Question

Force of attraction between the plates of a parallel plate capacitor is:
(A) $ \dfrac{{{q^2}}}{{2{\varepsilon _0}A}} $
(B) $ \dfrac{{{q^2}}}{{{\varepsilon _0}A}} $
(C) $ \dfrac{q}{{2{\varepsilon _0}A}} $
(D) $ \dfrac{{{q^2}}}{{2{\varepsilon _0}{A^2}}} $

Answer 1

Hint :To solve this question, we have to know what capacitor is and what electric field is. We know that Capacitor is an electronic part that stores electric charge. The capacitor is made of two close channels (generally plates) that are isolated by a dielectric material. The plates collect electric charge when associated with a power source. One plate amasses positive charge and the other plate gathers negative charge. We know that an electric field encompasses an electric charge, and applies power on different charges in the field, drawing in or repulsing them. Electric field is here and there contracted as E - field.

Complete Step By Step Answer:
We know that, electric field between the capacitors is,
 $ \dfrac{\sigma }{{{\varepsilon _0}}} $
Electric field due to one charge plate is $ \dfrac{q}{{2{\varepsilon _0}A}} $
 $ = q/A{\varepsilon _0} $
(here, we can say, the electric field is multiplied by charge present on the other plate)
Now, the force on plate is equal to $ \dfrac{1}{2}E.q $
 $ = \dfrac{q}{{2{\varepsilon _0}A}}.q $
 $ = \dfrac{{{q^2}}}{{2{\varepsilon _0}A}} $
Now, here we know that the factor half comes as we have to consider the electric field of only one plate acting on another.
So, the right option will be option number A.

Note :
We know that the unit of capacitance is Farad. We also have to know that, the capacitance is the measure of electric charge that is put away in the capacitor at voltage of one Volt. We know that, At whatever point a charge is set in an electric field; it encounters an electric power on it. The power felt by a unit good charge or test charge when it's kept close to a charge is called Electric Field. The electric field is likewise characterized as the district which pulls in or repulses a charge. The electric field is a vector amount and it is indicated by E.