A) \[17\times {{10}^{-22}}N/C\]
B) \[1.5\times {{10}^{-15}}N/C\]
C) \[1.9\times {{10}^{-10}}N/C\]
D) zero
Correct Answer: C
Solution :
(c) \[1.9\times {{10}^{-10}}N/C\] In region II or between the plates, the electric field \[{{E}_{||}}={{E}_{A}}-{{E}_{B}}=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}\] \[=\frac{\sigma \left( {{\sigma }_{A}}\,or\,{{\sigma }_{B}} \right)}{{{\varepsilon }_{0}}}=\frac{17.0\times {{10}^{-22}}}{8.85\times {{10}^{-12}}}\] \[E=1.9\times {{10}^{-10}}N{{C}^{-1}}\]
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