Question

A metallic sphere of radius $R$ is cut in two parts along a plane whose minimum distance from the sphere's centre is $h=\frac{R}{2}$ and the sphere is uniformly charged with a total electric charge $Q$. The minimum force that should be applied on each of the two parts, to hold the two parts of the sphere together is $\frac{3K{Q}^2}{p{R}^2}$. Then, find the value of $p$

• A. 32.0
• B. 32
• C. 32.00

Answer 1

Answer: (A), (B), (C)

Here, we use the concept of electric pressure.
$P_e=\frac{{\sigma }^2}{2{\epsilon }_0}$ at each point on a conducting surface.

The plane cutting the sphere is shown below (minimum distance from centre is $R/2$).


We know that, surface charge density $\sigma =\frac{Q}{4\pi R^2}$

We know that, minimum force necessary (to be applied on each of the two parts) to hold the two parts of the sphere together is,
$F=P_{e}A.....\left(1\right)$

Here, $A$ is the effective projected area = $\pi r^2$

Now, calculating $r$ from the figure,

$r=\sqrt{R^2-\frac{R^2}{4}}=\frac{\sqrt{3}R}{2}$

Now, substituting the values in $\left(1\right)$,
$F=PA=\frac{{\sigma }^2}{2{\epsilon }_0}\times (\pi R^2\times \frac{3}{4})$

Substituting the value of $\sigma$,
$\Rightarrow F=\frac{Q^2}{(4\pi R^2{)}^2}\times \frac{1}{2{\epsilon }_0}\pi R^2\left(\frac{3}{4}\right)$

$\Rightarrow F=\frac{3Q^2}{128\pi R^2{\epsilon }_0}$

$\Rightarrow F=\frac{3K{Q}^2}{32R^2}$ where $\frac{1}{4\pi {\epsilon }_0}=K$

Now, comparing this with the given value of force,
$F=\frac{3K{Q}^2}{p{R}^2}$,
we can get the value of $p=32$.