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A parallel plate capacitor consists of square plates of edge length l that are separated by a distance d where d%u300Al. A potential difference delta V is maintained between the plates. A material of dielectric constant k fills half of the space between the plates. The dielectric slab is now withdrawn from the capacitor

(a) Find the capacitance when the left edge of the dielectric is at a distance x from the center of the capacitor.

(b) If the dielectric is removed at a constant speed v, What is the current in the circuit?

Added Sun, 21 Jun '15

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(a) the capacitance when the left edge of the dielectric is at a distance x from the center of the capacitor as 2 capacitors connecting parallel:

C = C1 + C2

= e0 * (L/2 + x)/L * A/d + e0 * k * (L/2 - x)/L * a /d

= e0 * A/d * (1/2 + x/L + k/2 - kx/L)

= e0 * L^2/d*[(k+1)/2 + (1-k)x/L) where A = L^2

(b) Current Magnitude

I = |dq/dt|

= |DeltaV*dC/dt|

= DeltaV*e0*A/d*(k-1)/L*v ; dx/dt = v

Added Sun, 21 Jun '15

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