Find the change in electric potential energy , `Delta U` as a charge of (a) `2.20 xx 10^(-6) C.` (b) `- 1.10 xx 10^(-6) C ` moves from a point A to a point given that the change in electric potential between these points is `DeltaV= V_(B) -V_(A) = 24.0 V`

Explanation

(a) Solving `DeltaV= DeltaU//q_(0)` for `DeltaU`, we find
`DeltaU = q_(0) DeltaV = ( 2.20 xx 10^(-6) C) ( 24.0 V) = 5.28 xx 10^(-5) J`
(b) Similarly , using `q_(0) = - 1.10 xx 10^(-6) C`, we obtain
`DeltaU = q_(0) DeltaV = ( - 1.10 xx 10^(-6) C) ( 24.0V) = - 2.64 xx 10^(-5) J`