Question

A point charge with charge q1 = 2.10 μC is held stationary at the origin. A second point charge with charge q2 = -4.60 μC moves from the point ( 0.110 m , 0) to the point ( 0.275 m , 0.240 m ). How much work W is done by the electric force on the moving point charge? Express your answer in joules. Use k = 8.99×109 N⋅m2/C2 for Coulomb's constant: k=14πϵ0.

Answer 1

Work done is - 0.552 J, i.e 0.552 J of work is done by the force.

Solution:

As per the question:

Charge,
`q_(1) = 2.10\mu C`

Charge,
`q_(2) = - 4.60\mu C`

The point P is at (0.110 m, 0)

The point Q is at (0.275 m, .240 m )

k =
`8.99* 10^(9)\ N.m^(2)/C^(2)`

Now,

Work done on moving a charge is given by the change in its potential energy:

W =
`\Delta P`

Now, the electrostatic potential energy is given by:

`PE = (kq_(1)q_(2))/(R)`

where

R = Distance between the charges.

Now,

Initial PE,
`PE_(in)` at point P(0.110 m, 0) from origin:

`R_(1) = 0.110\ m`

`PE_(in) = (kq_(1)q_(2))/(R_(1))`

`PE_(in) = (8.99* 10^(9)* 2.10* 10^(- 6)* - 4.60* 10^(- 6))/(0.110) = - 0.789\ J`

Similarly, at point Q(0.275 m, 0.240 m) from origin:

`R_(2) = \sqrt{0.275^(2) + 0.240^(2)} = 0.365 m`

`PE_(f) = (kq_(1)q_(2))/(R_(2))`

`PE_(f) = (8.99* 10^(9)* 2.10* 10^(- 6)* - 4.60* 10^(- 6))/(0.365) = - 0.237\ J`

Work done, W =
`\Delta PE = PE_(f) - PE_(in) = -0.789 + 0.237 = - 0.552\ J`