Question

A spherical shell of radius r carries a uniformly distributed surface charge q on it. A hemispherical shell of radius R(> r) is placed covering it with its centre coinciding with that of the sphere of radius r. The hemisphere has a uniform surface charge Q on it. The charge distribution on the sphere and the hemisphere is not affected due to each other. Calculate the force that the sphere will exert on the hemisphere.

Answer 1

$\frac{qQ}{8\pi\epsilon_0 R^2}$

Answer 2

The electric field due to the sphere at the location of the hemisphere is $E_s = \frac{q}{4\pi\epsilon_0 R^2}$ radially outwards. The force on an infinitesimal charge element $dQ$ on the hemisphere is $dF = E_s dQ$. Due to symmetry, only the component of force along the hemisphere's axis of symmetry (e.g., z-axis) will be non-zero.

$F_z = \int E_s dQ \cos\theta$.

Substitute $dQ = \sigma dA = \frac{Q}{2\pi R^2} R^2 \sin\theta d\theta d\phi = \frac{Q}{2\pi} \sin\theta d\theta d\phi$.

$F_z = \int_0^{2\pi} \int_0^{\pi/2} \frac{q}{4\pi\epsilon_0 R^2} \frac{Q}{2\pi} \sin\theta \cos\theta d\theta d\phi$.

Integrating $\int_0^{2\pi} d\phi = 2\pi$ and $\int_0^{\pi/2} \sin\theta \cos\theta d\theta = \frac{1}{2}$.

$F_z = \frac{qQ}{8\pi^2\epsilon_0 R^2} (2\pi) (\frac{1}{2}) = \frac{qQ}{8\pi\epsilon_0 R^2}$.