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A charge + Q is uniformly distributed over a thin ring with radius R. A negative point charge – Q and mass m starts from rest at a point far away from the centre of the ring and moves towards the centre. Find the velocity of this particle at the moment it passes through the centre of the ring.

moidin afsan

12 Years agoGrade 11

3 Answers

Arun Kumar

Approved Tutor Answer12 Years ago

Draw the figure .
See we find the force by whole ring at a point where distance of point is x from the center.
Center is our origin here.
|Work done|=gain in KE
$cos(\theta)=x/\sqrt{x^2+R^2} \\W=\int_{-\infty}^{0}(\int_{0}^{2\pi R}KQQds/(2\pi R*(x^2+R^2))cos(\theta)dx \\=>W=\int_{-\infty}^{0}KQ^2*x/(\sqrt{x^2+R^2})^3dx \\=>W=[-KQ^2/\sqrt{x^2+R^2}]_{-\infty}^0 \\=>W=KQ^2/R \\\\ v^2=2KQ^2/Rm$
Arun Kumar
IIT Delhi
Askiitians Faculty

moidin afsan

12 Years ago

thank uuuuuuu

moidin afsan

12 Years ago

how did dx come ????? pls ans

Please send the full solution as soon as possible 🙏

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