Question

A point source S is placed midway between two converging mirrors having equal focal length f as shown in figure. Find the values of d for which only one image is formed.
Figure

Answer 1

Explanation:

A capacitor behaves like an open circuit for DC. So, the reactance (the AC analog of DC resistance) would be infinite once the capacitor reaches the steady state. However, the capacitor would conduct during the transition period (when the direct current grows from I(0) to I(t)), due to changing electric field between its plates.

Consider this:

Charge Q, stored in a capacitor of value C, when a potential V is applied across it's plate, is given by:

Q = C*V

Now if you differentiate this equation w.r.t. time, you'll get the time dependent current flowing through the capacitor:

I(t) = dQ / dt = C * dV / dt.

Also, the electrostatic field inside a capacitor can be represented by:

E = -(grad)V. In simple terms, |E| = dV / dx.

Since, the plate separation is constant (say d), we can represent the field as:

V = E * d

That gives you a time varying current I(t) as:

I(t) = -C * d * dE / dt.

So, there you have it. If you have a changing electrostatic field inside the capacitor, it would give rise to a time varying current. This means, you'll have a non-zero reactance for the setup. But once your circuit reaches the steady state, the electrostatic field derivative w.r.t time will cease to exist and hence, the current will become zero (infinite reactance).

There is a definitive relation between the capacitive reactance and the rate at which the field changes inside it:

Capacitive reactance, Xc is given by:

Xc = 1 / w * C; where w = angular frequency of the source and C = capacitance.

Higher the rate at which the field inside it changes, lower will be the capacitive reactance (and vice versa).