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
`        Part b please. Please show the diagram and give complete conceptual explanation as is the legacy of this website.`
7 months ago

Arun
23523 Points
```							Let Resultant force on charge Q at corner A due to charges at corner B & C be F1.F1 = [(√3Q2)/(4πε0a2)].Resultant force on charge Q at corner A due to charge q at centre O is F2. F2 = [[(3Qq)/(4πε0a2)] The system of charges will be starting if the force acting on each charge is zero. This happens when q is negative & has magnitude such that F1 = F2.∴ [(√3Q2)/(4πε0a2)] = [(3Qq)/(4πε0a2)]∴ √3Q = 3q∴ q = (Q/√3) ------- magnitude∴ q = – (Q/√3)
```
7 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### Course Features

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

### Course Features

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