Question
Physics Question on Capacitance
A parallel plate capacitor is charged by connecting a 2 Volt battery across it. It is then disconnected from the battery and a glass slab is introduced between plates. Which of the following pair of quantities decrease ?
Energy stored and capacitance
Charge and potential difference
Capacitance and charge
Potential difference and energy stored
Potential difference and energy stored
Solution
Now, let's analyze the given scenario step by step:
Step 1: The capacitor is charged by connecting a 2 Volt battery across it.
Initially, the potential difference (V) across the capacitor is 2 Volts.
The charge (Q) stored on the plates increases as the voltage is applied.
Step 2: The capacitor is disconnected from the battery.
After disconnecting the battery, the potential difference (V) across the capacitor remains the same as it was when charged, i.e., 2 Volts.
The charge (Q) remains the same because there is no current flow.
Step 3: A glass slab is introduced between the plates.
When a dielectric material (such as a glass slab) is introduced between the plates of a capacitor, it increases the capacitance (C) of the capacitor. The capacitance is given by the formula C=k⋅C0, where k is the dielectric constant of the material and Co is the original capacitance without the dielectric.
As the capacitance increases due to the glass slab, the charge (Q) stored on the plates also increases to maintain the same potential difference (V).
Based on the above analysis, we can conclude that the correct pair of quantities that decrease is:
(D) Potential difference and energy stored.