Parallel Plate Capacitor: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:

To plot a graph for the current

$i$

versus time

$t$

, let's break the problem into steps:

1. Capacitor with Dielectric Slab

• Initial Configuration: The capacitor has plates of total area
  $2A$ 

  and plate separation $d$ 

  . It is connected to a battery with emf $E$ 

  .

• Dielectric Slab: A dielectric slab of area
  $A$ 

  is inserted at a constant speed $v$ 

  .

2. Capacitance with Partial Dielectric

When a dielectric is partially inserted into the capacitor, the total capacitance is the sum of two capacitors:

• One part with the dielectric slab (
  $C_1$ 

  ).

• One part without the dielectric slab (
  $C_2$ 

  ).

Capacitance of the two regions:

1. With Dielectric Slab: 

   $$C_1 = \frac{\kappa \varepsilon_0 A}{d}$$where

   $\kappa$is the dielectric constant.

2. Without Dielectric Slab: 

   $$C_2 = \frac{\varepsilon_0 (2A - A)}{d} = \frac{\varepsilon_0 A}{d}$$ 

Thus, the total capacitance

$C_{\text{total}}$

is:

$$C_{\text{total}} = C_1 + C_2 = \frac{\kappa \varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{d}$$

$$C_{\text{total}} = \frac{\varepsilon_0 A}{d} (\kappa + 1)$$

3. Change in Capacitance with Time

As the slab moves with speed

$v$

, the area covered by the slab changes with time:

$$\text{Covered area } A_{\text{covered}} = v \cdot t$$

The effective capacitance changes as:

$$C(t) = \frac{\kappa \varepsilon_0 (v \cdot t)}{d} + \frac{\varepsilon_0 (2A - v \cdot t)}{d}$$

$$C(t) = \frac{\varepsilon_0}{d} \left[ \kappa (v \cdot t) + (2A - v \cdot t) \right]$$

Simplify:

$$C(t) = \frac{\varepsilon_0}{d} \left[ 2A + (\kappa - 1)(v \cdot t) \right]$$

4. Current in the Circuit

The current in the circuit is related to the rate of change of capacitance:

$$i(t) = E \cdot \frac{dC}{dt}$$

Differentiate

$C(t)$

with respect to

$t$

:

$$\frac{dC}{dt} = \frac{\varepsilon_0}{d} (\kappa - 1) v$$

Thus:

$$i(t) = E \cdot \frac{\varepsilon_0}{d} (\kappa - 1) v$$

5. Graph of $i$

vs. $t$

The current

$i(t)$

is constant because it does not depend on

$t$

(the rate of capacitance change is constant). Hence, the graph of

$i$

vs.

$t$

will be a horizontal line at:

$$i = E \cdot \frac{\varepsilon_0}{d} (\kappa - 1) v$$