Question

The plates of a parallel plate capacitor of capacity $C_1$ are moved closer together until they are at half their original separation. The new capacitance $C_2$ is

• A. $C_2 = \frac{C_1}{2}$
• B. $C_2 = C_1$
• C. $C_2 = 2C_1$
• D. $C_2 = 3C_1$

Hint:

Capacitance measures a system's ability to store electric charge via fields. Bringing opposing plates closer together increases the electrostatic attractive force across the gap, boosting field containment. Halving the gap ($1/2$) always doubles the storage performance ($2\times$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem details a geometric change in a standard parallel-plate capacitor configuration.
Its internal separation parameter $d$ is mechanically compressed down to exactly half its initial value. We must deduce the updated capacitance output $C_2$ relative to $C_1$.

Step 2: Key Formula or Approach:
The fundamental capacitance of two matching parallel plate surfaces separated by a distance $d$ in a medium is defined by:
$$C = \frac{\varepsilon_0 A}{d}$$ This demonstrates that capacitance is strictly inversely proportional to plate spacing: $C \propto \frac{1}{d}$.

Step 3: Detailed Explanation:
Let the initial arrangement be defined as:
$$C_1 = \frac{\varepsilon_0 A}{d_1}$$ The problem establishes that the spacing is halved: $d_2 = \frac{d_1}{2}$.
Substitute this updated spacing dimension into the geometric structural equation for $C_2$:
$$C_2 = \frac{\varepsilon_0 A}{d_2} = \frac{\varepsilon_0 A}{\left(\frac{d_1}{2}\right)}$$ Rearranging the fraction moves the denominator factor up to the numerator position:
$$C_2 = 2 \times \left(\frac{\varepsilon_0 A}{d_1}\right)$$ Substituting $C_1$ back into the bracketed statement yields:
$$C_2 = 2C_1$$

Step 4: Final Answer:
The updated capacitance configuration scales to $C_2 = 2C_1$, which corresponds to option (C).