30 th question in the attachment................................................
sri harshitha honey , 5 Years ago
Grade 12
Mechanics> 30 th question in the attachment............
2 AnswersArun
Last Activity: 5 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
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
Last Activity: 5 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
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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