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
        
Five point charges each of value +q are placed on five vertices of a regular hexagon of side L m. What is the magnitude of force on a point charge of value -q coulomb placed at the centre of hexagon?
2 years ago

Answers : (3)

ankush singh
30 Points
							
The answer is kq2/L2 towards the direction of empty vertics.  Because force due to other 4 charges cancel out each other.
2 years ago
Heet Mehta
26 Points
							There are five vertices so it must be a Pentagon,  four charges cancel each other out and there is one charge left on the remaining vertex so apply the regular formula F = kq2/r2
						
one year ago
Ashutosh
12 Points
							If there were 6 charges at each corner then the net electric field at the centre would be zero since it is a regular hexagon . the electric field due to 5 charges will be equal and opposite to the net electric field due to sixth charge so as to make the net electric field at the centre to be zero . Hence electric field at the centre due to 5 charges is will be equal to electric field at the centre due to sixth charge. F=(kq^2)/L^2
						
7 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