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:

                        

Sir answer for the question which I have attached as a image

5 months ago

Answers : (2)

Parth
58 Points
							
See in this case 
Work done = change in potential energy + K.E. here KE = 0
assuming charge At a = 1
so U1 – U2
U1 intial Potential Energy 
U1 = k*(q^2)/a + 2*k*q*1/a
for final configuration
U2 = k*(q^2)/a + 2*k*(q)/(a/sqrt(3))
so |U1- U2| = 
(2*k*q*(3^(1/2) - 1))/a
So option B
pls Approve if it helps
5 months ago
Ayush_Deep
120 Points
							
The correct answer will be option B .
Work done in moving a charge particle is equal to change in the potential energy of particle  = final PE - initial PE
The initial potential energy is 2kq/a
a is side of equilateral triangle.
Final potential energy is 2kq/(a/3½)=2√3kq/a
Work done = kq(2√3-2)/a= 2kq(√3-1)/a
Option B .
Thanks 
 
5 months 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