Question

Example 1.6 class 12th Physics Chapter 1.
Answer this only:---
Consider what would happen if the system was rotated through 60° about O.
See figure in attachment...
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Answer 1

Answer:

Your answer is the attachment

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Answer 2

Answer:

If  the system was rotated through 60° about O, net force on the charge would remain zero and the system will remain in equilibrium.

Explanation:

Before rotation:

Net force on the charge $Q$ = $\vec F_1 + \vec F_2 + \vec F_3$

$F_1 = \frac{kq_1 Q}{r^2} = \frac{kqQ}{r^2}$

$F_2 = \frac{kq_2 Q}{r^2} = \frac{kqQ}{r^2}$

$F_3 = \frac{kq_3 Q}{r^2} = \frac{kqQ}{r^2}$

$|\vec F_1+\vec F_2| = \sqrt{(\frac{kqQ}{r^2})^2+(\frac{kqQ}{r^2})^2 + 2\frac{kqQ}{r^2}\frac{kqQ}{r^2}.cos(120^o)}=\frac{kqQ}{r^2}$

$|\vec F_1+\vec F_2|$   and   $\vec F_3$  are in opposite direction and are of the same magnitude, so, they cancel out. So, net force acting on charge $Q$ is zero.

After rotation:

When, the systems of charges is rotated by $60^o$. The net force on charge $Q$ still remains zero as, the relative directions of each force to each other is not at all changed.

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