Consider a dielectric slab of length and width L , thickness D , and dielectric
ID: 1426693 • Letter: C
Question
Consider a dielectric slab of length and width L, thickness D, and dielectric constant . The slab is partially inserted a distance x between the two plates of a charged parallel plate capacitor as shown in the figure below. As indicated in the figure, the capacitor has the same dimensions (length L, width L, thickness D) as the the dielectric slab.
Let U be the potential energy of the slab capacitor system. Of course, this U will depend on the amount of overlap between the slab and the capacitor plates . If the slab is pulled out a small distance dx, then the amount of work done is given by
dU=Fpull·dx, where Fpull is the pulling force exerted on the slab to counteract the electrical force Fe. The electrical force is therefore given by Fe = Fpull = dU/dx.
(a) Suppose that when the slab has been inserted a distance x between the two plates of the capacitor, an electric potential difference V is applied between the plates, and then the source of the electric potential is disconnected from the capacitor. Thus, the charge on the capacitor plates must remain constant. Starting with the expression U=(1/2)Q2/C, and the information above, find an expression for the magnitude of the electrical force Fe on the dielectric slab. Your answer should be expressed only in terms of 0, , V, L, and D.
(b) Now assume that the potential V on the capacitor plates remains constant. Starting with the expression U=(1/2)CV2, (neglecting any contribution from the source of the constant potential) and the information above, find an expression for themagnitude of the electrical force Fe on the dielectric slab. Your answer should be expressed only in terms of 0, , V, L, and D.
Explanation / Answer
A) C_1 = k0l*x/D
C_2 = 0l*(l-x)/D
both are parallel
=> Ceq = C_1+C_2 = k0L*x/D + 0L*(L-x)/D = 0L*(L+(k-1)x)/D
Q1 = C1*V and Q2 = C2*V
=> Qeq = (C1 + C2)V
=> U = 0.5*Qeq^2/Ceq = 0.5*Ceq*v^2 =0.5*v^2*(0L*(L+(k-1)x))/D
Fe = -DU/Dx = -0.5*V^2*0L*(K-1)/D
magnitude = 0.5*V^2*0L*(K-1)/D Answer
B) => U = 0.5*Ceq*V^2 = 0.5*v^2*(0L*(L+(k-1)x))/D
Fe = -DU/Dx = 0.5*V^2*0L*(K-1)/D
magnitude = 0.5*V^2*0L*(K-1)/D Answer
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