A charge +q is revolving around a stationary charge +Q of radius r if

1.	A charge +q is revolving around a stationary charge +Q of radius r if force b/w carge is F then work done of this motion will be?
Answer» Answer: Given: +q CHARGE is revolving around a charge +Q in a radius "r". Force between the charge is F To find: Work done by the revolving charge in this kind of motion. Concept: Work done by any force is given as the DOT product of Force and DISPLACEMENT. Mathematically, $\boxed{ \sf{ \large{work = \vec F. \: \vec r}}}$ $\sf{ \implies \: work = f \times r \times \cos( \theta)}$ Now in a CIRCULAR motion , the force vector and the displacement vector is at 90° with ANOTHER with one another. We know that cos (90°) = 0 $\sf{ \implies \: work = f \times r \times \cos( 90 \degree)}$ $\sf{ \implies \: work = f \times r \times ( 0)}$ $\sf{ \implies \: work = 0 \: joules}$ So , no work is done .

A charge +q is revolving around a stationary charge +Q of radius r if force b/w carge is F then work done of this motion will be?

Answer»

Answer:

Given:

+q CHARGE is revolving around a charge +Q in a radius "r". Force between the charge is F

To find:

Work done by the revolving charge in this kind of motion.

Concept:

Work done by any force is given as the DOT product of Force and DISPLACEMENT.

Mathematically,

$\boxed{ \sf{ \large{work = \vec F. \: \vec r}}}$

$\sf{ \implies \: work = f \times r \times \cos( \theta)}$

Now in a CIRCULAR motion , the force vector and the displacement vector is at 90° with ANOTHER with one another.

We know that cos (90°) = 0

$\sf{ \implies \: work = f \times r \times \cos( 90 \degree)}$

$\sf{ \implies \: work = f \times r \times ( 0)}$

$\sf{ \implies \: work = 0 \: joules}$

So , no work is done .

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