Question

31. A light rod (of length 1) with two point charges q
and –q, both of mass m are attached at its end is
placed in a region of uniform electric field with
intensity E as shown in figure. The angular
acceleration of the rod at the instant shown is​

Answer 1

The angular  acceleration of the rod at the instant is​ ∝ =  2qEθ/ Ml

Explanation:

Torque = Force x perpendicular distance

Z = qE. l x sinθ

OR

Z = qElsinθ

Also ewe know that

Z = I∝

Here

∝ is angular acceleration.

So ∝ = Z / I

I = Ml^2 / 2

∝ = qElsinθ / Ml^2 / 2

∝ = 2qElsinθ  / Ml

∝ =  2qEθ/ Ml

Thus The angular  acceleration of the rod at the instant is​ ∝ =  2qEθ/ Ml

Also learn more

A particle completes 3 revolutions per second on a  circular path of radius 8 cm. Find the values of  angular velocity and centripetal acceleration of the  particle?

https://brainly.in/question/7331148