JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Electrostatics

As shown in the figure, a configuration of two equal point charges (

q_{0} = - 2 μ C

) is placed on an inclined plane. Mass of each point charge is 20 g. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height

h = x \times 1 0^{- 3} m

. The value of

x

is ______.
(Take

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}, g = 10 ms^{- 1}

)

Answer: 300

Solution

The upper charge rests on the frictionless incline at height

h

, so its distance along the slope from the lower charge is

r = \frac{h}{sin 3 0 ^{\circ}} = 2 h

. Equilibrium along the incline requires the gravity component to balance the Coulomb repulsion:

m g sin 3 0^{\circ} = \frac{1}{4 π ε _{0}} \frac{q _{0}^{2}}{( 2 h ) ^{2}}

. Solving,

h^{2} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 6} ) ^{2}}{4 ( 0.02 ) ( 10 ) ( 0.5 )} = \frac{0.036}{0.4} = 0.09

, so

h = 0.3

= 300 \times 1 0^{- 3}

m, giving

x = 300

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Electrostatics

Solution

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