Question

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on the x axis, and a 70-nc charge at y = –6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis?

Answer 1

Answer:48 V

Explanation:

Given

Three charged particle with charge

$q_1=50\ nC at y=6\ m$

$q_2=-80\ nC at x=-4\ m$

$q_3=70\ nC at y=-6\ m$

Electric Potential is given by

$V=\frac{kQ}{r}$

Distance of $q_1$ from $x=8\ m$

$d_1=\sqrt{6^2+8^2}$

$d_1=\sqrt{36+64}$

$d_1=10\ m$

similarly $d_2=8-(-4)$

$d_2=12\ m$

$d_3=\sqrt{(-6)^2+8^2}$

$d_3=\sqrt{36+64}$

$d_3=10\ m$

Potential at $x=8\ m$ is

$V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}$

$V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]$

$V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}$

$V_{net}=9\times 5.33$

$V_{net}=47.97\approx 48\ V$