Question
If the electric flux entering and leaving an enclosed surface respectively is $${\phi _1}$$ and $${\phi _2},$$ the electric charge inside the surface will be
A.
$$\left( {{\phi _2} + {\phi _2}} \right) \times {\varepsilon _0}$$
B.
$$\left( {{\phi _2} - {\phi _2}} \right) \times {\varepsilon _0}$$
C.
$$\left( {{\phi _1} + {\phi _2}} \right) \times {\varepsilon _0}$$
D.
$$\left( {{\phi _2} - {\phi _1}} \right) \times {\varepsilon _0}$$
Answer :
$$\left( {{\phi _2} - {\phi _1}} \right) \times {\varepsilon _0}$$
Solution :
$$\Delta \phi = \frac{{{\theta _{{\text{in}}}}}}{{{\varepsilon _0}}} \Rightarrow Q\,{\text{in}}\,\left( {\phi = {\phi _2} - \phi } \right){\varepsilon _0}$$