Question

A metal sphere of radius 0.1m is charged to a potential 36kv and kept at the corner B of a right angled triangle ABC in which AB=0.4m ,BC=0.5m and amgle B is 90, degree calculate the work required to shift a change of 3microcoulmb from c to corner A of the triangle

Answer 1

Answer :

Hey Mate,

Given,

AB = 0.4m

BC = 0.5m

∠B = 90°

#Figure is attach with this answer.

Q = $\frac{36*10^{3}*0.1}{K}$

KQ = $36*10^{3}*0.1$

$(\frac{KQ}{0.4}-\frac{KQ}{0.5})*3$μ

=> $3600(\frac{0.1}{0.4*0.5} )3*10^{-6}$

=> $\frac{2700*10^{-6}}{0.5}$

=> $5400*10^{-6}$

=> $5.4*10^{-3}J$

Answer 2

Answer: The work done is $W _{CA} = 5.4\times10^{-3}\ J$.

Explanation:

Given that,

Radius r = 0.1 m

AB $r_{AB}= 0.4 m$

BC$r_{BC}= 0.5 m$

Angle B = $90^{0}$

Charge $q = 3\times 10^{-6} C$

Charge potential $V = 36\times10^{3} V$

We know that,

The Electric potential is

$V = \dfrac{kQ}{r}$

$Q = \dfrac{36\times10^{3}\times0.1 m}{k}$

Now, the work done is

$W_{CA} = kQq\times[\dfrac{1}{r_{AB}} -\dfrac{1}{r_{BC}}]$

Put the value of all elements

$W_{CA} = 36\times10^{3}\ V\times0.1\ m\times3\times10_{-6}\ m\times[\dfrac{1}{0.4\ m}-\dfrac{1}{0.5\ m}]$

$W _{CA} = 5.4\times10^{-3}\ J$

Hence, the work done is $W _{CA} = 5.4\times10^{-3}\ J$.