**Force on a Dielectric in a Capacitor**are also related force on a dielectric slab in a capacitor, force on a dielectric slab inserted into a parallel‐plate capacitor, and on the dielectric in a parallel plate capacitor.

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Force on a Dielectric in a Capacitor ?

Just as a conductor is attached into an electric field so, too is dielectric and for essentially the same reason: the bound charge tends to accumulate near the free charge of opposite sign.

consider, for example the case of a slab of linear

**dielectric**material, partially inserted between the plates of a parallel plate capacitor, and zero outside. If this were literally true, there would be no net force on the dielectric at all, since the field everywhere would be perpendicular to the plates. However, there is in reality a fringing field around the edges, which for most purposes can be ignored but in this case is responsible for the whole effect.**Read Also useful Article:-**

*[ Indeed, the field could not terminate abruptly at the edge of the Capacitor, for if it did line integral of E*

**(electric field)**around the closed loop shown in the figure (a) would not be zero ]
It is this non uniform fringing field that pulls the dielectric into the capacitor. Fringing fields are notoriousaly difficult to calculate we can avoid this altogether, by the following ingenious method.

Also Read:

• What is Coulomb’s Law and some related important point.

• Write the properties of Conductor.

• What is charge and also described properties of charge.

• Write the properties of Conductor.

• What is charge and also described properties of charge.

**Case-1:**Let a capacitor is charged with a battery of emf V volt and battery is removed. Let W be energy of the system it depends, of course, on the amount of overlap. If I pull the dielectric out an infinitesimal distance dx, the energy is changed by an amount equal to the work done.

**dW**

**=Fme.dx .......(1)**

where F(me) is the force, must exert, to counteract the electrical force F on the dielectric force:

Thus the electrical force F on the slab is

**F(me)=-F**.Thus the electrical force F on the slab is

Now, the energy stored in the capacitor is

And the capacitance in this case is

Where 'l' is the length of the plates in the figure. since the total charge on the plates

**(Q=CV)**is held constant, as the dielectric moves. In terms of Q,**Case-2:**Let a capacitor is charged with a battery of emf V volt and battery remains connected. Hence, the capacitor is maintained at a fixed potential. In this case the battery also does work as the dielectric moves instead of

**equation (1) ,**we now have

**dW=Fme.dx + VdQ**

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**Fast2study**
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