 ×     #### Thank you for registering.

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

Click to Chat

1800-1023-196

+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
```
A charge Q is divide intointo parts and then they are placed at a fixed. The force between the two charges is always maximum when charges are

```
11 months ago

```							 let parts are sepereted at a distance of dso repulsion force F= k q(Q-q)/d2     here Q and d are constant so for maximum force differentiate the force w.r.t q dF/dq = k/d2 [Q -2q] for maximum force dF/dq=0                             Q-2q=0                              Q/q =2
```
11 months ago
```							Dear studentThe above ans is incorrect , Please follow this  Let q1 = qand q2 = Q-q r  be the distance or separation between them The force of repusion between them is F = k (Q-q)q/r^2 Differenciating F with respect to q and setting it to zero will give us extreme force.dF/dq = 0 so, q/Q = 1/2 For this value q/Q the force is maximum .The force will be maximum if the second derivative of F is less than 0 d^2F /dq^2= -2k/r^2 which is less than 0.Thus the force of repulsion is maximum when q=Q/2 so , q1 = Q/2 an q2 = Q/2 Hope this helps
```
11 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Electrostatics

View all Questions »  ### 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