Question

a point charge of 6.0×10^-8 coulomb is situated at the co-ordinate origin. how much work will be done to taking an electron from the point X =3 metre to x= 6m?​

Answer 1

Hope this help..........

Answer 2

Given,

Charge, $q=6\times 10^{-8}\ C$

To find,

Work done to take an electron from the point X = 3 metre to x = 6m

Solution,

Firstly, we will find the potential at x = 3 m and x = 6 m as :

$V_1=\dfrac{kq}{x}\\\\V_1=\dfrac{9\times 10^9\times 6\times 10^{-8}}{3}\\\\V_1=180\ V$

$V_2=\dfrac{kq}{x}\\\\V_2=\dfrac{9\times 10^9\times 6\times 10^{-8}}{6}\\\\V_2=90\ V$

The work done per unit charge is called potential difference. It is given by :

$W=q\times (V_2-V_1)\\\\W=-1.6\times 10^{-19}\times (90-180)\\\\W=1.44\times 10^{-17}\ J$

So, the work done is $1.44\times 10^{-17}\ J$.

Learn more,

Electric potential

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