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

Explanation :

It is given that,

Charge, $q=1.3\times 10^{-16}\ C$

Electric field, $E=3.2\times 10^2\ N/C$

Distance, $d=1.1\times 10^{-2}\ m$

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

$W=F.d$

$\because F=qE$

So, $W=qE\ d$

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

$W=4.576\times 10^{-16}\ J$

Hence, this is the required solution.

Answer 2

Answer:

$U=-4.58*10^{-16}J$

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:

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