badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

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
Menu
Grade: 11

                        

A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium of q is equal to (a) -(Q/2) (b) -(Q/4) (c) + (Q/4) (d) +(Q/2)

8 years ago

Answers : (2)

Anirban Mukherjee
47 Points
							

Here use this formula

U = 1/(4Πε0)[2qQ/r + Q2/2r] = 0

q=-Q/4

8 years ago
Jevin George
14 Points
							

well...you hv not mentioned what q is in ur ques....

,,solving ques assuming that q is the third charge that u mentioned

for the first Q charge to be in equilibrium...net force on it shud be zero..

i.e... force due to q + force due to Q =0

let the distance btw Q charge to Q charge be 2l.....so dat distance btw q and Q is l

kqQ/l2+kQ2/4l2=0

solving it we get..q=-Q/4...

..hope it helps..:)

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


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

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