Thank you for registering.

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

Please check your email for login details.
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

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.

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.

Grade:upto college level

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free