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A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2xx10^(-9) from a point P(0, 0, 3cm) to a point Q (0, 4cm, 0) via a point R(0, 6cm, 9cm).

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Answer Text

Solution :
charge located at the origin , q=8 mc =`8xx10^(-3)C` <br> Magnitude of a small charge, which is taken from a point p to point R to point Q,`q_(1)=-2xx10^(-9)C` <br> All the points are represented in the given figure, <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_V01_XII_C02_E01_012_S01.png" width="80%"> <br> point P ios at a distance `d_(1)=3 cm `, fromt he origin along z-axis <br> point Q is at a distance , `d_(2)=4 cm, ` from the origin along y-axis . <br> potential at point p , `V_(1)=q/(4pi in_(0)xxd_(1))` <br> Potential at point Q, `V_(2)=q/(4pi in_(0)d_(2))` <br> work done (W) by the electrostatic force is independent of the path. <br> `:. W=q_(1) [V_(2)-V_(1)]` <br> `=q_(1)[q/(4pi in_(0)d_(2))-q/(4pi in_(0)d_(1)]` <br> `=(qq_(1))/(4pi in_(0))[1/(d_(2))-1/(d_(1))]....(i)` <br> Where, `1/(4pi in_(0))=9xx10^(9) Nm^(2)C^(-2)` <br> `:. W=9xx10^(9) xx8xx10^(-3)xx(-2xx10^(-9))[1/0.04-1/0.03]` <br> `=-144xx10^(-3)xx((-25)/3)` <br> =1.27 J <br> Therefore, work done during the process is 1.27 J.

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