Please dear god help.. What I have so far... a. Q=C*(delta)V b. E=(delta)V/d c.

Question

Please dear god help..

What I have so far...

a. Q=C*(delta)V

b. E=(delta)V/d

c. IM NOT SURE HOW TO WRITE THIS INTO AN EQUATION TO RANK. I think I should use U = 1/2*Q*(deltaV) = 1/2*C*(deltaV)^2 = Q^2 / 2C

A.

B. Since the battery is removed, we know that the Q is the same, and voltage changes to accomodate for the capacitance, which must double. Therefore, voltage must double per Q=C*(delta)V

C. If the battery remains connected, then the voltage is constant. If the plates are brought to have the distance, then we know that since voltage is constant, charge must change since capicatance is doubled.

D. If the battery is removed, and a dielectric with k=2 is placed between the plates, then we know that voltage will change since battery is removed. The new C is equal to double the old C, since k=2. Thus, by the equation of Q=C*(delta)V, if Q is held constant since the battery is removed, and C doubles, then voltage must half.

E. If the battery is connected, V is constant, and the k=2 for the diaelectric, then we know that C doubles, and thus Q doubles also.

Explanation / Answer

Capacitance C0 = e0*A/d

(a)

charge Qo = C0*dVo = e0*A*dv0/d

(b)

Electric field Eo = dVo/d

(c)

(A)

Uo = (1/2)*Co*dVo^2 = (1/2)*Q0^2/C0 = (1/2)*Qo*dVo

(B)

as the battery is removed the charge remains same Q = Qo

Capacitance C = eo*A/(d/2) = 2*eo*A/d = 2*Co

energy U = (1/2)*Q^2/C = (1/2)*Qo^2/2Co = Uo/2

(C)

the battery remains connected

potential remains same

dV = dVo

C = C0/2

U = (1/2)*C*V^2 = (1/2)*2*Co*dVo^2 = 2*Uo

------------------------------------------

(D)

battery removed

charge remains same Q = Q0

C = k*eo*A/d = K*Co

U = (1/2)*Q^2/C = (1/2)*Qo^2/(k*Co) = Uo/K = U0/2

(E)

battery remains connected

potential remains same V = dVo

C = k*eo*A/d = K*Co

U = (1/2)*C*V^2 = (1/2)*(k*Co)*dvo^2 = k*Uo = 2*Uo

C = E > A > B = D

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