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30 th question in the attachment................................................

30 th question in the attachment................................................

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Grade:12

2 Answers

Arun
25750 Points
4 years ago
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
Vikas TU
14149 Points
4 years ago
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 
 

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