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

A, B, C, D

Explanation

a. The work that must be done on `q_3` by an external force is equal to the difference of of potential energy `U` when the charge is at `x=2m` and the potential energy when it is at infinity.
`:. W=U_f-U_i`
`=1/(4piepsilon_0) [(q_3q_2)/((r_32)_f)+(q_3q_1)/((r_31)_f)+(q_2q_1)/((r_21)_f)]-1/(4piepsilon_0)[(q_3q_2)/((r_31)_i)+(q_3q_1)/((r_31)_i)+(q_2q_1)/((r_21)_i)]`
Here `(r_21)_i=(r_21)_f`
and `(r_32)_i=(r_31)_i=oo`
`:. W=1/(4piepsilon_0)[(q_3q_2)/((r_32)_f-(q_3q_1)/((r_31)_f)]`
Substituting the values, we have
`W=(9.0xx10^9)(10^-12)[((1)(1))/((1.0))+((1)(-1))/((2.0))]`
`=4.5xx10^-3J`
b. The total potential energy of the three charges in given by
`U=1/(4piepsilon_0)((q_3q_2)/r_32+(q_3q_1)/r_31+(q_2q_1)/r_21)`
`=(9.0xx10^9)[((1)(1))/((1.0))+((1)(-1))/(2.0))+((1)(-1))/((1.0))](10^-12)`
`=-4.5xx10^-3J`