Question

If the charge on the capacitor is increased by 3C, the energy stored in it increases by 21%. The original charge on the capacitor is

• A. 6 C
• B. 3 C
• C. 30 C
• D. 90 C

Hint:

The energy stored in a capacitor is proportional to the square of the charge. Therefore, a small increase in charge leads to a much larger increase in energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a relationship between the increase in charge and the resulting percentage increase in stored energy. We need to find the initial charge \( Q \).
Step 2: Key Formula or Approach:
Energy stored in a capacitor: \( U = \frac{Q^2}{2C} \).
For a fixed capacitance \( C \), \( U \propto Q^2 \).
Step 3: Detailed Explanation:
Initial energy \( U_1 = \frac{Q^2}{2C} \).
Final charge \( Q_2 = Q + 3 \).
Final energy \( U_2 = U_1 + 21% \text{ of } U_1 = 1.21 U_1 \).
Using proportionality:
\[ \frac{U_2}{U_1} = \left( \frac{Q_2}{Q_1} \right)^2 \]
\[ 1.21 = \left( \frac{Q + 3}{Q} \right)^2 \]
Take the square root of both sides:
\[ \sqrt{1.21} = \frac{Q + 3}{Q} \]
\[ 1.1 = 1 + \frac{3}{Q} \]
\[ 0.1 = \frac{3}{Q} \implies Q = \frac{3}{0.1} = 30\text{ C} \]
Step 4: Final Answer:
The original charge on the capacitor is 30 C.