Question

If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

The Correct Option is A

To solve this problem, let's analyze the relationship between the charge on a capacitor and the energy stored in it. The energy (\(E\)) stored in a capacitor is given by the formula:

E = \frac{Q^2}{2C}

where:

• E is the energy stored.
• Q is the charge on the capacitor.
• C is the capacitance of the capacitor (constant in this case).

We are given that the energy increases by 44% when the charge is increased by 2 C. Let's define the original charge as Q_0. The original energy stored is:

E_0 = \frac{{Q_0}^2}{2C}

When the charge is increased by 2 C, the new charge becomes (Q_0 + 2). The new energy E_1 is then:

E_1 = \frac{(Q_0 + 2)^2}{2C}

According to the problem, E_1 is 44% more than E_0, so:

E_1 = E_0 + 0.44E_0 = 1.44E_0

Substituting the values:

\frac{(Q_0 + 2)^2}{2C} = 1.44 \times \frac{{Q_0}^2}{2C}

Remove \(\frac{1}{2C}\) from both sides (since capacitance is constant and not zero), we get:

(Q_0 + 2)^2 = 1.44 \cdot {Q_0}^2

Expanding both sides:

Q_0^2 + 4Q_0 + 4 = 1.44Q_0^2

Rearrange the equation:

4Q_0 + 4 = 0.44Q_0^2

The above can be written as:

0.44Q_0^2 - 4Q_0 - 4 = 0

Multiply the whole equation by 100 to simplify:

44Q_0^2 - 400Q_0 - 400 = 0

Divide the entire equation by 4:

11Q_0^2 - 100Q_0 - 100 = 0

Use the quadratic formula where a = 11, b = -100, c = -100:

Q_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute here:

Q_0 = \frac{100 \pm \sqrt{(-100)^2 - 4 \times 11 \times (-100)}}{2 \times 11}

Q_0 = \frac{100 \pm \sqrt{10000 + 4400}}{22}

Q_0 = \frac{100 \pm \sqrt{14400}}{22}

Q_0 = \frac{100 \pm 120}{22}

Calculate the positive value:

Q_0 = \frac{220}{22} = 10

Therefore, the original charge on the capacitor is 10 C, which matches the correct answer option.