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Charges Q1 and Q2 lie inside and outside...

Charges `Q_1 and Q_2` lie inside and outside, respectively, of a closed surface S. Let E be the field at any point on S and `phi` be the flux of E over S.

A

If `Q_1` changes both E and `phi` will change

B

If `Q_2` changes, will changes but `phi` will not change

C

If `Q_1=0` and `Q_2!=0`, then `E!=0` but `phi=0`

D

If `Q_1!=0` and `Q_2=0,` the `E=0` but `phi !=0`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the relationship between the electric field \( E \) and the electric flux \( \Phi \) through a closed surface \( S \) when charges \( Q_1 \) and \( Q_2 \) are present inside and outside the surface, respectively. ### Step-by-Step Solution: 1. **Understanding Electric Flux**: The electric flux \( \Phi \) through a closed surface is given by Gauss's law: \[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \] where \( Q_{\text{enc}} \) is the total charge enclosed by the surface \( S \) and \( \epsilon_0 \) is the permittivity of free space. 2. **Identifying Charges**: In this scenario, \( Q_1 \) is located inside the closed surface \( S \) and \( Q_2 \) is located outside the surface. 3. **Calculating Flux**: Since \( Q_2 \) is outside the surface, it does not contribute to the electric flux through surface \( S \). Therefore, the flux \( \Phi \) through surface \( S \) depends only on \( Q_1 \): \[ \Phi = \frac{Q_1}{\epsilon_0} \] 4. **Effect of Changing \( Q_1 \)**: If \( Q_1 \) changes, the enclosed charge changes, which directly affects the electric flux \( \Phi \). Thus, both the electric field \( E \) at any point on the surface and the flux \( \Phi \) will change when \( Q_1 \) changes. 5. **Effect of Changing \( Q_2 \)**: If \( Q_2 \) changes, since it is outside the closed surface, it does not affect the enclosed charge \( Q_{\text{enc}} \). Therefore, the electric flux \( \Phi \) remains unchanged. However, the electric field \( E \) at the surface may change due to the influence of \( Q_2 \). 6. **Special Cases**: - If \( Q_1 = 0 \) (no charge inside), then \( \Phi = 0 \) regardless of \( Q_2 \). - If \( Q_2 = 0 \) (no charge outside), the electric field \( E \) at the surface will still be due to \( Q_1 \) and will not be zero. ### Conclusion: - The electric flux \( \Phi \) through the surface \( S \) depends only on \( Q_1 \). - The electric field \( E \) at the surface depends on both \( Q_1 \) and \( Q_2 \).
To solve the problem, we need to analyze the relationship between the electric field \( E \) and the electric flux \( \Phi \) through a closed surface \( S \) when charges \( Q_1 \) and \( Q_2 \) are present inside and outside the surface, respectively. ### Step-by-Step Solution: 1. **Understanding Electric Flux**: The electric flux \( \Phi \) through a closed surface is given by Gauss's law: \[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} ...
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