A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

Question

A 5-ohm resistor is connected to a 10 V battery. Calculate the current flowing through the resistor.

Answer 1

To solve this problem, let's understand the situation with the capacitor:

The initial setup involves an air-filled capacitor with capacitance \( C \) connected to a cell with emf \( V \). When connected, the energy stored in the capacitor is given by the formula:
The initial energy stored is \( U_1 = \frac{1}{2} CV^2 \).
Since the capacitor is disconnected from the cell, the charge on the capacitor is conserved. The charge \( Q \) on the capacitor is given by: \( Q = CV \).
Now, when a dielectric slab of dielectric constant \( K \) is inserted, the new capacitance becomes \( C' = KC \) because the dielectric increases the capacitance \( K \) times.
However, because the capacitor is disconnected from the battery, the charge remains the same \( Q = CV \). Hence, the statement "The charge on the capacitor is not conserved" is incorrect.
Let’s calculate the change in energy:
- New potential difference is \( V' = \frac{Q}{C'} = \frac{CV}{KC} = \frac{V}{K} \).
- New energy stored is \( U_2 = \frac{1}{2} C' (V')^2 = \frac{1}{2} KC \left(\frac{V}{K}\right)^2 = \frac{1}{2} \frac{CV^2}{K} \).
The change in energy stored is: \( \Delta U = U_2 - U_1 = \frac{1}{2}CV^2\left(\frac{1}{K} - 1\right) \).
Since \( \Delta U \) is the given change, this statement is correct.
Thus, the potential difference between the plates decreases by a factor of \( K \), confirming that another statement is correct.
Finally, the energy stored decreases by a factor of \( K \), which is also correct as shown by \( U_2 \).

In conclusion, the option "The charge on the capacitor is not conserved" is indeed the incorrect statement.

Answer 2

To solve this problem, let's understand the situation with the capacitor:

The initial setup involves an air-filled capacitor with capacitance \( C \) connected to a cell with emf \( V \). When connected, the energy stored in the capacitor is given by the formula:
The initial energy stored is \( U_1 = \frac{1}{2} CV^2 \).
Since the capacitor is disconnected from the cell, the charge on the capacitor is conserved. The charge \( Q \) on the capacitor is given by: \( Q = CV \).
Now, when a dielectric slab of dielectric constant \( K \) is inserted, the new capacitance becomes \( C' = KC \) because the dielectric increases the capacitance \( K \) times.
However, because the capacitor is disconnected from the battery, the charge remains the same \( Q = CV \). Hence, the statement "The charge on the capacitor is not conserved" is incorrect.
Let’s calculate the change in energy:
- New potential difference is \( V' = \frac{Q}{C'} = \frac{CV}{KC} = \frac{V}{K} \).
- New energy stored is \( U_2 = \frac{1}{2} C' (V')^2 = \frac{1}{2} KC \left(\frac{V}{K}\right)^2 = \frac{1}{2} \frac{CV^2}{K} \).
The change in energy stored is: \( \Delta U = U_2 - U_1 = \frac{1}{2}CV^2\left(\frac{1}{K} - 1\right) \).
Since \( \Delta U \) is the given change, this statement is correct.
Thus, the potential difference between the plates decreases by a factor of \( K \), confirming that another statement is correct.
Finally, the energy stored decreases by a factor of \( K \), which is also correct as shown by \( U_2 \).

In conclusion, the option "The charge on the capacitor is not conserved" is indeed the incorrect statement.

The Correct Option is B