Question

Two point charges of $30 \mu C$ and $40 \mu C$ are $20 cm$ 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. $5 cm$

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

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