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`        30 th question in the attachment................................................`
one month ago

Arun
23520 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/r²100 × 10⁻³ × 10 × 1/2 = 9 × 10⁹ × (2 × 10⁻⁶)²/r² 1/2 = 9 × 10⁹ × 4 × 10⁻¹²/r² r² = 72 × 10⁻³r² = 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
```
one month ago
Vikas TU
10070 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
```
one month ago
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### Course Features

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