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 by a total electric charge $Q$. The minimum force necessary (to 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 $p$?

Answer 1

Here we use concept of electric pressure, $P_e=\frac{{\rho }^2}{2{ϵ}_0}$, since we know that ,$F=PA$

Also we kno that surface charge density $\rho =\frac{Q}{4\pi R^2}$

and area should be the effective area , i.e, $A=\pi \sqrt{(}{R}^2-h^2{)}^2=\pi (R^2-h^2)$

Hence , $F=P_e\times A=\frac{{\rho }^2}{2{ϵ}_0}\times \pi (R^2-h^2)$

$=\frac{Q^2}{2{ϵ}_0(4\pi R^2{)}^2}\times \pi (R^2-h^2)$

simplifying it we get , $F=\frac{k{Q}^2}{8{ϵ}_0{R}^4}(R^2-h^2)$ , here $k=\frac{1}{4\pi {ϵ}_0}$

using $h=\frac{R}{2}$ , we get value of $F=\frac{k{Q}^2}{8{ϵ}_0{R}^4}\times (R^2-(\frac{R}{2}{)}^2)$

this gives , $F=\frac{3k{Q}^2}{32R^2}$

Now comparing this from the given value of force we can get the value of $p=32$