Electrostatics

JEE Main level, single correct, medium

310 of 311

Question 310 of 311

If the electric flux entering and leaving an enclosed surface respectively is ϕ1\phi_1 and ϕ2\phi_2, the electric charge inside the surface will be

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Answer: (A) (ϕ2−ϕ1)ε0\left(\phi_2-\phi_1\right) \varepsilon_0

According to Gauss's Law
∫( E.dA )=q0/ε0⇒q=ε0(ϕ2−ϕ1)[ since ϕ=∫E⋅dA]\begin{aligned} & \int(\text { E.dA })=\mathrm{q}_0 / \varepsilon_0 \Rightarrow \mathrm{q}=\varepsilon_0\left(\phi_2-\phi_1\right) \\ & {\left[\text { since } \phi=\int \mathrm{E} \cdot \mathrm{dA}\right]} \end{aligned}

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