| A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1m. If the electric field between the plates is \[100\,\,N\,\,{{C}^{-1}},\]the magnitude of charge on each plate is- [JEE Main 12-Jan-2019 Evening] |
| \[\left( \text{Take}\,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}\frac{{{C}^{2}}}{N{{m}^{2}}} \right)\] |
A) \[8.85\times {{10}^{-10}}C\]
B) \[7.85\times {{10}^{-10}}C\]
C) \[9.85\times {{10}^{-10}}C\]
D) \[6.85\times {{10}^{-10}}C\]
Correct Answer: A
Solution :
| [a] The electric field between two plates is |
| \[E=\frac{\sigma }{{{\varepsilon }_{0}}}=\frac{q}{A{{\varepsilon }_{0}}}\Rightarrow q=EA{{\varepsilon }_{0}}\] |
| \[=(100)(1)(8.85\times {{10}^{-12}})\] |
| \[q=8.85\times {{10}^{-10}}C\] |
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