Question

Two point charges $4\mu c$ and $9\mu c$ are separated by 50 cm. the potential at the point between them where the field has zero strength is

• A. $4.5\times {10}^{5}V$
• B. $9\times {10}^{5}V$
• C. $9\times {10}^{4}V$
• D. Zero

Answer 1

The correct option is A $4.5\times {10}^{5}V$

REF.Image

Consider at point P, the electric field is zero,

as Electric field,

$E=\frac{kq}{d^2}$

at P,

$\frac{k{q}_1}{x^2}=\frac{k{q}_2}{(50-x{)}^2}\Rightarrow \frac{4}{x^2}=\frac{9}{(50-x{)}^2}\Rightarrow (\frac{50-x}{x}{)}^2=\frac{9}{4}$

$\Rightarrow \frac{50-x}{x}=\pm \frac{3}{2}\Rightarrow x=20cm$ OR $x=-150cm$ (Not Applicable)

Hence, at a distance = 20 cm from A, E=0

Also, potential at that point,

$V=\frac{k{q}_1}{x}+\frac{k{q}_2}{50-x}=9\times {10}^9\times (\frac{4\times {10}^{-6}}{0.2}+\frac{9\times {10}^{-6}}{0.3})=4.5\times {10}^{5}v$ (Ans)