Electrostatics

JEE Main level, numerical answer, medium

17 of 311

Question 17 of 311

Three point charges q,−2qq,-2q and 2q2q are placed on xx axis at a distance x=0,x=34Rx=0,x=\dfrac{3}{4}R and x=Rx=R respectively from origin as shown. If q=2×10−6Cq=2\times {10}^{-6}C and R=2cmR=2cm, the magnitude of net force experienced by the charge −2q-2q is _____ NN.

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Answer: 5440

The net force on the given charge will be the resultant of the forces due to the other two charges.
Hence, the net force (Fn)\left({F}_{n}\right) can be calculated as follows:
Fn=14πϵ04q2(R4)2−14πϵ02q2(3R4)2=9×109×16q2R2[4−29]=9×109×16×(2×10−6C)2(0.02m)2×349=5440N\begin{matrix}{F}_{n} & = & \dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{4{q}^{2}}{{\left(\dfrac{R}{4}\right)}^{2}}-\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{2{q}^{2}}{{\left(\dfrac{3R}{4}\right)}^{2}} \\ & = & 9\times {10}^{9}\times \dfrac{16{q}^{2}}{{R}^{2}}\left[4-\dfrac{2}{9}\right] \\ & = & 9\times {10}^{9}\times \dfrac{16\times {\left(2\times {10}^{-6}C\right)}^{2}}{{\left(0.02m\right)}^{2}}\times \dfrac{34}{9} \\ & = & 5440N\end{matrix}

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