Answer to Question #304858 in Electric Circuits for Bluebalda

Question

Two charges Q1=+3.5 nC and Q2=-5.3 nC are separated by 10 cm. Calculate the electric potential at point B

Answer 1

Electric potential

$V_1=\frac{kq_1}{r}$

$V_1=\frac{9\times10^9\times3.5\times10^{-9}}{5\times10^{-2}}=630V$

$V_2=\frac{kq_2}{r}$

$V_2=-\frac{9\times10^9\times5.3\times10^{-9}}{5\times(10)^{-2}}=-954V$

$V_{net}=V_1+V_2$

$V_{net}=630-954=-324V$