Answer to Question #312490 in Electric Circuits for dan

Question

two charges Q1=+3nC and Q2=-5nC are separated by a distance of 6cm and 2cm for point A amd 2cm and 10cm for point B (a) Calculate the electric potential at point A and point B.

(b) How much work is done by the electric field in a moving 4.00nC particle from point A and point B.

Answer 1

Potential

$V_A=\frac{kq_1}{r}$

$V_A=\frac{9\times10^9\times3\times10^{-9}}{0.02^2}=67500V$

$V_B=-\frac{9\times10^9\times5\times10^{-9}}{0.1^2}=-4500V$

(2)

Work done

$W=q(V_A-V_B)$

$$W=4\times10^{-9}\times(76500+4500)=3.24\times10^{-4}J$$