a capacitor c=4 microfarad is charged by a battery 10 volt . how will charge , capacity and energy associated with capacitor will change if distance between it's plate is doubled with battery disconnected
Answers
Answer:
CHARGE REMAINS CONSTANT ENERGY DOUBLES
Explanation:
when capacitor is initially charges is accquires a charge of q= cv
=4 * 10⁻⁶ * 10
=4 * 10 ⁻⁵ COULOMBS
NOW THE RELATION BETWEEN CAPACITANCE OF 2 PLATES = Aε/D
where A is the area of the plates , D is the distance between them and, ε is the permittivity of free space
when the battery is disconnected the charge already in the capacitor remains same until we discharge it by some means. (law of conservation of charge)
we can see from the above relation that capacitance and distance D are inversely proportional.. as distance increases charge decreases
when distance is doubled new capacitance C₀ = Aε/2D
= C/2 (C IS THE OLD CAPACITANCE )
INITIAL ENERGY OF THE CAPACITOR = Q²/2C = E
FINAL ENERGY OF CAPACITOR =Q²/2C₀
=Q²/C = 2E