Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# A Particle a with a Charge of 2.0 × 10−6 C is Held Fixed on a Horizontal Table. a Second Charged Particle of Mass 80 G Stays in Equilibrium on the Table at a Distance of 10 Cm from the First Charge. - Physics

Short Note

A particle A with a charge of 2.0 × 10−6 C is held fixed on a horizontal table. A second charged particle of mass 80 g stays in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is μ = 0.2. Find the range within which the charge of this second particle may lie.

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#### Solution

Given:
Magnitude of charge of particle A, q1 =  2.0 × 10−5 C
Separation between the charges, r = 0.1 m
Mass of particle B, m = 80 g = 0.08 kg
Let the magnitude of charge of particle B be q2.
By Coulomb's Law, force on B due to A,

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

Force of friction on B =μmg

Since B is at equilibrium,
Coulomb force = Frictional force

⇒ F = μmg

$\Rightarrow \frac{9 \times {10}^9 \times 2 \times {10}^{- 5} \times q_2}{\left( 0 . 1 \right)^2} = 0 . 2 \times 0 . 08 \times 9 . 8$

$\Rightarrow q_2 = 8 . 71 \times {10}^{- 8} C$

∴ The range of the charge = ∓ 8.71×10-8 C

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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 7 Electric Field and Potential
Q 29 | Page 122
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