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Force on dielectric slab in capacitor

Force on dielectric slab in capacitor :- Capacitor is a device to store electric charge. To increase the efficiency of a capacitor, we use a non conducting material like a dielectric (insulator) in between the plates of a capacitor.

The dielectric helps in increasing the charge on the capacitor plates. In this article we are going to find the force acting on the dielectric placed between the plates of the capacitor.

Let us consider a capacitor having two place of length “l” and width “b” separated by a distance “d”.

Let a dielectric slab of dielectric constant K is inserted between the plates up to a length of “x” :-

Let us connect the upper plate of the capacitor with positive terminal of a battery and the lower one with the negative terminal. An electric field is set up between the plates of the capacitor from upper plate (positive plate) to the lower plate (negative plate) which is shown by parallel arrows from positive to negative plate. But at the edges of the capacitor the direction of electric field is curved due to fringing effects or edge effect.

Due to applied electric field, opposite charges develop in dielectric slab. The charges on the surface of dielectric slab experience a force due to electric field lines which are curved on the edges of capacitor. The force on the “positive induced charges” is in the direction of electric field and on the “negative charges”, it is in the opposite direction of electric field as shown in figure below :-

We can resolve the forces in the vertical and horizontal components. The vertical components nullifies each others and a resultant horizontal force acts on the dielectric in the direction shown in figure below :-

This net horizontal force acting on the dielectric slab, pulls it inside the capacitor.

Now let us find an expression for this net horizontal force. We will find the value of this force in two cases :-

When there is an external voltage source connected to the capacitor
When there is no external voltage source connected to the capacitor

1. Force on dielectric slab in capacitor when there is an external voltage source connected to the capacitor

Let a batter of V volt is connected across the terminals of the capacitor. It can be considered as a parallel combination of two capacitors, one with air core (of length l – x) and the other with dielectric core (of length x).

Let C₁ be the capacitance of capacitor with air core, then

$\displaystyle {{C}_{1}}=\frac{{{\varepsilon }_{0}}b(l-x)}{d}$

And let C₂ be the capacitance of capacitor with dielectric core, so

$\displaystyle {{C}_{2}}=\frac{K{{\varepsilon }_{0}}bx}{d}$

The equivalent capacitance C of the combination is given by

C = C₁ + C₂

$\displaystyle C={{C}_{1}}+{{C}_{2}}=\frac{{{\varepsilon }_{0}}b(l-x)}{d}+\frac{K{{\varepsilon }_{0}}bx}{d}$

$\displaystyle C=\frac{{{\varepsilon }_{0}}b}{d}\left[ l+x(K-1) \right]$

The total electric potential energy U stored in the capacitor is given by

$\displaystyle U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\frac{{{\varepsilon }_{0}}b}{d}\left[ l+x(K-1) \right]{{V}^{2}}$

Now we know that the negative derivative of potential energy function(gradient) is equal to the electrostatic force acting on the dielectric, i.e.,

$\displaystyle F=-\frac{dU}{dx}$

So the magnitude of force F, is given by

$\displaystyle F=\left| \frac{dU}{dx} \right|$

$\displaystyle F=\left| \frac{dU}{dx} \right|=\left( \frac{1}{2}\frac{{{\varepsilon }_{0}}b}{d} \right)\frac{d}{dx}\left[ l+x(K-1) \right]{{V}^{2}}$

$\displaystyle F=\left( \frac{1}{2}\frac{{{\varepsilon }_{0}}b}{d} \right)\left[ 0+(K-1) \right]{{V}^{2}}$

$\displaystyle \Rightarrow F=\frac{{{\varepsilon }_{0}}b{{V}^{2}}}{2d}(K-1)$ ……….(1)

This force is independent of the value of x, i.e., the force remains the same whatever be the position of slab. So it moves with constant acceleration.

2. Force on dielectric slab in capacitor when there is no external voltage source connected to the capacitor

The electric potential energy stored in the capacitor is given by :-

$\displaystyle U=\frac{1}{2}\frac{{{q}^{2}}}{C}$

So putting the value of equivalent capacitance in the above formula, we get

$\displaystyle U=\frac{1}{2}\frac{{{q}^{2}}}{\frac{{{\varepsilon }_{0}}b}{d}\left[ l+x(K-1) \right]}$

Again taking the magnitude of negative derivative of potential energy function(gradient) with respect to position x to find the electrostatic force acting on the dielectric :-

$\displaystyle F=\left| \frac{dU}{dx} \right|=\frac{d}{dx}\left[ \frac{1}{2}\frac{{{q}^{2}}}{\frac{{{\varepsilon }_{0}}b}{d}\left[ l+x(K-1) \right]} \right]$