If the charge on a capacitorn is increased by 2C, then the energy stored in it increases by 21 %. The original charge on the capacitor is

Options

A. 10 C
B. 20 C
C. 30 C
D. 40 C

Answer

To solve the problem, we need to determine the original charge on the capacitor given that an increase of 2C in charge results in a 21% increase in the energy stored in the capacitor.

### Step-by-Step Solution:

1. **Understanding the Energy Stored in a Capacitor**:
The energy (U) stored in a capacitor can be expressed as:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
where \( Q \) is the charge on the capacitor and \( C \) is the capacitance.

2. **Initial Energy**:
Let the initial charge on the capacitor be \( Q \). The initial energy stored in the capacitor is:
\[ U_i = \frac{1}{2} \frac{Q^2}{C} \]

3. **Final Charge and Energy**:
When the charge is increased by 2C, the new charge becomes \( Q + 2 \). The energy stored in the capacitor now is:
\[ U_f = \frac{1}{2} \frac{(Q + 2)^2}{C} \]

4. **Energy Increase**:
According to the problem, the energy increases by 21%. Therefore:
\[ U_f = U_i + 0.21 U_i = 1.21 U_i \]

5. **Setting Up the Equation**:
Now, we can set up the equation:
\[ \frac{1}{2} \frac{(Q + 2)^2}{C} = 1.21 \left(\frac{1}{2} \frac{Q^2}{C}\right) \]
We can cancel \(\frac{1}{2C}\) from both sides:
\[ (Q + 2)^2 = 1.21 Q^2 \]

6. **Expanding and Rearranging**:
Expanding the left side:
\[ Q^2 + 4Q + 4 = 1.21 Q^2 \]
Rearranging gives:
\[ 4Q + 4 = 1.21 Q^2 - Q^2 \]
\[ 4Q + 4 = 0.21 Q^2 \]
Rearranging further:
\[ 0.21 Q^2 - 4Q - 4 = 0 \]

7. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula:
\[ Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 0.21 \), \( b = -4 \), and \( c = -4 \):
\[ Q = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 0.21 \cdot (-4)}}{2 \cdot 0.21} \]
\[ = \frac{4 \pm \sqrt{16 + 3.36}}{0.42} \]
\[ = \frac{4 \pm \sqrt{19.36}}{0.42} \]
\[ = \frac{4 \pm 4.4}{0.42} \]

8. **Calculating the Values**:
This gives us two potential solutions:
\[ Q = \frac{8.4}{0.42} \quad \text{or} \quad Q = \frac{-0.4}{0.42} \]
The negative value is not physically meaningful, so we take:
\[ Q = 20 \text{ C} \]

### Final Answer:
The original charge on the capacitor is \( Q = 20 \text{ C} \).

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