Question

Two identical balls carrying charges $+5\mu C$ and $-2\mu C$ attracts each other with force $F$ in air. After keeping them into contact they are placed at same distance in water $\left({\in }_r=81\right)$ the new force will be

• A. $\frac{9}{40}F$
• B. $F$
• C. $\frac{F}{81}$
• D. $\frac{F}{360}$

Answer 1

The correct option is C $\frac{F}{81}$

Two identical balls carrying charges $=+5\mu e$ and $-2\mu e$

Force $=F$ (is air)

${\in }_r=81$

Using coulomb's law,

$F=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$

$F=\frac{1}{4\pi {\in }_0}\times \frac{5\times 2}{r^2}$

$F^1=\frac{1}{K4\pi {\in }_0}\times \frac{5\times 2}{r^2}=\frac{1}{81}\times \left(F\right)$

or, $F^1=\frac{F}{81}$.