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

30 th question in the attachment................................................
3 months ago

Arun
23760 Points

A charged particle charge Q = 2 × 10^-6C and mass m = 100g is placed at bottom of a smooth inclined plane another charged particle of same charge and mass are placed r distance from first particle as shown in figure.
Now, at equilibrium
mgsin30° = Coulombs force = Kq²/r²
100 × 10⁻³ × 10 × 1/2 = 9 × 10⁹ × (2 × 10⁻⁶)²/r²
1/2 = 9 × 10⁹ × 4 × 10⁻¹²/r²
r² = 72 × 10⁻³
r² = 0.072
taking square root both sides,
r = 0.268 m or, 26.8 cm
Hence, charge B should placed at 26.8 cm from charge A
3 months ago
Vikas TU
10469 Points

Dear student
The above ans is wrong
Mass = 0.1 kg
charge = 2*10^-6 C
Let distance between particle be r .
Fg = Fe
mgcos30 = kq^2 /r^2
r = 20.6 cm

3 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

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