A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is:

We begin by recalling the energy-charge-capacitance relation for a capacitor whose charge remains constant. The electrostatic energy stored is given by the formula

$$U=\frac{Q^{2}}{2C},$$

where $$Q$$ is the charge on the plates and $$C$$ is the capacitance. Because the battery is disconnected before the dielectric slab is introduced, the charge $$Q$$ already on the plates cannot change.

At the very start, the given parallel-plate capacitor has a capacitance

$$C_{0}=12\;\text{pF}=12\times10^{-12}\,\text{F}$$

and is charged to a potential difference

$$V_{0}=10\;\text{V}.$$

Using the definition $$Q=CV,$$ the initial charge stored on the plates is

$$Q=C_{0}V_{0}=12\times10^{-12}\,\text{F}\times10\;\text{V}=120\times10^{-12}\;\text{C}=120\;\text{pC}.$$

Squaring this charge for later use gives

$$Q^{2}=(120\;\text{pC})^{2}=120^{2}\times10^{-24}\;\text{C}^{2}=14400\times10^{-24}\;\text{C}^{2}=1.44\times10^{-20}\;\text{C}^{2}.$$

Now we compute the initial energy stored in the capacitor:

$$U_{1}=\frac{Q^{2}}{2C_{0}}=\frac{1.44\times10^{-20}\;\text{C}^{2}}{2\times12\times10^{-12}\,\text{F}}=\frac{1.44\times10^{-20}}{24\times10^{-12}}\;\text{J}.$$

Carrying out the division,

$$\frac{1.44}{24}=0.06,\qquad 10^{-20+12}=10^{-8},$$

so

$$U_{1}=0.06\times10^{-8}\;\text{J}=6.0\times10^{-10}\;\text{J}.$$

Because $$1\;\text{pJ}=10^{-12}\;\text{J},$$ this is

$$U_{1}=600\;\text{pJ}.$$

Next, a porcelain slab with dielectric constant $$k=6.5$$ is inserted. For a completely filled capacitor the capacitance multiplies by the dielectric constant, giving

$$C_{2}=kC_{0}=6.5\times12\;\text{pF}=78\;\text{pF}=78\times10^{-12}\,\text{F}.$$

The charge is still $$Q=120\;\text{pC},$$ so the new energy becomes

$$U_{2}=\frac{Q^{2}}{2C_{2}}=\frac{1.44\times10^{-20}}{2\times78\times10^{-12}}\;\text{J}=\frac{1.44\times10^{-20}}{156\times10^{-12}}\;\text{J}.$$

Simplifying the numerical factor,

$$\frac{1.44}{156}=0.00923077,\qquad 10^{-20+12}=10^{-8},$$

hence

$$U_{2}=0.00923077\times10^{-8}\;\text{J}=9.23\times10^{-11}\;\text{J}.$$

Converting to picojoules,

$$U_{2}=92.3\;\text{pJ}.$$

The capacitor’s energy has decreased, and by conservation of energy that decrease appears as work done by the capacitor on the dielectric slab. Therefore the work done is

$$W=U_{1}-U_{2}=600\;\text{pJ}-92.3\;\text{pJ}=507.7\;\text{pJ}\approx508\;\text{pJ}.$$

Hence, the correct answer is Option C.