Question

Two point charges of $30\mu C$ and $40\mu C$ are $20cm$ apart from each other. Where will be the electric field zero on the line joining the charges from $30\mu C$ charge?

• A. $\frac{20\sqrt{3}}{\sqrt{2}+1}cm$
• B. $\frac{20\sqrt{3}}{\sqrt{3}+2}cm$
• C. $\frac{20\sqrt{3}}{\sqrt{5}}cm$
• D. $5cm$

Answer 1

Answer: (B)

Both charges are of same nature so the neutral point will be obtained anywhere midway between the line joining the charge

$\frac{K\times 30\times 10^{-6}}{(x{)}^2}=\frac{K\times 40\times 10^{-6}}{(20-x{)}^2}$
$\frac{3}{x^2}=\frac{4}{(20-x{)}^2}$
$\frac{\sqrt{3}}{x}=\frac{2}{20-x}$
$20\sqrt{3}-\sqrt{3}x=2x$
$20\sqrt{3}=2x+\sqrt{3}x$
$\frac{20\sqrt{3}}{\sqrt{3}+2}cm=x$