Question

A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105. If the electric potential energy of the charge is 75 J, what is the distance between the charge and the source of the electric field? Round your answer to the nearest tenth.

Answer 1

The answer is 2.2 meters.

Answer 2

The distance between the charge and the source of the electric field is 2.2 m.

The potential energy U of a charge q placed in an electric field created by a source charge Q, at a distance r from the source charge is given by,

`U=(kQq)/(r)` ...... (1)

Here, k is the Coulomb constant.

The electric field E at a distance r from the source charge is given by,

`E =(kQ)/(r^2)` ......(2)

From equations (1) and (2)

`U=E*q*r`

Rewrite the expression for ri.

`r=(U)/(Eq)`

Substitute 75 J for U,
`4.5*10^5 V/m` for E and
`7.5*10^(-5) C` for q.

`r=(U)/(Eq) \\ =(75 J)/((4.8*10^(5) V/m)(7.2*10^(-5)C)) \\ =2.17 m`

Rounding off to the nearest tenth, the the distance between the charge and the source charge is 2.2 m