ELECTROSTATICS O29

Step 1: Conservation of Energy Principle

When the disk is released, the electrostatic repulsion from the fixed hemispherical shell pushes it away. At infinity, the interaction energy becomes zero. The work done by the electrostatic forces converts the initial interaction potential energy into kinetic energy.

$$ K_{max} = U_{interaction} $$

We need to find the interaction energy between a charged disk and a charged hemispherical shell that closes it.

Step 2: Calculate Interaction Energy

The interaction energy can be calculated by considering the potential due to the hemisphere at the location of the disk.

From electrostatics, we know that the potential at the center $O$ of a uniformly charged hemispherical shell (charge $Q_{hemi}$) is:

$$ V_O = \frac{1}{4\pi\epsilon_0} \frac{Q_{hemi}}{R} $$

Where $Q_{hemi} = (2\pi R^2)\sigma$. Thus,

$$ V_O = \frac{1}{4\pi\epsilon_0} \frac{2\pi R^2 \sigma}{R} = \frac{\sigma R}{2\epsilon_0} $$

For a “closed” spherical shell system (Disk + Hemisphere), the calculation of interaction energy simplifies to:

$$ U_{int} = Q_{disk} \times V_{effective} $$

the potential due to the hemisphere everywhere on the circular base is same as that at the centre because the electric field does not have any component along the circular base. This can be understood by recalling that net field inside a spherical shall is zero if upper hemisphere has a component of field radially outward, the lower hemisphere will also have component in same direction. This means there is no component of electric field due to hemisphere along the circular base..

$$ U_{int} = Q_{disk} \left( \frac{\sigma R}{2\epsilon_0} \right) $$

Charge on the disk $Q_{disk} = \pi R^2 \sigma$.