A parallel plate capacitor is comprised of two metal plates with area A and sepa
ID: 1772211 • Letter: A
Question
A parallel plate capacitor is comprised of two metal plates with area A and separated by distance d. This parallel plate capacitor is connected to a battery with voltage AVo. Your answer should depend on A, d, 0, and any other physical constants. a. Determine the charge stored on the plates of the capacitor and the energy stored in the capacitor. Determine the strength of the electric field between the plates of the capacitor. An experimenter has five of these identical capacitors charged with the same battery. After being charged, each capacitor undergoes a different experiment: (A) remains unchanged (B) the battery is removed and the plates are brought together to half their initial b. c. separation (C) the battery remains connected to the plates of the capacitor and the plates are brought together to half their initial separation (D) the battery is removed and a dielectric with K = 2 is placed between the plates (E) the battery remains connected to the plates of the capacitor and a dielectric with K = 2 is placed between the plates The motion of the plates are accomplished using insulating contacts. Rank the energy stored in the three capacitors after the experiments are conducted. Explain your reasoning.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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