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
$$\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}$$
$$\left( {{\phi _2} + {\phi _1}} \right)/{\varepsilon _0}$$
$$\left( {{\phi _1} - {\phi _2}} \right)/{\varepsilon _0}$$
$$\left( {{\phi _1} + {\phi _2}} \right){\varepsilon _0}$$
설명
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 $$q = { \varepsilon _0}\,\left( {{\phi _2} - {\phi _1}} \right)$$