A) \[\frac{q}{3A{{\varepsilon }_{0}}}\]
B) \[\frac{q}{2A{{\varepsilon }_{0}}}\]
C) \[\frac{q}{A{{\varepsilon }_{0}}}\]
D) \[\frac{2q}{A{{\varepsilon }_{0}}}\]
Correct Answer: B
Solution :
Charge distribution will be as shown Field at a point in between two plates, Due to \[{{Q}_{B}},\] \[{{E}_{Q}}=\frac{q-q'}{2A{{\varepsilon }_{0}}}\] Net field at that point, \[E={{E}_{Q}}+{{E}_{P}}\] \[=\frac{q-q'}{2A{{\varepsilon }_{0}}}+\frac{q'}{2A{{\varepsilon }_{0}}}\] \[\Rightarrow \] \[E=\frac{q}{2A{{\varepsilon }_{0}}}\]You need to login to perform this action.
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