Question

A charge q of 1.3 × 10-16 coulombs moves from point A to a lower potential at point B in an electric field of 3.2 × 102 newtons/coulomb. If the distance traveled is parallel to the field is 1.1 × 10-2 meters, what is the difference in the potential energy?

Answer 1

`U=-4.58*10^(-16)J`

Step-by-step explanation:

The electrostatic potential energy of one point charge in the presence of an electric field is defined as the negative of the work done by the electrostatic force:

`U=-W(1)`

The work done by the electrostatic force is:

`W=Fd\\F=qE\\W=qEd(2)`

Replacing (2) in (1) and solving:

`U=-qEd\\U=-(1.3*10^(-16)C)(3.2*10^2(N)/(C))(1.1*10^(-2)m)\\U=-4.58*10^(-16)J`

Answer 2

Explanation :

It is given that,

Charge,
`q=1.3* 10^(-16)\ C`

Electric field,
`E=3.2* 10^2\ N/C`

Distance,
`d=1.1* 10^(-2)\ m`

The work done is stored in the form of potential energy.

`W=F.d`

`\because F=qE`

So,
`W=qE\ d`

`W=1.3* 10^(-16)\ C* 3.2* 10^2\ N/C* 1.1* 10^(-2)\ m`

`W=4.576* 10^(-16)\ J`

Hence, this is the required solution.