1.

Two point charges q_1 and q_2 are placed in an external uniform electric field as shown in figure. The potential at the location of q_1 and q_2 are V_1 and V_2, i.e., V_1 and V_2 are potentials at location of q_1 and q_2 due to external unspecified charges only. Then electric potential energy for this configuration of two charged particle is

Answer»

`(q_1V_1+q_2V_2)/2`
`q_1V_1 + q_2V_2`
`q_1V_1+q_2V_2+(q_1q_2)/(4piepsilon_0r)`
`(q_1q_2)/(4piepsilon_0r)`

Solution :The potential energy of the system can be found by computing the work done by enternal AGENT in ASSEMBLING the system without changing the kinetic energy of the system. Let us first bring `q_(1)` from infinity to the desired location, then in doing so we have to do work against enternal electric FIELD which is equal to `W_(1)=q_(1)V_(1)` (potential at infinity is considered as zero). Now, we will bring `q_(2)` from infinity to the desired location, to do so, we have to do work against external electric field and against electric FORCE of `q_(1)`, so work done is
`W_(2)=q_(2)V_(2)+(q_(1)q_(2))/(4piepsilon_(0)r)`
So total work done is `U`,
`q_(1)V_(1)=q_(2)V_(2)+(q_(1)q_(2))/(4piepsilon_(0)r)`


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