Question

A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. 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

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

$KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i$

$k$ is Coulomb's constant.

#The electric field,$E$ at radius r is expressed as:

$E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii$

From i and ii, we have:

$KE=Eqr$

$r=(KE)/Eq$

#Substitute actual values in our equation:

$r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m$

Hence, the distance between the charge and the source of the electric field is 2.2 meters

Answer 2

Answer:

2.2

Explanation: