Question

A system of two conductors is placed in air and they have net charges of +80µC and −80µC which causes a potential difference of 16V between them.
1. Find the capacitance of the system.
2. If the air between the capacitor is replaced by a dielectric medium of di electric constant K, what will be the potential difference between the two conductors?
3. If the charges on two conductors are changed to +160µC and −160µC, will the capacitance of the system change? Give reason for your answer.

Hint:

The capacitance of a system is determined by its geometry and the dielectric constant of the medium between the conductors, and it is independent of the amount of charge.

Answer 1

1. Finding the Capacitance: 

The capacitance \( C \) of a system is given by the formula:

\[ C = \frac{Q}{V} \]

Where:

• \( C \) is the capacitance,
• \( Q \) is the charge on each conductor (the charges are equal in magnitude but opposite in sign),
• \( V \) is the potential difference between the conductors.

From the given data:

• \(Q = 80 \, \mu C = 80 \times 10^{-6} \, \text{C}\),
• \( V = 16 \, \text{V} \).

Substituting the values into the formula:

\[ C = \frac{80 \times 10^{-6}}{16} = 5 \times 10^{-6} \, \text{F} = 5 \, \mu F \]

2. Capacitance with Dielectric Medium:

When a dielectric material of dielectric constant \( K \) is placed between the conductors, the capacitance increases by a factor of \( K \). The new capacitance \( C' \) is given by:

\[ C' = K \cdot C \]

Since the initial capacitance is \( C = 5 \, \mu F \), the new capacitance with the dielectric becomes:

\[ C' = K \cdot 5 \, \mu F \]

Now, the potential difference between the conductors will decrease because the capacitance has increased, while the charge remains the same. The new potential difference \( V' \) is given by:

\[ V' = \frac{Q}{C'} \]

Since \( C' = K \cdot C \), we have:

\[ V' = \frac{Q}{K \cdot C} = \frac{V}{K} \]

Therefore, the potential difference decreases by a factor of \( K \). If \( K \) is the dielectric constant of the material, the new potential difference is:

\[ V' = \frac{16}{K} \]

3. Effect of Changing the Charges on the Conductors:

When the charges on the conductors are changed to \( +160 \, \mu C \) and \( -160 \, \mu C \), the charge on each conductor becomes twice the original charge. However, the capacitance of a system depends only on the geometry of the system and the dielectric constant of the medium. The capacitance is independent of the charge on the conductors.

Therefore, the capacitance of the system remains the same at \( 5 \, \mu F \), and the potential difference will increase due to the increase in charge. The new potential difference \( V' \) can be found by:

\[ V' = \frac{Q'}{C} = \frac{160 \times 10^{-6}}{5 \times 10^{-6}} = 32 \, \text{V} \]

Conclusion:

• The capacitance of the system is \( 5 \, \mu F \).
• If the air between the capacitor is replaced with a dielectric medium of dielectric constant \( K \), the potential difference will decrease to \( \frac{16}{K} \) volts.
• The capacitance will not change if the charges on the conductors are changed, but the potential difference will double when the charge is doubled, making the new potential difference \( 32 \, \text{V} \).