JEE Main 12 Jan 2019 Evening Question 13
Question: 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 $ 100N{C^{-1}}, $ the magnitude of charge on each plate is $ ( Take,{\varepsilon_0}=8.85\times {10^{-12}}\frac{C^{2}}{Nm^{2}} ) $
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
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 $
Show Answer
Answer:
Correct Answer: A
Solution:
- 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 $
