freddieishawt
contestada

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?

Respuesta :

Explanation :

It is given that,

Charge, [tex]q=1.3\times 10^{-16}\ C[/tex]

Electric field, [tex]E=3.2\times 10^2\ N/C[/tex]

Distance, [tex]d=1.1\times 10^{-2}\ m[/tex]

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

[tex]W=F.d[/tex]

[tex]\because F=qE[/tex]

So, [tex]W=qE\ d[/tex]

[tex]W=1.3\times 10^{-16}\ C\times 3.2\times 10^2\ N/C\times 1.1\times 10^{-2}\ m[/tex]

[tex]W=4.576\times 10^{-16}\ J[/tex]

Hence, this is the required solution.

sebasacevedop

Answer:

[tex]U=-4.58*10^{-16}J[/tex]

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:

[tex]U=-W(1)[/tex]

The work done by the electrostatic force is:

[tex]W=Fd\\F=qE\\W=qEd(2)[/tex]

Replacing (2) in (1) and solving:

[tex]U=-qEd\\U=-(1.3*10^{-16}C)(3.2*10^2\frac{N}{C})(1.1*10^{-2}m)\\U=-4.58*10^{-16}J[/tex]

Otras preguntas