JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 10)
A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is :
(Take $$\varepsilon $$0 = 8.85 $$ \times $$ 10$$-$$12 $${{{C^2}} \over {N - {m^2}}}$$)
(Take $$\varepsilon $$0 = 8.85 $$ \times $$ 10$$-$$12 $${{{C^2}} \over {N - {m^2}}}$$)
9.85 $$ \times $$ 10–10 C
8.85 $$ \times $$ 10–10 C
6.85 $$ \times $$ 10–10 C
7.85 × 10–10 C
Explanation
$$E = {\sigma \over {{ \in _0}}} = {Q \over {A\,{ \in _0}}}$$
Q = AE$$ \in $$0
Q = (1) (100) (8.85 $$ \times $$ 10$$-$$12)
Q = 8.85 $$ \times $$ 10$$-$$10C
Q = AE$$ \in $$0
Q = (1) (100) (8.85 $$ \times $$ 10$$-$$12)
Q = 8.85 $$ \times $$ 10$$-$$10C
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