You are to bring two charges `q_(1)` and `q_(2)` from infinity to the points represented by the potentials `V_(1)` and `V_(2)` in an electric field E. If the distance between `q_(1)`, and `q_(2)` within the field E is r, find the total work done in assembling the configuration. Imagine an electric field `E=(20hati+30hatj)NC^(-1)` in a space. The potential at the origin is zero. Find the potential at the point (2, 2).

Explanation

Suppose `q_(1)` and `q_(2)` are two point charge at position vectors `r_(1)`, and `r_(2)` respectively, in a uniform external electric field of intensity E.
Now, work done in bringing charge `q_(1)` from infinity to position `r_(1)` is `W_(1)=q_(1).V(r_(1))`, where `V(r_(1))` is potential at `r_(1)` due to external field.
Again, work done in bringing charge `q_(2)` from infinity to position `r_(2)` against the external field is `W_(2)=q_(2).V(r_(2))`, where `V(r_(2))` is potential at `r_(2)` due to extemal field.
While bringing `q_(2)` from infinity to position `r_(2)`, work has also to be done against the field due to `q_(1)`. This is `W_(3)=(q_(1)q_(2))/(4piin_(0)r)`, where r is the distance between `q_(1)` and `q_(2)` (given).
By the superposition principle, total work done in assembling the charge configuration
`W_("net")=W_(1)+W_(2)+W_(3)`
`W_("net")=q_(1).V(r_(1))+q_(2).V(r_(2))+(q_(1)q_(2))/(4piepsi_(0)r)`
Given, electric field, `E=(20hati+30hatj)NC^(-1)`
As, `E=-(dV)/(dr)` or dV=-E.dr
`impliesdV=-(20hati+30hatj).(dxhati+dyhatj)`
`impliesdV=-20dx-30dy`
On integration both sides within the given limits, we get
`int_(0)^(v)dV=-20int_(0)^(2)dx-30int_(0)^(2)dy`
`[V]_(0)^(v)=-20[x]_(0)^(2)-30[y]_(0)^(2)`
`V-0=-20(2-0)-30(2-0)`
`:.` Potential, `V=-40-60=-100V`