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Q. What is the electric potential at a distance of 9 cm from 3 nC ?

KCETKCET 2014Electrostatic Potential and Capacitance

Solution:

Given: $\quad r=9 cm =0.09 m$
Charge $Q =3 nC =3 \times 10^{-9} C$
Potential at a distance
$9 cm , \quad V=\frac{ kQ }{ r } \quad$ where $k =9 \times 10^{9}$
$\therefore V =\frac{9 \times 10^{9} \times 3 \times 10^{-9}}{0.09}=300 V$