Answer to Question #110209 in Electric Circuits for Mukelo Madolo

Question

Test 2

A 200-V battery is connected to a 0.50-uF parallel-plate air filled capacitor.Now the battery is disconnected, with care taken not to discharge the plate. Some Pyrex glass is then inserted between the plates,completely filling up the space.What is the final (a) potential difference (b) electric field between the plates?

Answer 1

The initial capacitance of the parallel plates capacitor

$$C_0=\frac{\epsilon_0 A}{d}=50\times 10^{-6}\:\rm F.$$

When the capacitor filled by Pyrex glass with dielectric constant $k=4.84$, the capacitance increases

$$C=\frac{k\epsilon_0 A}{d}=4.84\times 50\times 10^{-6}=2.42\times 10^{-4}\:\rm F.$$

(a) Since the capacitor is disconnected from source, so charge at the plates is constant. Hence,

$$q_0=q\\ C_0V_0=CV$$

The final potential difference

$$V=\frac{C_0V_0}{C}=\frac{200\:\rm V}{4.84}=41.3\:\rm V.$$