Question

a small spheres each having mass ‘m' kg and charge ‘q’ coulomb are suspended from a point by insulating threads each ‘l' metres long of negligible mass. if theta is the angle of thread making with the vertical when equilibrium has been attained show that q = √mg tan theta 4πefsilentnot​

Answer 1

EXPLANATION.

• STEP = 1.

Draw F. B. D of both the block

we get,

$t \sin( \theta) = f_{e} \: .......(1)$

$t \cos( \theta) = mg \: .....(2)$

On dividing both equation we get,

$\boxed{\bold{\tan( \theta) = \frac{ f_{e} }{mg}}}$

$f_{e} \: = \: electrostatic \: \: force$

$\bold{f_{e} \: = \frac{k {q}^{2} }{4 {l}^{2} \sin {}^{2} ( \theta) }}$

$radius \: = \: 2l \sin( \theta)$

$\bold{m \: = \frac{ f_{e} }{g \tan( \theta) }}$

substitute the value of Fe in equation

we get,

$\bold{m \: = \frac{ {q}^{2} }{16\pi {l}^{2} e_{0} \sin {}^{2} ( \theta) \times g \tan( \theta) }}$

$\bigstar\green{\boxed{\bold{{q}^{2} \: = 16\pi e_{0} {l}^{2} \sin {}^{2} ( \theta) \times m \times g \tan( \theta)}}}$

HENCE PROVED

note = also see the image attachment.