question_answer

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. Now study the following statements.

(i) If \[{{Q}_{1}}\] changes, both E and \[\phi \] will change,

(ii) If \[{{Q}_{2}}\] changes, E will change but \[\phi \] will not change.

(iii) If \[{{Q}_{1}}=0\] and \[{{Q}_{2}}\ne 0,\] then \[E\ne 0\] but \[\phi =0\]

(iv) If \[{{Q}_{1}}\ne 0\] and \[{{Q}_{2}}=0,\] then \[E=0\]but \[\phi =0\].

The correct statements are

 

A) (i), (iii) and (iv)      

B) (i), (ii) and (iv)    

C) (i), (ii) and (iii) 

D) (i), (iii) and (iv)

Correct Answer: C

Solution :

   [c]  Total electric field at surface S is \[{{\vec{E}}_{1}}+{{\vec{E}}_{2}}\] where \[{{\vec{E}}_{1}}\] and \[{{\vec{E}}_{2}}\] are field due to \[{{Q}_{1}}\] and \[{{Q}_{2}}\] and the flux of E over S is due to the charge enclosed by surface. According to Gauss theorem \[\Phi =\frac{{{Q}_{in}}}{{{\in }_{0}}}=\frac{{{Q}_{1}}}{{{\in }_{0}}}\]