Question

(a) Calculate the energy of the system when a length $x$ of the slab is introduced into the capacitor.

(b) What force should be applied on the slab to ensure that goes slowly into the capacitor? Neglect any effect of friction or gravity.

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Similarly, the capacitance of the part without the dielectric is $C_{2}=dΞ΅_{0}b(ββx)β$

These two parts are connected in parallel. The capitance of the system is therefore,

$C=C_{1}+C_{2}=dΞ΅_{0}bβ[β+x(Kβ1)]$Β Β Β Β Β ...(1)

The energy of the capacitor is

$U=21βCV_{2}=2dΞ΅_{0}bV_{2}β[β+x(Kβ1)]$

(b) Suppose, the electric field attracts the dielectric slab with a force $F$. An external force of equal magnitude $F$ should be applied in opposite direction so that the plate moves slowly (no acceleration).

Consider the part of motion in which the dielectric moves a distance $dx$ further inside the capacitor. The capacitance increases to $C+dC$. As the potential difference remains constant at $V$, the battery is therefore,

$dW_{b}=VdQ=(dC)V_{2}$

The external force $F$ does a work

$dW_{e}=(βFdx)$

during the displacement. The total work done on the capacitor is

$dW_{b}+dW_{e}=(dC)V_{2}βFdx$.

This should be equal to the increase $dU$ in the stored energy. Thus,

$21β(dC)V_{2}=(dC)V_{2}βFdx$

$F=21βV_{2}(dC/dx)=2dΞ΅_{0}bV_{2}[Kβ1]β$

Thus, the electric field attracts the dielectric into the capacitor with a force $2dΞ΅_{0}bV_{2}[Kβ1]β$ and this minimum force should be applied in the opposite direction, so as to remove it from the capacitor.

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