Question

If the charge on a capacitor is increased by $2\ \text{C}$, the energy stored in it increases by $21\%$. The total original charge on the capacitor is

• A. $10\ \text{C}$
• B. $5\ \text{C}$
• C. $20\ \text{C}$
• D. $15\ \text{C}$

Hint:

Whenever a problem states that a squared quantity increases by $21\%$, remember that its base variable must increase by exactly $10\%$ (since $1.1^2 = 1.21$). Therefore, the $2\ \text{C}$ addition represents exactly $10\%$ of the original charge: $0.10 \times Q = 2 \implies Q = 20\ \text{C}$. This relationship can be worked out mentally in under five seconds!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem relates a change in the electrical charge of a capacitor to its total stored potential energy. Increasing the charge by $2\ \text{C}$ results in a $21\%$ increase in stored energy. We need to find the initial charge $Q$ on the capacitor.

Step 2: Key Formula or Approach:
The electrostatic energy ($U$) stored inside a capacitor of capacitance $C$ carrying a charge $Q$ is given by: $$U = \frac{Q^2}{2C}$$ Since the physical capacitance $C$ is a fixed geometric constant, the stored energy is directly proportional to the square of the charge: $$U \propto Q^2 \implies \frac{U_2}{U_1} = \left(\frac{Q_2}{Q_1}\right)^2$$

Step 3: Detailed Explanation:
Let's define the initial and final states from the problem data: Initial charge, $Q_1 = Q$ Final charge, $Q_2 = Q + 2$ Initial energy, $U_1 = U$ Final energy after a $21\%$ increase, $U_2 = U + 0.21U = 1.21U$ Substitute these terms into the proportional ratio equation: $$\frac{1.21U}{U} = \left(\frac{Q + 2}{Q}\right)^2$$ $$1.21 = \left(\frac{Q + 2}{Q}\right)^2$$ Take the square root on both sides of the equation ($\sqrt{1.21} = 1.1$): $$1.1 = \frac{Q + 2}{Q}$$ Cross-multiplying by $Q$: $$1.1Q = Q + 2$$ Group the $Q$ variables on the left side: $$1.1Q - Q = 2$$ $$0.1Q = 2$$ $$Q = \frac{2}{0.1} = 20\ \text{C}$$

Step 4: Final Answer:
The original charge on the capacitor is $20\ \text{C}$, matching option (C).