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

Answer:

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 = $8.99\times 10^{9}\ N.m^{2}/C^{2}$

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:

$PE = \frac{kq_{1}q_{2}}{R}$

where

R = Distance between the charges.

Now,

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

$R_{1} = 0.110\ m$

$PE_{in} = \frac{kq_{1}q_{2}}{R_{1}}$

$PE_{in} = \frac{8.99\times 10^{9}\times 2.10\times 10^{- 6}\times - 4.60\times 10^{- 6}}{0.110} = - 0.789\ J$

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

$R_{2} = \sqrt{0.275^{2} + 0.240^{2}} = 0.365 m$

$PE_{f} = \frac{kq_{1}q_{2}}{R_{2}}$

$PE_{f} = \frac{8.99\times 10^{9}\times 2.10\times 10^{- 6}\times - 4.60\times 10^{- 6}}{0.365} = - 0.237\ J$

Work done, W = $\Delta PE = PE_{f} - PE_{in} = -0.789 + 0.237 = - 0.552\ J$