Question

A light rod (of length ) with two point charges qand -q, both of mass m are attached at its end isplaced in a region of uniform electric field withintensity E as shown in figure. The angularacceleration of the rod at the instant shown isCora6qE cose69E sino(1)(2)mlml2qEsine(3)2qE coseml-44 COSO(4)ml​

Answer 1

Answer:

Angular acceleration of the dipole system is given as

$\alpha = \frac{2qEsin\theta}{mL}$

Explanation:

As we know that the system of two charges will behave like a dipole

So here the torque on the dipole is given as

$\tau = PE sin\theta$

$\tau = qLE sin\theta$

now moment of inertia of the system about its mid point is given as

$I = \frac{mL^2}{4} + \frac{mL^2}{4}$

$I = \frac{mL^2}{2}$

now we have

$\tau = I \alpha$

$\alpha = \frac{qLE sin\theta}{\frac{mL^2}{2}}$

$\alpha = \frac{2qEsin\theta}{mL}$

#Learn

Topic : Dipole

https://brainly.in/question/1890275

Answer 2

Answer : 2qEsinθ /ml

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