An uncharged parallel plate capacitor is connected to a battery. The electric field between the plates is 10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entries space. The electric field between the plates now is

Options

A. 5 V/m
B. 20 V/m
C. 10 V/m
D. none of these

Answer

To solve the problem, we need to understand how inserting a dielectric affects the electric field in a parallel plate capacitor. Here's the step-by-step solution:

### Step 1: Understand the initial conditions
The initial electric field (E₀) between the plates of the capacitor is given as 10 V/m. The capacitor is connected to a battery, which means the voltage (V) across the plates remains constant.

**Hint:** Remember that the electric field (E) in a capacitor is related to the voltage (V) and the distance (d) between the plates by the formula \( E = \frac{V}{d} \).

### Step 2: Introduce the dielectric
When a dielectric material is inserted between the plates of the capacitor, it affects the electric field. The dielectric constant (κ) of the material is given as 2.

**Hint:** The dielectric constant reduces the electric field in the capacitor. The new electric field (E) can be calculated using the formula \( E = \frac{E_0}{\kappa} \), where \( E_0 \) is the initial electric field.

### Step 3: Calculate the new electric field
Using the formula mentioned above:
\[ E = \frac{E_0}{\kappa} = \frac{10 \, \text{V/m}}{2} = 5 \, \text{V/m} \]

**Hint:** Make sure to divide the initial electric field by the dielectric constant to find the new electric field.

### Step 4: Conclusion
The electric field between the plates after inserting the dielectric is 5 V/m.

### Final Answer
The electric field between the plates now is **5 V/m**.

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