#### Thank you for registering.

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

Click to Chat

1800-1023-196

+91-120-4616500

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

# 1.Two point charges placed at a distances of 20 cm in air repel each other with a certain force. When a dielectric slab of thickness 8 cm and dielectric constant K is introduced between these point charges, force of interaction becomes half of it’s previous value. Then K is approximately

noogler
489 Points
6 years ago
= 4
on introducing slab distance btwn charges reduces to (20-8)+8(k)1/2 =rl
F=1/2F
Akshat Jain
74 Points
5 years ago
can you please tell me the whole solution
Sowmya yarnagula
15 Points
2 years ago
• We know F=1/4pi£0 q1q2/r^2
• F/2=1/4pi£0 q1q2/(distance bw charges without dielectric+thickness of slab under root of dielectric constant-1)^2
• Substituting nd solving gives the value 4