Question

Two identical metal sphere carrying charges of $9\times {10}^{-9}C$ and $-3\times {10}^{-9}$ C are brought in contact and then they are separated by a distance of 3 m and kept in a medium of dielectric constant 5. what is the electrostatic force between them ?

• A. $1.25\times {10}^{-9}N$
• B. $1.8\times {10}^{-9}N$
• C. $3\times {10}^{-9}N$
• D. $2.5\times {10}^{-9}N$

Answer 1

The correct option is B $1.8\times {10}^{-9}N$

As per the property of charge the charge spread all around a conducting body. So as soon as the identical metallic spheres will be brought together, the total charge will get divided equally on both the spheres as $Q/2$ where $Q$ is the total charge.

i.e, $Q/2=\frac{9\times {10}^{-9}+(-3\times {10}^{-9})}{2}=3\times {10}^{-9}C$

$\therefore$ Q/2 is the charge on both spheres now.

Now as we know

Electrostatic force $F=\frac{K{Q}_1{Q}_2}{r^2}$

Where $K=\frac{1}{4\pi E}$ and $E=E_r\times E_0$

A/Q $E_r=5$ So, $K^1=\frac{1}{4\pi \left(5E_0\right)}\Rightarrow K^1=\frac{K}{5}\Rightarrow K=\frac{1}{2\pi E_0}=9\times {10}^9$

So, $\Rightarrow F=\frac{K(Q/2{)}^2}{5r^2}\Rightarrow F=\frac{9\times {10}^9\times (3\times {10}^{-9}{)}^2}{5\times (3{)}^2}=\frac{9}{5}\times {10}^{-9}F=1.8\times {10}^{-9}N$