Question

Four identical charges (+2.0 ?C each) are brought from infinity and fixed to a straight line. The charges are located 0.40 m apart. Calculate the amount of work done in assembling the charges.

Answer 1

Let all four charges be a, b, c and d.

Consider charge a is moved first from infinity. The line looks like

a-------

Now as there is electrostatic field or other charge to apply force on charge 'a' , so work done for moving it 0.

Now charge 'b' is moved from infinity at a distance 0.40 m from charge 'a' . If potential energy at infinity is zero then work done by charge 'a' on charge 'b' is

U_ab = q₁ × q₂ / ( 4×?×e₀×r )

= 2×10^-6 × 2×10^-6/ ( 4×3.14×8.86×10^-12×.4)

U_ab = 0.089 J

The line looks like

a-----b-----

Now as charge 'c' is moved from infinity to point 0.40 m away from charge 'b' , there are two force acting on it. Therefore two work done needs to be calculated

Work done by charge 'a' on charge 'c',

U_ac = q₁ × q₂/( 4×?×e₀×r)

Here r is double i.e. 0.8 m

Calculating we get, U_ac = 0.449 J

Now work done by charge 'b' on charge 'c'

U_bc = q₂ × q₃ / (4×?×e₀×r)

We get U_bc = 0.089 J

Now the line looks like

a------b------c------

Now the final charge 'd' is moved from infinity to .4 m away from charge 'c'

Here three work done needs to be calculated

The calculated work done just like shown above are

U_ad = 0.0299

U_bd = 0.0449

U_cd = 0.089

So total work done in moving charges from infinity to straight line is

W = U_ab +U_ac +U_bc +U_ad +U_bd + U_cd

W = 0.089 + 0.0449 + 0.089 + 0.0299 + 0.0449 + 0.089

W = 0.3867 J