Question

Find the expression of the capacity of the parallel plate capacitor partly filled with dielectric substance. If the capacitor is charged by 100 μC and dielectric constant of the slab putting in it is 2.0, then find the induced charge on the dielectric slab in the capacitor.

Hint:

The induced charge on the dielectric slab in a capacitor is proportional to the total charge and the dielectric constant of the material inserted.

Answer 1

Step 1: Expression for the capacitance of a parallel plate capacitor.
The capacitance of a parallel plate capacitor without a dielectric is given by: \[ C_0 = \frac{\varepsilon_0 A}{d} \] where: - \( \varepsilon_0 \) is the permittivity of free space, - \( A \) is the area of each plate, - \( d \) is the separation between the plates. When a dielectric material with dielectric constant \( \kappa \) is inserted between the plates, the capacitance increases by a factor of \( \kappa \), and the capacitance becomes: \[ C = \kappa C_0 = \frac{\kappa \varepsilon_0 A}{d} \]
Step 2: Induced charge on the dielectric slab.
Let the total charge on the capacitor be \( Q = 100 \, \mu \text{C} \). The electric field \( E \) inside the capacitor is related to the charge \( Q \) and the capacitance \( C \) by: \[ E = \frac{Q}{C} \] For the dielectric material inserted, the induced charge \( Q_{\text{induced}} \) on the dielectric slab is given by: \[ Q_{\text{induced}} = \left( \kappa - 1 \right) \times Q \] Substituting the given values: \[ Q_{\text{induced}} = \left( 2.0 - 1 \right) \times 100 \, \mu\text{C} = 1 \times 100 \, \mu\text{C} = 100 \, \mu\text{C} \]
Step 3: Conclusion.
The induced charge on the dielectric slab in the capacitor is \( 100 \, \mu\text{C} \).