A) \[\frac{{{Q}_{1}}-{{Q}_{2}}}{2\,{{\varepsilon }_{0}}A},\,\,\frac{{{Q}_{1}}-{{Q}_{2}}}{2\,{{\varepsilon }_{0}}A}\]
B) \[\frac{{{Q}_{1}}+{{Q}_{2}}}{2\,{{\varepsilon }_{0}}A},\,\,\frac{{{Q}_{1}}+{{Q}_{2}}}{2\,{{\varepsilon }_{0}}A}\]
C) \[\frac{{{Q}_{1}}+{{Q}_{2}}}{2\,{{\varepsilon }_{0}}A},\,\,\frac{{{Q}_{1}}-{{Q}_{2}}}{2\,{{\varepsilon }_{0}}A}\]
D) \[\frac{{{Q}_{1}}+{{Q}_{2}}}{{{\varepsilon }_{0}}A},\,\,\frac{{{Q}_{1}}-{{Q}_{2}}}{{{\varepsilon }_{0}}A}\]
Correct Answer: C
Solution :
At 1, \[{{\operatorname{E}}_{1}} = \frac{{{\sigma }_{2}}}{2{{\varepsilon }_{0}}}+\frac{{{\sigma }_{1}}}{2{{\varepsilon }_{0}}} towards left\] At 2, \[{{\operatorname{E}}_{2}} = \frac{{{\sigma }_{2}}}{2{{\varepsilon }_{0}}}+\frac{{{\sigma }_{1}}}{2{{\varepsilon }_{0}}} towards right\] \[{{\sigma }_{2}}=\frac{{{Q}_{2}}}{A}\,and\,\,{{\sigma }_{1}}=\frac{{{Q}_{1}}}{A}\] \[\therefore \,\,\,\,\,\,\,\,\,{{E}_{1}}=\frac{{{Q}_{1}}+{{Q}_{2}}}{2\,{{\varepsilon }_{0}}A}\] and \[{{E}_{2}}=\frac{{{Q}_{2}}-{{Q}_{1}}}{2\,{{\varepsilon }_{0}}A}\]You need to login to perform this action.
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