Question 571: Electrostatics & Capacitance

Question

If the charge on the capacitor is increased by 2 C the energy stored in it increased by 21%. Total original charge on the capacitor is

Accepted Answer

Let's start by recalling the formula for the energy stored in a capacitor. The energy $U$ stored in a capacitor with capacitance $C$ and charge $Q$ is given by:
$$U = \frac{Q^2}{2C}$$
Since we do not know the capacitance $C$, we will manipulate the formula using given information. If the charge on the capacitor is increased by 2 C, the new charge $Q'$ becomes $Q + 2$. Thus, the new energy $U'$ becomes:
$$U' = \frac{(Q + 2)^2}{2C}$$
We are given that this new energy is 21% greater than the original energy. Therefore, we can write:
$$U' = 1.21U$$
Substituting the expressions for $U$ and $U'$, we get:
$$\frac{(Q + 2)^2}{2C} = 1.21 \cdot \frac{Q^2}{2C}$$
Since $C$ is a common factor in the numerator and denominator, it cancels out. Hence, we can simplify to:
$$ (Q + 2)^2 = 1.21Q^2 $$
Expanding and rearranging the equation, we obtain:
$$ Q^2 + 4Q + 4 = 1.21Q^2 $$
$$ 0.21Q^2 - 4Q - 4 = 0 $$
Solving this quadratic equation using the quadratic formula, where $a = 0.21$, $b = -4$, and $c = -4$, we get:
$$ Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
Substitute the values of $a$, $b$, and $c$:
$$ Q = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 0.21 \cdot (-4)}}{2 \cdot 0.21} $$
$$ Q = \frac{4 \pm \sqrt{16 + 3.36}}{0.42} $$
$$ Q = \frac{4 \pm \sqrt{19.36}}{0.42} $$
$$ Q = \frac{4 \pm 4.4}{0.42} $$
Therefore, we have two possible solutions for $Q$:
$$ Q = \frac{4 + 4.4}{0.42} = 20 $$
$$ Q = \frac{4 - 4.4}{0.42} = -0.95 \ (not \ practical) $$
Since a negative charge is not practical in this context, the original charge on the capacitor is:
Option C: 20 C

Answer

10 C

Answer

5 C

Answer

20 C

Answer

15 C

Electrostatics & Capacitance

If the charge on the capacitor is increased by 2 C the energy stored in it increased by 21%. Total original charge on the capacitor is

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