Question

An electrostatic force of attraction between two point charges $A$ and $B$ is$1000N$. If the charge on $A$ is increased by $25$% and that on $B$ is reduced by $25$% and the initial distance between them is decreased by $25$%, the new force of attraction between them is ______$N.$

• A. $1666.67$
• B. $3256.33$
• C. $1253.45$
• D. $3333.3$

Answer 1

The correct option is A $1666.67$

Given, $1000=\frac{q_A{q}_B\times k}{r^2}$ ...(i)

Now, $F=\frac{(q_A\times \frac{5}{4})(q_B\times \frac{3}{4})\times k}{r^2{\left(\frac{3}{4}\right)}^2}$ ...(ii)

Dividing (ii) by (i), we get,

$\frac{F}{1000}=\frac{\frac{5}{4}\times \frac{3}{4}}{\frac{9}{16}}$

$F=1000\times \frac{15\times 16}{16\times 9}$

$F=1666.67N$