Chapter 33, Problem 29A

Suppose that the strength of the electric field about an isolated point charge has a certain value at a distance of 1 m. How will the electric field strength compare at a distance of 2 m from the point charge? What law guides your answer?

To determine

To explain: Comparison of electric field strength at a distance of 2 m from the point charge.

Answer to Problem 29A

  $\frac{E_1}{E_2}=\frac{1}{4}$

And the law is inverse square law.

Explanation of Solution

Introduction:

A force field that is surrounding an electric charge or a group of charges can be defined as an electric field. Mathematically, the electric field strength is given as:

  $E=\frac{F}{q}$

Hence, the electric field is strongest when the force is greatest and vice versa.

According to Coulomb’s law, the electric force is given as: $F=\frac{kqQ}{r^2}$

Putting the value of electric force in the expression of electric field we get:

  $E=\frac{\left(\frac{kqQ}{r^2}\right)}{q}=\frac{kQ}{r^2}$

At a distance, $r=1\text{ m}$ , the electric field strength is given by:

  $E_1=\frac{kQ}{{\left(\text{1 m}\right)}^2}=\frac{kQ}{1}\mathrm{...............................}(1)$

At a distance, $r=2\text{ m}$ , the electric field strength is given by:

  $E_2=\frac{kQ}{{\left(\text{2 m}\right)}^2}=\frac{kQ}{4}\mathrm{...............................}(2)$

Hence, by comparing both the equations (1) and (2), we get:

  $\frac{E_1}{E_2}=\frac{1}{4}$

Conclusion:

Hence, the electric field strength will be $\frac{1}{4}$ of the field if it would be 1 m distance apart.