(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 \[\Phi =\frac{{{Q}_{in}}}{{{\in }_{0}}}=\frac{{{Q}_{1}}}{{{\in }_{0}}}\]You need to login to perform this action.
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