Question

Two small identical conducting balls $A$ and $B$ of charges $+10\mu \text{C}$ and $+30\mu \text{C}$ respectively are kept at a separation of $50\text{cm}$. These balls have been connected by a wire for a short time.

$\left(\text{ii}\right)$ The electrostatic force between the balls will be

​​​​​​​$[$1 Mark$]$

• A. $28.8\text{N}$
• B. $32.6\text{N}$
• C. $14.4\text{N}$
• D. $72\text{N}$

Answer 1

The correct option is C $14.4\text{N}$

The electrostatic force between two charges $q_1$ and $q_2$ seperated by a distance $r$ is given by
$F=\frac{k{q}_1{q}_2}{r^2}$
Where $k=9\times {10}^9$
After connection, charges on both the balls will be $+20\mu \text{C}$
Here $r=50\text{cm}=0.5\text{m}$

$F=\frac{9\times {10}^9\times (20\times {10}^{-6}{)}^2}{(0.5{)}^2}$

$F=14.4\text{N}$

$\therefore$ option $\left(c\right)$ is correct