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Q. Figure shows a triangular array of three point charges. The electric potential V of these source charges at the midpoint P of the base of the triangle isPhysics Question Image

VITEEEVITEEE 2006

Solution:

The net electric potential is algebraic sum of potential due to individual point charges.
$V=\frac{1}{4\pi\in_0}\bigg[\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\bigg]$
$V=\frac{1}{4\piε_0}\bigg[\frac{1 X 10^{-6}}{0.2}-\frac{2 X 10^{-6}}{0.2}+\frac{3 X 10^{-6}}{0.3}$
$=9 X10^{9}[5-10+10]X10^{-6}$
$=9X10^3[5]=45X10^3V=45kV$