#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# two parallel plates (each of area 200cm2) of a capacitor are given equal and opposite charge of magnitude 1.75microC.The space between the plates is filled with dielectric.If the electric field in the dielectric is 1.5*10^6 V/m what will be the magnitude of chare induced on each dielelectric plate

Susmita
425 Points
3 years ago
Induced charge on dielectric is given by the formula
Q'=(k-1)Q/kA
Q is charges on capacitor plate,A area,k dielectric constant.so basically you have to find out k.
From the continuity of normal component of electric field across the boundary
Eo=kE where Eo is the applied field and E is the resultant field.
E=Eo-Ep,where Ep is the polarization field inside the dielectric,that is given in this question.
Eo  is the applied field,or field between the plates when the dielectric is absent.You can find it out from the given data as
$E_{0}=Q/A\epsilon_{0}$
I hope now you can solve it.