A) \[\frac{{{Q}_{1}}+{{Q}_{2}}}{2{{\varepsilon }_{0}}A},\frac{{{Q}_{2}}-{{Q}_{1}}}{2{{\varepsilon }_{0}}A}\]
B) \[\frac{{{Q}_{1}}-{{Q}_{2}}}{2{{\varepsilon }_{0}}A},\frac{{{Q}_{1}}+{{Q}_{2}}}{2{{\varepsilon }_{0}}A}\]
C) \[\frac{{{Q}_{1}}+{{Q}_{2}}}{{{\varepsilon }_{0}}A},\frac{{{Q}_{1}}-{{Q}_{1}}}{{{\varepsilon }_{0}}A}\]
D) \[\frac{{{Q}_{1}}-{{Q}_{2}}}{{{\varepsilon }_{0}}A},\frac{{{Q}_{1}}+{{Q}_{2}}}{{{\varepsilon }_{0}}A}\]
Correct Answer: A
Solution :
At 1; \[{{E}_{1}}=\frac{{{\sigma }_{2}}}{2{{\varepsilon }_{0}}}+\frac{{{\sigma }_{1}}}{2{{\varepsilon }_{0}}}\] At 2; \[{{E}_{2}}=\frac{{{\sigma }_{2}}}{2{{\varepsilon }_{0}}}-\frac{{{\sigma }_{1}}}{2{{\varepsilon }_{0}}}\]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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