Q.Derive the expression for the capacitance of spherical capacitor. How its capacitance can be increased? OR Using Gauss's law, derive the expression for electric field intensity due to a uniformly charged solid cylinder at a point which lies:

Capacitance of a Spherical Conductor

The Core Intuition

Imagine you have a metal sphere sitting alone in space. You connect it briefly to a battery — some charge $Q$ flows onto it. The sphere now sits at some potential $V$ relative to infinity (where we take potential as zero).

What determines how much the potential rises for a given charge? The sphere's size. A large sphere can "spread out" the charge over a bigger surface, so the charge density is lower, and the potential near it is smaller. A small sphere concentrates the same charge, producing a much higher potential.

Capacitance is simply the ratio that captures this: how much charge you need to raise the potential by one volt. A larger capacitance means you can store more charge for the same potential rise.

Deriving the Formula

We need two things: the potential of a charged sphere, and the definition of capacitance.

Potential of a charged conducting sphere

A conducting sphere of radius $R$ carrying charge $Q$ behaves, for points outside it, exactly as if all the charge were concentrated at its centre. The potential at its surface (which is the same as the potential of the whole conductor) is:

This comes directly from the formula for potential due to a point charge, applied at distance $R$ from the centre.

Definition of capacitance

Capacitance is defined as:

Substitute the expression for $V$:

That is the capacitance of an isolated conducting sphere of radius $R$.

What This Tells You

• Capacitance is directly proportional to radius. Double the radius, double the capacitance. A bigger sphere can hold more charge at the same potential.
• The constant $4\pi\varepsilon_0 \approx 1.11 \times 10^{-10}\ \text{F/m}$. So a sphere of radius $1\ \text{m}$ has a capacitance of about $111\ \text{pF}$ — a very small number. That is why real capacitors use two close plates: to get far larger capacitance in a small volume.
• The formula depends only on $R$, not on the material of the sphere (as long as it is a conductor). The charge always resides on the outer surface, and the field outside is the same regardless of what the sphere is made of. …