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A dielectric slab is introduced between the plates of a capacitor. If the charge on the capacitor is q and the magnitude of the induced charge on the dielectric surface is q?, then (a) q? q (always) (c) q? = q (always) (d) q? = 0

A dielectric slab is introduced between the plates of a capacitor. If the charge on the capacitor is q and the
magnitude of the induced charge on the dielectric surface is q?, then
(a) q? < q (always) (b) q? > q (always) (c) q? = q (always) (d) q? = 0

Grade:10

1 Answers

Vasantha Kumari
askIITians Faculty 38 Points
7 years ago

A dielectric slab is introduced between the plates of a capacitor. If the charge on the capacitor is q and the
magnitude of the induced charge on the dielectric surface is q?

Q’ must be smaller than q

The induced charges on the faces of the dielectric decrease the electric field. Thus, q’<q.

Electric field produced due to charge q’ is in a direction just oppostite to that produced by charge q.

The induced surface charges on the dielectric cause an electric field just opposing the original field and these fields tend to cancel each other resulting in a reduction of the original field.

Thanks & Regards,

Vasantha Sivaraj,

askIITians faculty


181-1263_Capacitor_dielectric.JPG

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