Question 468: Electrostats & Capacitance

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

Accepted Answer

Step 1. Recall capacitor energy formula
The energy in a capacitor in terms of charge $Q$ and capacitance $C$ is:
$ U = \frac{Q^2}{2C}. $
Step 2. Let original charge = $Q$
Original energy:
$ U = \frac{Q^2}{2C}. $
New charge = $Q+3$.
New energy:
$ U' = \frac{(Q+3)^2}{2C}. $
Step 3. Relating energies
We are told:
$ U' = U (1 + 0.21) = 1.21 U. $
So:
$ \frac{(Q+3)^2}{2C} = 1.21 \cdot \frac{Q^2}{2C}. $
Cancel $	frac{1}{2C}$:
$ (Q+3)^2 = 1.21 Q^2. $
Step 4. Expand
$ Q^2 + 6Q + 9 = 1.21 Q^2. $
$ 6Q + 9 = 1.21Q^2 - Q^2 = 0.21 Q^2. $
$ 0.21 Q^2 - 6Q - 9 = 0. $
Step 5. Solve quadratic
Multiply through by 100 to eliminate decimals:
$ 21 Q^2 - 600Q - 900 = 0. $
Divide by 3:
$ 7Q^2 - 200Q - 300 = 0. $
Quadratic:
$ Q = \frac{200 \pm \sqrt{200^2 + 4\cdot7\cdot300}}{14}. $
$ = \frac{200 \pm \sqrt{40000 + 8400}}{14} = \frac{200 \pm \sqrt{48400}}{14}. $
$\sqrt{48400} = 220$.
So:
$ Q = \frac{200 \pm 220}{14}. $
Case 1: $Q = \frac{200+220}{14} = \frac{420}{14} = 30.$
Case 2: $Q = \frac{200-220}{14} = \frac{-20}{14} = -	frac{10}{7}$ (not possible).
Final Answer:
$ \boxed{30 \, 	ext{C}} $
So the correct option is
C (30Ã¢ÂÂ¯C)
.

Answer

6 C

Answer

3 C

Answer

30 C

Answer

90 C

Electrostats & Capacitance

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

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