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October 20, 2017

Removed Charge Problem (Kvpy '10)

Let's discuss a problem based on Removed Charge from Kishore Vaigyanik Protsahan Yojana, KVPY, 2010. First, try the problem yourself, then read the solution here.

The Problem:

(12) positive charges of magnitude (q) are placed on a circle of radius (R) in a manner that they are equally spaced. A charge (Q) is placed at the center. If one of the charges (q) is removed, then the force on (Q) is
(A) zero
(B)$$ \frac{qQ}{4\pi\epsilon_0R^2}$$ away from the position of the removed charge

(C) $$ \frac{11qQ}{4\pi\epsilon_0R^2}$$ away from the position of the removed charge
(D)
$$ \frac{qQ}{4\pi\epsilon_0R^2}$$ towards from the position of the removed charge

Discussion:

If one of the charges is removed, then the net force on Q is $$ \frac{qQ}{4\pi\epsilon_0R^2}$$ towards the position of removed charge

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