A capacitor is constructed from two square, metallic plates of sides and separat

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 is inserted a distance x into the capacitor as shown in the figure below. Assume d is much smaller than x. Suggestion: The system can be considered as two capacitors connected in parallel. (Use the following as necessary: 0, , , Q, d, and x.)

(a) Find the equivalent capacitance of the device.
Ceq =  

(b) Calculate the energy stored in the capacitor.
U =  

(c) Find the direction and magnitude of the force exerted by the plates on the dielectric.

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Explanation / Answer

given capacitor, made of two square metallic plates of side l, seperation d
Charge on plates = Q, -Q
dielectric material of constant k is inserted for distance x

a. considering this as equivalent capacitance of two capacitors connected in parallel,
   Ceq = C1 + C2
   C1 = k*l*x*epsilon/d [ where epsilon is permittivity of free space]
   C2 = l*(l - x)*epsilon/d

   so Ceq = k(lx)epsilon/d + l(l-x)*epsilon/d = l(k(x) + (l-x))epsilon/d

b. energy stored in Ceq = 0.5*Q^2/Ceq
   U = Q^2*d/2l*epsilon(kx + l - x)

c. force exerted on the dielectric be F
   then F = -dU/dx = Q^2*d(k - 1)/2l*epsilon(kx + l - x)^2
   so the dielectric slab will be acted upon a force to decrease total energy store in the capacitor, so in +ve x direction
   this force is given by F = Q^2*d(k - 1)/2l*epsilon(kx + l - x)^2

d. x = l/2
   l = 5.3 cm = 0.053 m
   d = 2.1 mm = 2.1*10^-3 m
   k = 4.5
   Q = l^2*epsilon*1.9*10^3/d
   epsilon = 1/4*pi*8.98*10^9 = 8.86*10^-12

   so, Q = 2.253*10^-8 C
   hence
   F = 1.87*10^-4 N

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