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 problem, we need to understand how the electric field (E) changes when the charge (Q) on an object is doubled.

### Step-by-Step Solution:

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

**Hint**: Remember that the electric field is directly proportional to the charge and inversely proportional to the square of the distance.

2. **Doubling the Charge**:
If the charge on the object is doubled, we can denote the new charge as \(Q' = 2Q\).

3. **Calculate the New Electric Field**:
The new electric field \(E'\) when the charge is doubled can be calculated using the same formula:
\[ E' = \frac{1}{4 \pi \epsilon_0} \frac{Q'}{R^2} = \frac{1}{4 \pi \epsilon_0} \frac{2Q}{R^2} \]

4. **Relate the New Electric Field to the Original**:
We can express \(E'\) in terms of the original electric field \(E\):
\[ E' = 2 \left(\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2}\right) = 2E \]

5. **Conclusion**:
Therefore, when the charge is doubled, the electric field also doubles. The answer is:
\[ \text{The electric field becomes doubled.} \]

### Final Answer:
The electric field becomes doubled (Option B).