A capacitor is constructed from two square, metallic plates of sides and separat
ID: 1659550 • Letter: A
Question
A capacitor is constructed from two square, metallic plates of sides and separation d. Charges +Q and-Q are placed on the plates, and the power supply is then removed. A material of dielectric constant K is inserted a distance x into the capacitor as shown in the figure below. Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (Use the following as necessary: EO, K. , d, and .) 20 (b) Calculate the energy stored in the capacitor. (Use the following as necessary: Eo, K, f, Q, d, and x.) (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (Use the following as necessary: Eo K, t, Q, d, and x.) Direction (d) Obtain a numerical value for the force when x-/ 2, assuming - 4.70 cm, d 2.00 mm, the dielectric is glass (x- 4.5), and the capacitor was charged to 1800 V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel. IN DirectionExplanation / Answer
(a)
Use capacitence of parallel plate capacitor to solve the given problem
The length and breadth of the square plate is l
the area of the plate is A = l^2
The separation between two plates is d
Let
x = width of the plate
l = length of the slab
The area of portion of the dielectric slab is A = l*x
the capacitance of the dielectric slab region is
C1 = (k*e0*l*x)/d
the capacitance of the empty region is
C2 = (k*e0*l*(l-x)) / d
the net capacitance is
C = C1+ C2
= (e0*l*(l-x))/d + (k*e0*l*x)/d
= e0*l/d*[(l-x+k*x)]
= e0*l/d*[(l + x (k-1))]
(b)
The energy stored in the capacitor is
U = Q^2 / 2*C
= 1/2 *[Q^2 / e0*l/d*[(l + x (k-1))]
= 1/2 [Q^2*d / e0*l(l + x(k - 1)]
(c)
The force exoerienced by the dielectric is
F = -Fex
= - dU/dx
dU/dx = d/dx(1/2*(Q^2/C)
= - 1/2*(Q^2/C)dC/dx
= - 1/2*(deal v)^2dC/dx
= - 1/2*(deal v)^2d/dx[ e0*l/d*[(l + x (k-1))]
= -1/2*(deal v)^2*(e0*l/d)*d/dx [(l + x (k-1))]
= -1/2*(deal v)^2*(e0*l/d)* [(0 + (k-1))]
(d)
the magnitude of the force is
F = 1/2*(deal v)^2*(e0*l/d) [(k-1)]
= (1800)^2*(8.85*10^-12)*(0.047)*(4.5 - 1) / 2*(2.00*10^-3)
= 1.179*10^-3 N
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