Question

If the potential difference across a capacitor is increased from $5 \text{ V}$ to $15 \text{ V}$, then the ratio of final energy to initial energy stored in the capacitor is

• A. $1 : 3$
• B. $27 : 1$
• C. $3 : 1$
• D. $9 : 1$

Hint:

Whenever a quantities relation follows a square law like $U \propto V^2$, look at the scaling factor of the independent variable. Here, the voltage triples ($15/5 = 3$). Squaring that scale factor instantly gives the energy modification factor: $3^2 = 9$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem presents a capacitor whose terminal voltage is increased from an initial value to a final value. We need to determine the ratio of the electrostatic potential energy stored in its electric field at the final state compared to the initial state.

Step 2: Key Formula or Approach:
The electrostatic potential energy $U$ stored inside a capacitor of capacitance $C$ charged to a potential difference $V$ is given by:
$$U = \frac{1}{2}CV^2$$ Since the capacitance $C$ of a given physical capacitor remains constant during the charging process, the stored energy is directly proportional to the square of the potential difference:
$$U \propto V^2$$

Step 3: Detailed Explanation:
Let the initial potential difference be $V_1 = 5 \text{ V}$ and the final potential difference be $V_2 = 15 \text{ V}$.
Set up the energy ratio between the final state ($U_2$) and the initial state ($U_1$):
$$\frac{U_2}{U_1} = \left(\frac{V_2}{V_1}\right)^2$$ Substitute the given values into the equation:
$$\frac{U_2}{U_1} = \left(\frac{15}{5}\right)^2$$ Simplify the fraction inside the parentheses:
$$\frac{U_2}{U_1} = (3)^2 = \frac{9}{1}$$ This yields a final ratio of $9 : 1$.

Step 4: Final Answer:
The ratio of final energy to initial energy stored in the capacitor is $9 : 1$, which matches option (D).