An ideal air-filled parallel-plate capacitor has round plates and carries a fixe

Question

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its plates. All the geometric parameters of the capacitor (plate diameter and plate separation) are now DOUBLED. If the original energy stored in the capacitor was u_0, how much energy does it now store? U_0/2 U_0/4 4u_0 When two or more capacitors are connected in series across a potential difference each capacitor carries the same amount of charge. the potential difference across the combination is the algebraic sum of the potential differences across the individual capacitors. the equivalent capacitance of the combination is less than the capacitance of any of the capacitors. All of the above choices are correct. None of the above choices are correct. A 6.00-pF parallel-plate capacitor has charges of plusminus 40.0 mu C on its plates. How much potential energy is stored in this capacitor? 123 mu J 133 mu J 113 mu J

Explanation / Answer

(6)
Energy stored in Capacitor is given by , Uo = Q^2/2*C
Where, C = A*eo/d

Cnew = A'*eo/d' = (4*A*eo)/(2*d) = 2*A*eo/d = 2*C

Unew = Q^2/2*Cnew
Unew = Q^2/(2*2*C)
Unew = Uo/2

(7)
In Series, Charge remians same but Potential is sum of individual potential difference !!
Ceq = 1/(1/C1 + 1/C2)

Correct option :-
(D) All the Options are correct.

(8)
C = 6.0 uF
Q = 40.0 uC
Potential Energy , U = Q^2/(2*C)
U = 40^2 / 2* 6.0
U = 133 uJ

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