 cm forV and 4,7TE for V in (1) Απε,η Substitute 1 1 AV: 4πε, | (2)

1 20cm for r and 43cmfor, in (2). 47tE Substitute 58.0 nC for q,, 9x10'Nm2/C2 for- 1 (58.0nC) (9x109 Nm2/C2' 1 AV 20 cm 43 cm 100 cm 1C - (58.0 nC)10 nC 1 9x10' Nm2/c2) 1 43 cm 20 cm 1m 23 =(5220x43 00)V Rewrite the above equation. (3) AV 1396.04 V Write the equation for the change in potential energy. AU =qAV (4) Here, AU is the change in potential energy and q is the charge on the test particle Substitute 8.00 nC for q and 1396.04V for AV in (4) AU (8.00 nC) (1396.04 V) 1C (1396.04V) =(8.00 nC) 10 nC =1.12X10J Hence, the change in potential energy is 1.12x105 J

(b) The potential difference between the final position of the test charge in part A and the initial position of the test charge in part A caleulated in equation (3) Hence, the potential difference between the final position of the test charge in part A and the initial position of the test charge in part A is 1396.04 V (e) Write the equation of Gauss law. ЕA- (5) At the center of the spherical shell, charge is zero Substitute 0C for q, in (5) OC E - Ав, 0 N/C Thus, potential at the center of the spherical shell is zero Write the formula for the potential difference between center and surface of the shell AV -V (6) Here, AV is the potential difference, V is the potential at the center and V is the potential at the surface Я. Substitute 0V forV, and -for V, in (6) Απε,

AV" =0 V -_4 4πεJ, Rewrite the above equation. AV". (7) 4πε. 1 and 20cm forr, in (7) Απε, Substitute 58.00 nC for q.. 9x10° Nm2/C2for' (58.00nC) (9x10 Nm2/C2) AV 20 cm 1C (58.001C)10 nC (9x10°Nm2/C2 1m (20 em) 100cm Rewrite the above equation AV2610 V (8) Write the equationfor the change in potential energy AU qAV (9) Substitute 2610V for AV and 8.00 nC forq in (9) AU (8.00 nC)2610V) 1C (8.00nC10nC /-2610V) =2.09x105 J

Hence, the change in potential energy is-1.73x10 J (d) The potential difference betweenthe center of the shell and the surface of the shell calculated in equation (3) Hence, the potential difference between the center of the shell and the surface of the shell is 2610V