A) \[\text{1}\text{.5}\,\text{N/C}\]
B) \[\text{1}\text{.5 }\!\!\times\!\!\text{ }\,\text{1}{{\text{0}}^{\text{-10}}}\,\text{N/C}\]
C) \[3\,\text{N/C}\]
D) \[\text{3 }\!\!\times\!\!\text{ }\,\text{1}{{\text{0}}^{\text{-10}}}\text{N/C}\]
Correct Answer: C
Solution :
The situation is shown in the figure. Plate 1 has surface charge density \[\sigma \] and plate 2 has surface charge density\[\sigma \]. The electric field at point \[P\] due to two charged plates add up, giving \[E=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{\sigma }{{{\varepsilon }_{0}}}\] Given, \[\sigma =2.64\times {{10}^{-12}}C/{{m}^{2}},\] \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N\text{-}{{m}^{2}}\] Hence, \[E=\frac{26.4\times {{10}^{-12}}}{8.85\times {{10}^{-12}}}\approx 3\,\,N/C\]You need to login to perform this action.
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