Question

Two points charges are located on the x-axis, q_1=-1muC at x=0 and q_2=+1muc at x=1m. a. Find the work that must be done by an external force to bring a third point charge q_3=+1muC from infinity to x=2m. b. Find the total potential energy of the system of three charges.

Answer 1

1. Work done= $4.5 * 10^{-3} J$

2. Total Energy= $-4.5 * 10^{-3} J$

For work done,

1. Refer attached diagram
2. Work done= Change in potential energy
3. Work done= Potential Energy(final)- Potential Energy(Initial)
4. Potential Energy= $\frac{1}{4\pi E} \frac{q1 * q2}{r}$
5. Final potential energy=$\frac{1}{4\pi E } (\frac{q1*q2}{r_{12} } + \frac{q1*q3}{r_{13}} + \frac{q2*q3}{r_{23}} )$
6. Initial potential energy=$\frac{1}{4\pi E } (\frac{q1*q2}{r_{12} } + \frac{q1*q3}{r_{13}} + \frac{q2*q3}{r_{23}} )$

        However, initially charge q3 is at infinity. Therefore, r₁₃ and r₂₃ is ∞.  Substituting in initial potential energy, two terms become 0. Also, r₁₂ is same initially and finally. Therefore terms with q1 and q2 cancel out since there is no change. Substituting the given values of charges and distances in formula, we get

   Work done= $4.5 * 10^{-3} J$

For Total Energy,

1. Refer attached diagram
2. Total Energy= Potential energy caused due to combinations of 3 charges(i.e q1.q2 ; q2.q3 ; q1q3)
3. Potential Energy formula is given above.
4. ∴ Total Energy=$\frac{1}{4\pi E } (\frac{q1*q2}{r_{12} } + \frac{q1*q3}{r_{13}} + \frac{q2*q3}{r_{23}} )$  
5. Total Energy= $\frac{1}{4\pi E } (\frac{1*-1}{1} + \frac{1*-1}{2} + \frac{1*1}{1} )$
6. Total Energy = $-4.5 * 10^{-3} J$

Answer 2

The total potential energy of the system of three charges is U = −4.5 × 10^−3 J

Explanation:

The work done is equal to the difference of potential energy U when the charge is at x = 2 m and the potential energy when it is at infinity.

W = Uf − Ui

W = 1 / 4πε0 [q3q2 / (r32)f + q3q1 / (r31)f  + q2q1 / (r21)f − 1 / 4 πε0 [q3q2 / (r31)i  + q3q1 / (r31) i + q2q1 / (r21)i]

Here (r21)i = (r21)f

and (r32)i = (r31)i = ∞

W = 1 / 4πε0 [ q3q2 / (r32)f − q3q1 / (r31)f

Substituting the values, we have

W = (9.0 × 10^9) (10^−12) [(1)(1) / (1.0) + (1)(−1) / (2.0) ]

= 4.5×10^−3 J  

b. The total potential energy of the three charges in given by

U = 1 / 4πε0 (q3q2 / r32 + q3q1 / r31 + q2q1 / r21)

U = (9.0 × 10^9) [(1)(1) / (1.0) + (1)(−1) / 2.0) + (1)(−1) / (1.0)] (10^−12)

U = −4.5 × 10^−3 J

Hence the total potential energy of the system of three charges is U = −4.5 × 10^−3 J