Question

Positive charges of $2\mu C$ and $8\mu C$ are placed $15cm$ apart. At what distance from the smaller charge and electric field due to them will be zero ?

• A. 3 cm
• B. 5 cm
• C. 7 cm
• D. 10 cm

Answer 1

Answer: (B)

$\frac{1}{4\pi {\epsilon }_0}\frac{2\times 10^{-6}}{x^2}=\frac{1}{4\pi {\epsilon }_0}\frac{8\times b10^6}{(15-x{)}^2}$
On solving the obtained Quadratic equation, $x^2+10x-75=0$, we get x = 5 cm.