JEE MAIN - Physics (2003 - No. 48)
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
$$\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}$$
$$\left( {{\phi _2} + {\phi _1}} \right)/{\varepsilon _0}$$
$$\left( {{\phi _1} - {\phi _2}} \right)/{\varepsilon _0}$$
$$\left( {{\phi _1} + {\phi _2}} \right){\varepsilon _0}$$
Paaiškinimas
The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention. Therefore the net flux leaving the enclosed surface $$ = {\phi _2} - {\phi _1}$$
$$\therefore$$ the change enclosed in the surface by Gauss's law is $$q = { \varepsilon _0}\,\left( {{\phi _2} - {\phi _1}} \right)$$
$$\therefore$$ the change enclosed in the surface by Gauss's law is $$q = { \varepsilon _0}\,\left( {{\phi _2} - {\phi _1}} \right)$$