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: Energy stored in a capacitor.
The energy stored in a capacitor is given by the formula:
\[ E = \frac{Q^2}{2C}, \]
where \( Q \) is the charge on the capacitor, and \( C \) is the capacitance.

Step 2: New energy when charge is increased.
When the charge is increased by 3C, the new charge on the capacitor becomes \( Q + 3 \). The new energy is:
\[ E' = \frac{(Q + 3)^2}{2C}. \]

Step 3: Energy increase by 21 %.
We are told that the energy increases by 21 %. Therefore, we have the equation:
\[ E' = E + 0.21 E. \]
Substituting the expressions for \( E' \) and \( E \), we get:
\[ \frac{(Q + 3)^2}{2C} = \frac{Q^2}{2C} + 0.21 \cdot \frac{Q^2}{2C}. \]
Simplifying: \[ (Q + 3)^2 = Q^2 + 0.21 Q^2 = 1.21 Q^2. \]

Step 4: Solving for \( Q \).
Expanding \( (Q + 3)^2 \):
\[ Q^2 + 6Q + 9 = 1.21 Q^2. \]
Rearranging: \[ 0.21 Q^2 - 6Q - 9 = 0. \]
Solving this quadratic equation for \( Q \), we find:
\[ Q = 30 \, \text{C}. \]
Final Answer:
Thus, the original charge on the capacitor is:
\[ \boxed{30 \, \text{C}}. \]