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Step-by-Step Explanation
Step 1: Identify the Relevant Physical Quantity
A parallel plate capacitor has a certain charge $Q$ when connected to a battery. Once the battery is disconnected, no additional charge flows into or out of the capacitor. Thus, the total charge $Q$ on the plates remains constant.
Step 2: Recall the Relationship Between Capacitance, Charge, and Potential
The fundamental relation governing a capacitor is given by:
$Q = C V$
where $Q$ is the charge on the capacitor, $C$ is the capacitance, and $V$ is the potential difference across the capacitor plates.
Step 3: Analyze Effect of Changing Plate Separation on Capacitance
For a parallel plate capacitor (with vacuum or air between the plates), the capacitance $C$ is inversely proportional to the distance $d$ between the plates:
$C = \frac{\epsilon_0 A}{d}$
where $A$ is the area of each plate and $\epsilon_0$ is the permittivity of free space (for air, it is approximately the same as vacuum). As the distance $d$ increases, the value of $C$ decreases.
Step 4: Relate Decrease in Capacitance to Potential Difference
Since $Q$ remains the same (battery disconnected), but $C$ decreases when the plate separation increases, we look at $V$ from $Q = CV$:
$V = \frac{Q}{C}$
With $Q$ constant and $C$ decreasing, $V$ must increase in order to maintain $Q = C V.$
Step 5: Conclusion
Therefore, when the distance between the plates is increased after disconnecting the battery, the potential difference increases.
