A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)

We have a capacitor of capacitance $$C_1 = 60\ \text{pF} = 60 \times 10^{-12}\ \text{F}$$ that is first charged to a potential difference $$V_1 = 20\ \text{V}$$ by a battery.

For a capacitor charged to a voltage $$V$$, the energy stored is given by the formula

$$U = \dfrac{1}{2} C V^{2}.$$

Substituting the given values, the initial electrostatic energy is

$$ U_{\text{initial}} = \dfrac{1}{2}\,(60 \times 10^{-12})\,(20)^{2} = \dfrac{1}{2}\,(60 \times 10^{-12})\,(400) = 30 \times 10^{-12} \times 400 \times \dfrac{1}{2}. $$

Simplifying step by step,

$$ \dfrac{1}{2}\,(60 \times 10^{-12})\,(400) = 30 \times 10^{-12} \times 400 = 12 \times 10^{-9}\ \text{J}. $$

Thus $$U_{\text{initial}} = 12\ \text{nJ}.$$

Now the battery is disconnected, so the total charge $$Q$$ on the first capacitor remains fixed. Using $$Q = C V$$ we get

$$ Q = C_1 V_1 = (60 \times 10^{-12})(20) = 1200 \times 10^{-12}\ \text{C} = 1.2 \times 10^{-9}\ \text{C}. $$

This charged capacitor is then connected in parallel to a second identical but uncharged capacitor $$C_2 = 60\ \text{pF}$$. In a parallel combination the equivalent capacitance is the sum, so

$$ C_{\text{eq}} = C_1 + C_2 = 60\ \text{pF} + 60\ \text{pF} = 120\ \text{pF} = 120 \times 10^{-12}\ \text{F}. $$

Since no charge can escape the system, the total charge $$Q$$ now spreads over both capacitors. The final common voltage $$V_f$$ is obtained from

$$ V_f = \dfrac{Q}{C_{\text{eq}}} = \dfrac{1.2 \times 10^{-9}}{120 \times 10^{-12}} = \dfrac{1.2}{120} \times 10^{3} = 0.01 \times 10^{3} = 10\ \text{V}. $$

The final energy stored in both capacitors together is again given by $$U = \tfrac{1}{2}CV^{2}$$, now with the equivalent capacitance and the common voltage, so

$$ U_{\text{final}} = \dfrac{1}{2}\,(120 \times 10^{-12})\,(10)^{2} = \dfrac{1}{2}\,(120 \times 10^{-12})\,(100) = 60 \times 10^{-12} \times 100 = 6 \times 10^{-9}\ \text{J}. $$

Thus $$U_{\text{final}} = 6\ \text{nJ}.$$

The energy lost in the process is the difference between the initial and final energies:

$$ \Delta U = U_{\text{initial}} - U_{\text{final}} = 12\ \text{nJ} - 6\ \text{nJ} = 6\ \text{nJ}. $$

So, the answer is $$6\ \text{nJ}$$.