Question

At one corner of a 3cm by 4cm rectangle is placed a charge of -20pc and at the two adjacent corner are charge of 10pc .calculate the potential at the fourth corner.

Answer 1

Potential due to point charge

$V=k_c\frac{q}{r}$

Where $V=$ Electric potential

$q=$ point charge

$r=$ Distance between any point around the charge to the point charge

$k_c=$ Coulombs constant $k_c=9×10^9N$

Potential due to $+10pc$ at A

$V_{+10A}=\frac{9×10^9×10×10^{-12}}{3×10^{-2}}$

$=3V$

Potential due to +10pc at C

$V_{+10C}=\frac{9×10^9×10×10^{-12}}{4×10^{-2}}=2.25V$

PB$=\sqrt{3^2+4^2}=5$

Potential due to -20pc at B

$V_{-20B}=\frac{-9×10^9×20×10^{-12}}{5×10^{-2}}$ $=-3.6V$

Potential at P$=3+2.25-3.6$

$=1.65V$