Question

Please explain step-by-step how you got your answer. Thank you!

A parallel-plate capacitor is connected to a battery and stores 11.1 nC of charge. Then, while the battery remains connected, a sheet of Pyrex glass is inserted between the plates. By how much does the charge change? (Express your answer in nC)

Answer 1

For a parallel plate capacitor, the expression for the capacitance is -

C = ε₀εr(A/d) in Farads
ε₀ is 8.8542e-12 F/m
εr is dielectric constant (vacuum = 1) [This is our case, where the space between the plates is vaccum]

A and d are area of plate in m² and separation in m

From the above, C is proportional to dielectric constant, so C1/C2 = a/b

From Q = CV, charge is proportional to C,
so Q1/Q2 = a/b ---------------------------------------------------(i)

Now, when consulting your text book you will find that

dielectric constant of pyrex glass, b = 4.6

and dielectric constant of vaccum, a = 1

and Q1 = 11.1 nC

put these values in (i) -

11.1 / Q2 = 1/4.6

=> Q2 = 11.1 x 4.6 = 51.06 nC

So, the new charge = 51.06 nC.