Problem Analysis

We have two isolated conducting bodies, A and B, placed far apart (so their mutual capacitance is negligible). Let their self-capacitances be $C_1$ and $C_2$ respectively.

We are given three scenarios involving charging and connecting these bodies, and we need to relate the work done by electrical forces in the third scenario to the work done in the first two.

Key Physics Principles:

• The energy stored in a charged capacitor (or conductor) is $U = \frac{q^2}{2C}$.
• When two conductors are connected, charge redistributes until they reach a common potential $V$. Charge conservation applies.
• Work done by electrical forces is equal to the decrease in the system’s electrostatic potential energy: $W_{electric} = -\Delta U = U_{initial} – U_{final}$.

Step-by-Step Derivation

Scenario 1

Body A is given charge $q$, body B is neutral ($q_B = 0$). Then B is brought in contact with A.

• Initial State: A has charge $q$, B has 0. Energy $U_{1i} = \frac{q^2}{2C_1}$.
• Final State: They are connected. Total charge $q$ redistributes. The combined system has capacitance $C_{eq} = C_1 + C_2$ (since they acquire common potential $V$ and are far apart, capacities add). The final potential is $V = \frac{q}{C_1+C_2}$.
• Final Energy: $U_{1f} = \frac{q^2}{2(C_1+C_2)}$.
• Work Done ($W_1$): $$ W_1 = U_{1i} – U_{1f} = \frac{q^2}{2C_1} – \frac{q^2}{2(C_1+C_2)} $$ $$ W_1 = \frac{q^2}{2} \left( \frac{1}{C_1} – \frac{1}{C_1+C_2} \right) $$

Scenario 2

Body B is given charge $q$, body A is neutral ($q_A = 0$). Then A is brought in contact with B.

• Initial State: B has charge $q$, A has 0. Energy $U_{2i} = \frac{q^2}{2C_2}$.
• Final State: Connected. Total charge $q$. Capacitance $C_{eq} = C_1 + C_2$.
• Final Energy: $U_{2f} = \frac{q^2}{2(C_1+C_2)}$. (Same as scenario 1).
• Work Done ($W_2$): $$ W_2 = U_{2i} – U_{2f} = \frac{q^2}{2C_2} – \frac{q^2}{2(C_1+C_2)} $$ $$ W_2 = \frac{q^2}{2} \left( \frac{1}{C_2} – \frac{1}{C_1+C_2} \right) $$

Scenario 3 (Target Problem)

Both bodies are given charge $q$ each. Then they are brought in contact.

• Initial State: A has charge $q$, B has charge $q$. $$ U_{3i} = \frac{q^2}{2C_1} + \frac{q^2}{2C_2} $$
• Final State: Connected. Total charge $Q_{total} = q + q = 2q$. $$ U_{3f} = \frac{(2q)^2}{2(C_1+C_2)} = \frac{4q^2}{2(C_1+C_2)} = \frac{2q^2}{C_1+C_2} $$
• Work Done ($W$): $$ W = U_{3i} – U_{3f} = \left( \frac{q^2}{2C_1} + \frac{q^2}{2C_2} \right) – \frac{2q^2}{C_1+C_2} $$ $$ W = \frac{q^2}{2} \left( \frac{1}{C_1} + \frac{1}{C_2} – \frac{4}{C_1+C_2} \right) $$

Relating W to W1 and W2

Let’s sum $W_1$ and $W_2$:

$$ W_1 + W_2 = \frac{q^2}{2} \left[ \left( \frac{1}{C_1} – \frac{1}{C_1+C_2} \right) + \left( \frac{1}{C_2} – \frac{1}{C_1+C_2} \right) \right] $$ $$ W_1 + W_2 = \frac{q^2}{2} \left[ \frac{1}{C_1} + \frac{1}{C_2} – \frac{2}{C_1+C_2} \right] $$

Now look at the expression for $W$:

$$ W = \frac{q^2}{2} \left[ \frac{1}{C_1} + \frac{1}{C_2} – \frac{4}{C_1+C_2} \right] $$

We can rewrite the bracketed term in $W$ using the sum we just found:

$$ W = \frac{q^2}{2} \left[ \left( \frac{1}{C_1} + \frac{1}{C_2} – \frac{2}{C_1+C_2} \right) – \frac{2}{C_1+C_2} \right] $$ $$ W = (W_1 + W_2) – \frac{q^2}{2} \cdot \frac{2}{C_1+C_2} $$ $$ W = W_1 + W_2 – \frac{q^2}{C_1+C_2} $$

Wait, let’s look closer at the problem statement. Is there a simpler relation? Let’s express the potentials. Let $V$ be the final potential in scenario 1. Since the final state has charge $q$, $V = q/(C_1+C_2)$. Then $\frac{q^2}{C_1+C_2} = qV$. So $W = W_1 + W_2 – qV$.

Another way to group terms in $W$: $$ W = \left( \frac{q^2}{2C_1} – \frac{q^2}{2(C_1+C_2)} \right) + \left( \frac{q^2}{2C_2} – \frac{q^2}{2(C_1+C_2)} \right) – \frac{2q^2}{2(C_1+C_2)} $$ Wait, the arithmetic is: $U_{3f} = \frac{4q^2}{2(C_1+C_2)}$. Sum of final energies of 1 and 2 is $\frac{q^2}{2(C_1+C_2)} + \frac{q^2}{2(C_1+C_2)} = \frac{2q^2}{2(C_1+C_2)}$. The difference is $\frac{2q^2}{2(C_1+C_2)} = \frac{q^2}{C_1+C_2}$. So $W = (U_{1i} + U_{2i}) – U_{3f} = (U_{1i} + U_{2i}) – (U_{1f} + U_{2f} + \frac{q^2}{C_1+C_2})$. $W = (U_{1i} – U_{1f}) + (U_{2i} – U_{2f}) – \frac{q^2}{C_1+C_2}$. $W = W_1 + W_2 – \frac{q^2}{C_1+C_2}$. Since $V = \frac{q}{C_1+C_2}$ is the potential acquired in the first process (as given in the problem statement), we can write: $$ \frac{q^2}{C_1+C_2} = q \left( \frac{q}{C_1+C_2} \right) = qV $$