If the charge on an object is doubled then electric field becomes

Options

A. half
B. double
C. unchanged
D. thrice

Explanation

To solve the question of how the electric field changes when the charge on an object is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Electric Field Formula**:
The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{r^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Identify the Initial Charge**:
Let's assume the initial charge is \( q \). Thus, the electric field \( E_1 \) at a distance \( r \) from this charge is:
\[ E_1 = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{r^2} \]

3. **Double the Charge**:
If the charge is doubled, the new charge becomes \( 2q \).

4. **Calculate the New Electric Field**:
Now, we can find the new electric field \( E_2 \) due to the doubled charge \( 2q \):
\[ E_2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{2q}{r^2} \]

5. **Relate the New Electric Field to the Old One**:
We can express \( E_2 \) in terms of \( E_1 \):
\[ E_2 = 2 \cdot \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{r^2} = 2E_1 \]

6. **Conclusion**:
Therefore, when the charge is doubled, the electric field also doubles. Thus, if the original electric field was \( E \), the new electric field becomes \( 2E \).

### Final Answer:
When the charge on an object is doubled, the electric field becomes **doubled**.

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