Question

Explain the following statement giving reason :
The potential difference between the plates of a charged parallel plate capacitor decreases when its plates are brought closer.

Hint:

If the question had mentioned that the capacitor remained actively connected to a battery while the plates were moved, the potential difference $V$ would remain strictly constant, and the battery would have forcibly pumped more charge $Q$ onto the plates instead! Always check if it's isolated or connected.

Answer 1

Concept:
• The capacitance $C$ of a parallel plate capacitor is fundamentally dependent on its precise geometric configuration.
• For an isolated charged capacitor, the total charge $Q$ stored on the plates remains strictly conserved because there is no external path for the electrons to flow.
• The macroscopic relation between voltage, charge, and capacitance is universally given by $V = \frac{Q}{C}$.

Step 1: Establish the geometric capacitance formula
The theoretical formula for the capacitance of an empty parallel plate capacitor is:
\[ C = \frac{\epsilon_0 A}{d} \]
Where:
$A$ is the overlapping surface area of the plates.
$d$ is the physical separation distance between the two metallic plates.
$\epsilon_0$ is the permittivity of free space.

Step 2: Analyze the physical change
The problem specifically states that the plates are physically brought closer to each other.
This action means that the separation distance $d$ actively decreases.
Looking directly at our established formula, the capacitance $C$ is inversely proportional to the separation distance $d$ ($C \propto \frac{1}{d}$).
Therefore, a mathematical decrease in $d$ strictly causes a proportional increase in the overall capacitance $C$.

Step 3: Apply the voltage-charge relation
Assume the capacitor is completely disconnected from any charging battery (it is isolated).
Under this highly specific condition, the stored charge $Q$ is completely trapped and must remain strictly constant.
Now, substitute the changing variables into the voltage definition:
\[ V = \frac{Q}{C} \]

Step 4: Conclusion
Since the numerator (charge $Q$) is perfectly constant, and the denominator (capacitance $C$) has significantly increased due to the plates moving closer, the resulting fraction must mathematically decrease.
Consequently, the electric potential difference $V$ strictly decreases.