Question

A 30 nC charge experiences a 0.035 N electric force. What is the magnitude of electric field at the position of this charge?

Answer 1

Concepts and reason

The concepts required to solve the problem are electric force, electric field, and charge. First, using the relation between electric force, electric field, and charge, find the electric field equation. Then, from the given values, find the electric field.

Fundamentals

The electric force is the force experienced between two charged particles. The electric force is directly proportional to the charges' product and inversely proportional to the square of the distance between them. The electric field is a region where other charged objects will experience a force. The expression for the electric force is as follows: \(F=q E\)

Here, \(F\) is the electric force, \(q\) is the charge, and \(E\) is the electric field.

The relation between electric force, electric field, and the charge are as follows: \(F=q E\)

Rewrite the electric field equation. \(E=\frac{F}{q}\) Electric field is defined as the electric force per unit charge.

Therefore, the electric field can be found from the electric force and the charge. The electric field is the source of the electric force. The electric field equation is given below: $$ E=\frac{F}{q} $$ Substitute \(30 \mathrm{nC}\) for \(q\) and \(0.035 \mathrm{~N}\) for \(F\) $$ E=\frac{0.035 \mathrm{~N}}{(30 \mathrm{nC})\left(\frac{10^{-9} \mathrm{C}}{1 \mathrm{nC}}\right)} $$ \(=11.67 \times 10^{5} \mathrm{~N} / \mathrm{C}\)

Thus, the magnitude of the electric field experienced by the charge is \(11.67 \times 10^{5} \mathrm{~N} / \mathrm{C}\). For a charge of \(30 \mathrm{nC}\) to experience an electric force of \(0.035 \mathrm{~N}\), the magnitude of the electric field is \(11.67 \times 10^{5} \mathrm{~N} / \mathrm{C}\), calculated using the relation between electric force and charge.

Thus, the magnitude of electric field experienced by the charge is \(11.67 \times 10^{5} \mathrm{~N} / \mathrm{C}\).