Click to Chat

1800-2000-838

+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
Menu
Shane Macguire Grade: upto college level
`        A charged particle having a charge of – 2.0 × 10^-6 is placed closed to a non-conducting plate having a surface charge density 4.0 × 10^-6 C/m^2. Find the force of attraction between the particle and the plate.`
3 years ago

## Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
```										Q = –2.0 × 10–6C Surface charge density = 4 × 10–6 C/m2
We know □(→┬E ) due to a charge conducting sheet = σ/2ε base 0
Again Force of attraction between particle & plate
= Eq = σ/2ε base 0 * q = 4 * 10^-6 * 2 * 10^-6/2 * 8 * 10^-12 = 0.452N

```
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Electromagnetic Induction

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Electricity and Magnetism
• OFFERED PRICE: Rs. 1,696
• View Details

## Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details