A point charge Q1-300 - C is located at P1- -1- -3 experiences an attractive force F-1-8-x-8-4- N due to point charge Q2 at 3- -3- -2 m- The value of charge Q2 will be

The position vector, R⃗_⃗2⃗1⃗=(1 3)â_̂x̂+( 1+3)â_̂ŷ+( 3+2)â_̂ẑ nbsp; nbsp; nbsp; nbsp; = 2â_̂x̂+2â_̂ŷ â_̂ẑ m The magnitude of force, nbsp; nb ...

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A point charg...

Question

A point charge $Q_{1} = 300 μ C$ is located at P(1, -1, -3) experiences an attractive force $_{1} = 8_{x} - 8_{y} + 4_{z}$ N due to point charge $Q_{2}$ at (3, -3, -2) m. The value of charge $Q_{2}$ will be

- 12 μ C

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- 40 μ C

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- 36 μ C

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- 20 μ C

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Solution

The correct option is B $- 40 μ C$
The position vector, $_{21} = (1 - 3)_{x} + (- 1 + 3)_{y} + (- 3 + 2)_{z}$
$= - 2_{x} + 2_{y} -_{z} m$

The magnitude of force,
$∣ ∣_{1} ∣ ∣ = \sqrt{8^{2} + (- 8)^{2} + (4)^{2}} = 12 N$

$∣ ∣_{1} ∣ ∣ = \frac{Q_{1} Q_{2}}{4 π ϵ_{0} {| R_{21} |}_{21}^{2}} = \frac{(300 \times 10^{- 6}) \times Q_{2}}{4 π (\frac{10^{- 9}}{36 π}) 3^{2}} = 3 \times 10^{5} Q_{2}$

$Q_{2} = \frac{12}{3 \times 10^{5}} = - 40 μ C$
$[∵$ attractive force so $Q_{2}$ have opposite polarity]

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A point charge Q1=300 μC is located at P(1, -1, -3) experiences an attractive force →F1=8^ax−8^ay+4^az N due to point charge Q2 at (3, -3, -2) m. The value of charge Q2 will be

A point charge $Q_{1} = 300 μ C$ is located at P(1, -1, -3) experiences an attractive force $_{1} = 8_{x} - 8_{y} + 4_{z}$ N due to point charge $Q_{2}$ at (3, -3, -2) m. The value of charge $Q_{2}$ will be