A capacitor with air as a dielectric is connected to a voltmeter of infinite impedance. The voltmeter reads a nonzero value. If the distance between the plates is doubled and the effective area of the plates is made in half, what happens to the voltmeter reading?

Concept:

Capacitance for a parallel plate capacitor is given by:

 \(C = {ϵ_o ϵ_r A\over d}\)

where, C = capacitance

ϵ_o = absolute permittivity 

ϵr = relative permittivity 

A = cross-sectional area

d = distance between the plates

Also, Q = CV

where, Q = stored charge

C = capacitance

V = voltage

Calculation:

When a parallel plate capacitor is connected across the voltmeter the charge remains constant.

Therefore, V α \( {1 \over C}\)

Case 1: When air as a dielectric is filled between the capacitor, then ϵ_r = 1

 \(C_1={ϵ_o (1) A_1\over d_1}\)

Case 2: When d₂ = 2d₁ and A₂ = \( {A_1\over 2}\)

 \(C_2= {ϵ_o (1) A_2\over d_2}\)

 \(C_2={ϵ_o {(A_1/2)}\over 2d_1}\)

\(C_2={ϵ_o {A_1}\over 4d_1}\)

\({V_2\over V_1} ={C_1\over C_2}\)

\({V_2\over V_1} = {\epsilon_oA_1 \over d_1} \times {4d_1 \over \epsilon_oA_1} \)

\({V_2\over V_1} = 4\)

\({V_2} = 4V_1\)

Voltmeter reading becomes four times of initial reading.