If the charge on a capacitor 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 find 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. **Understand the relationship between charge and energy**:
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{q^2}{2C} \]
where \( q \) is the charge and \( C \) is the capacitance.

2. **Set up the equation for the initial energy**:
Let the initial charge on the capacitor be \( q \). The initial energy stored in the capacitor is:
\[ U_i = \frac{q^2}{2C} \]

3. **Determine the new charge and energy**:
When the charge is increased by 2C, the new charge becomes:
\[ q_f = q + 2 \]
The new energy stored in the capacitor is:
\[ U_f = \frac{(q + 2)^2}{2C} \]

4. **Express the percentage increase in energy**:
According to the problem, the energy increases by 21%, so we can write:
\[ U_f = U_i + 0.21 U_i = 1.21 U_i \]
Substituting the expressions for \( U_f \) and \( U_i \):
\[ \frac{(q + 2)^2}{2C} = 1.21 \left(\frac{q^2}{2C}\right) \]

5. **Eliminate the common factor**:
Since \( \frac{1}{2C} \) is common on both sides, we can cancel it out:
\[ (q + 2)^2 = 1.21 q^2 \]

6. **Expand and rearrange the equation**:
Expanding the left side:
\[ q^2 + 4q + 4 = 1.21 q^2 \]
Rearranging gives:
\[ 0 = 1.21 q^2 - q^2 - 4q - 4 \]
Simplifying:
\[ 0 = 0.21 q^2 - 4q - 4 \]

7. **Multiply through by 100 to eliminate decimals**:
\[ 0 = 21 q^2 - 400q - 400 \]

8. **Use the quadratic formula**:
The quadratic formula is given by:
\[ q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 21 \), \( b = -400 \), and \( c = -400 \):
\[ q = \frac{400 \pm \sqrt{(-400)^2 - 4 \cdot 21 \cdot (-400)}}{2 \cdot 21} \]

9. **Calculate the discriminant**:
\[ = \frac{400 \pm \sqrt{160000 + 33600}}{42} \]
\[ = \frac{400 \pm \sqrt{193600}}{42} \]
\[ = \frac{400 \pm 440}{42} \]

10. **Find the two possible solutions**:
- First solution:
\[ q = \frac{840}{42} = 20 \]
- Second solution:
\[ q = \frac{-40}{42} \text{ (not physically meaningful)} \]

11. **Conclusion**:
The original charge on the capacitor is:
\[ q = 20C \]