Question

Two points A and B are near a point charge as shown below. ke = 9.00 x 10°N-m2/C2 80 = 8.85 x 10-12 CZ/N-m2 А 15.0 cm 25.0° B -4.00 PC 25.0 cm a) What is the difference in electric potential VB VA? b) What is the change in potential energy for moving a-3.00 pC charge moved from A to B ? A. The difference in potential is 1.02 x 106 B. The difference in potential is -1.02 x 106 The difference in potential is 9.60 x 104 v The difference in potential is -9.60 x 104 v E. The difference in potential is none of these answers The change in potential energy is - 3.06 ) C. D G. Н. The change in potential energy is 3.06 ) The change in potential energy is none of these answers The change in potential energy is 0.288) J. The change in potential energy is .0.288)

Answer 1

The potential of the electric field created by a charge is given by:

$V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$

where r is the distance from the charge.

The potential at A is thus:

1 9 VA 9 x 10 - 4 x 10-6 0.15 = -240000V 41TE0 0.15

and $V_B=\frac{1}{4\pi\epsilon_0}\frac{q}{0.25} =9\times 10^9\frac{-4\times 10^{-6}}{0.25}=-144000V$

Thus,

$V_B-V_A = -144000-(-240000)=+96000 V=+9.6\times 10^4 V$

The other charge is

$q_2 = -3.0\mu C$

The potential energy is given by:

$U=qV\implies \Delta U=q\Delta V=-3.0\times 10^{-6}(96000)=-0.288\text{ J}$