Air filled capacitor is charged by a battery and after charging battery is removed. A slab of dielectric material is inserted in it to fill the space completely . The electric field in the capacitor is

Options

A. Increased
B. Decreased
C. Remains constant
D. First increased then decreased

Answer

To solve the problem of finding the electric field in a capacitor after inserting a dielectric slab, we can follow these steps:

### Step 1: Understand the Initial Conditions
Initially, we have an air-filled parallel plate capacitor that is charged by a battery. The capacitance of the capacitor can be expressed as:
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
where:
- \(C_0\) is the initial capacitance,
- \(\varepsilon_0\) is the permittivity of free space,
- \(A\) is the area of the plates,
- \(d\) is the separation between the plates.

### Step 2: Charge on the Capacitor
When the capacitor is charged by the battery, the charge \(Q_0\) on the capacitor can be expressed as:
\[ Q_0 = C_0 V_0 \]
where \(V_0\) is the voltage of the battery.

### Step 3: Remove the Battery
After charging, the battery is removed. The charge \(Q_0\) on the capacitor plates remains constant because there is no external circuit connected to allow charge to flow.

### Step 4: Insert the Dielectric Slab
Now, a dielectric slab with dielectric constant \(k\) is inserted to fill the space completely between the plates. The presence of the dielectric affects the capacitance and the electric field.

### Step 5: New Capacitance with Dielectric
The new capacitance \(C\) of the capacitor with the dielectric slab is given by:
\[ C = k C_0 = k \frac{\varepsilon_0 A}{d} \]

### Step 6: Electric Field in the Capacitor
The electric field \(E\) in a capacitor is related to the charge and the area of the plates:
\[ E = \frac{\sigma}{\varepsilon} \]
where \(\sigma\) is the surface charge density given by:
\[ \sigma = \frac{Q_0}{A} \]
Since the charge \(Q_0\) remains constant, we can express the electric field in terms of the new permittivity:
\[ E' = \frac{\sigma}{\varepsilon} = \frac{Q_0/A}{\varepsilon_0 k} = \frac{Q_0}{\varepsilon_0 k A} \]

### Step 7: Relate to the Initial Electric Field
The initial electric field \(E_0\) without the dielectric is:
\[ E_0 = \frac{Q_0}{\varepsilon_0 A} \]
Thus, the new electric field \(E'\) can be expressed as:
\[ E' = \frac{E_0}{k} \]

### Conclusion
The electric field in the capacitor after inserting the dielectric slab is:
\[ E' = \frac{E_0}{k} \]
This indicates that the electric field decreases when the dielectric is introduced.