A) \[{{\operatorname{E}}_{2}}={{E}_{4}}>{{E}_{1}}={{E}_{3}}\]
B) \[{{\operatorname{E}}_{1}}={{E}_{2}}>{{E}_{3}}>{{E}_{4}}\]
C) \[{{\operatorname{E}}_{4}} >{{E}_{1}}= {{E}_{3}} > {{E}_{2}}~\]
D) \[{{\operatorname{E}}_{1}} > {{E}_{2}} > {{E}_{3}} > {{E}_{4}}\]
Correct Answer: C
Solution :
As electric field, \[E=-\frac{dV}{dx}\] For region 1 \[{{V}_{1}}\,\,=constant\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d({{V}_{1}})}{dx}\,\,\,=\,\,0\] \[\Rightarrow \,\,\,\,\,\,\,{{E}_{1}}=0\] For region 2 \[{{\operatorname{V}}_{2}} =+ve=+f\left( x \right)\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{E}_{2}}=-\frac{d({{V}_{2}})}{dx}=-ve\] For region 3 \[{{V}_{3}}=constant\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d({{V}_{3}})}{dx}=0\,\,\,\Rightarrow \,\,{{E}_{3}}=0\] For region 4 \[{{\operatorname{V}}_{4}}=-ve=-f\left( x \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,{{E}_{4}}=\,\,\frac{d({{V}_{4}})}{dx}=+\,ve\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\operatorname{E}}_{4}}>{{F}_{1}}={{F}_{3}}>{{E}_{2}}\]bYou need to login to perform this action.
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