Question

If possible can you answer the different if the battery is connected or disconnected. Thank you

Battery gets disconnected

A capacitor is connected to a 12-volt battery. The initial capacitance is C; the initial charge stored is Q; the initial field between the plates is E; and the initial stored energy is U.

Then, the distance between the plates is reduced by a factor of 2. The battery is then disconnected. Following this, the space between the plates is filled with a material of dielectric constant 2.5 (there was originally air between the plates, with a dielectric constant of 1.0).

A) After these changes, what is the charge stored, in terms of the original charge? Express your answer in terms of Q.

B) After these changes, what is the capacitance, in terms of the original capacitance? Express your answer in terms of C.

C) After these changes, what is the voltage across the capacitor, in volts?

D) After these changes, what is the magnitude of the electric field between the plates, in terms of the original field? Express your answer in terms of E.

E) After these changes, what is the energy stored in the capacitor, in terms of the original energy? Express your answer in terms of U.

Now Battery gets connected

F) After these changes, what is the energy stored in the capacitor, in terms of the original energy? Express your answer in terms of U

Answer 1

6, Q.,E,U. Let 'd be the distance between the plates initially with given, C, Q, E U values. Vo=12v By connecting the battery distance between the plates is reduced by a factor 2. Then a=d is distance bin plates now. . Capacitance becomes, cr EA = 2(6A) = 26. =)=2C. potential to = 12V remaus same. Then charge becomes a = C, V = q vo= 20 Vo = 2Q Energy = Q = 2 Q. Electry field becomes out = Ic, v = (2c) v = 2 u=20 Electric field becomes E= Vo = 2100) = 2E. constant Now if battery is disconnected and dielectric of 2.55kes filled betines the plates thes C2, Q., Ęų be the changes respectively. (B. C₂ = K C, = (2.5) C = (2,5) (20) = 50. v A After Removal of battery, there is no additional - Charge added. Hence charge remaug same. il Q₂ = 9, = 2Q ~

(C) Voltage changes to V. V 2 (%) -(V). =) 2 = 2 (12) = 24 volt (①) Electric field between the plates is g - (- 2004 =)E, = 2() () E = 4(E) (E) Energy Stred, U = £C V₂ > U = $150 (2) => U2 = 514 / ) ( 2 (v²) UE4U