A parallel-plate capacitor with a plate area of 20 cm^2 and a...

Let the initial capacitance be $C_i = \dfrac{\epsilon_0 A}{d}$. When the separation is increased to $2d$, the final capacitance becomes $C_f = \dfrac{\epsilon_0 A}{2d} = \dfrac{C_i}{2}$.

The initial stored energy in the capacitor is $U_i = \dfrac{1}{2} C_i V^2$.

The final stored energy is $U_f = \dfrac{1}{2} C_f V^2 = \dfrac{1}{4} C_i V^2$.

The initial charge on the capacitor is $Q_i = C_i V$ and the final charge is $Q_f = C_f V = \dfrac{C_i V}{2}$.

Since the charge decreases, charge flows back to the battery. The energy absorbed by the battery is $\Delta Q \times V = \left(C_i V - \dfrac{C_i V}{2}\right) V = \dfrac{1}{2} C_i V^2$.

The attractive force between the plates at a separation $x$ is $F = \dfrac{Q^2}{2 \epsilon_0 A} = \dfrac{\epsilon_0 A V^2}{2 x^2}$.

The work done by the external agent (person) to pull the plates apart is:

$W_{ext} = \int_d^{2d} F dx = \int_d^{2d} \dfrac{\epsilon_0 A V^2}{2 x^2} dx = \dfrac{\epsilon_0 A V^2}{2} \left[ -\dfrac{1}{x} \right]_d^{2d} = \dfrac{\epsilon_0 A V^2}{4 d} = \dfrac{1}{4} C_i V^2$

Applying the principle of conservation of energy:

$W_{ext} + U_i = U_f + \text{Energy absorbed by battery} + H$

$\dfrac{1}{4} C_i V^2 + \dfrac{1}{2} C_i V^2 = \dfrac{1}{4} C_i V^2 + \dfrac{1}{2} C_i V^2 + H$

Solving for $H$, we get $H = 0$.

Thus, no heat is produced because the work done by the person plus the initial stored energy equals the final stored energy plus the energy absorbed by the battery.