A) \[1.5\,N/C\]
B) \[1.5\times {{10}^{-10}}\,N/C\]
C) \[3\,N/C\]
D) \[3\times {{10}^{-10}}\,N/C\]
Correct Answer: C
Solution :
The situation is shown in the figure. Plate 1 has surface charge density o and plate 2 has surface charge density \[-\sigma .\] The electric fields 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, \[\sigma =26.4\times {{10}^{-12}}C/{{m}^{2}},\] \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N-{{m}^{2}}.\] Hence, \[E=\frac{26.4\times {{10}^{-12}}}{8.85\times {{10}^{-12}}}\approx 3\,N/C\] Note The direction of electric field is from the Positive to the negative plate.You need to login to perform this action.
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