Chapter 21, Problem 64A

To determine

Magnitude of electric field intensity.

Answer to Problem 64A

  $E=3.497\times {10}^{19}\text{N/C}$ in the same direction of force.

Explanation of Solution

Given:

Positive test charge, $q\text{'}=4.005\times {10}^{-19}\text{C}$

Force experienced, $F=14.005\text{N}$.

Formula Used:

Strength of electric field is given by,

  $E=\frac{F}{q}$

Where, $E$ is electric field strength, $F$ is force exerted by electric field and $q$ is the test charge.

Calculation:

On putting the values, the magnitude of electric field intensity is

  $\begin{array}{c}E=\frac{F}{q\text{'}}\\ E=\frac{14.005}{4.005\times {10}^{-19}}\text{N/C}\\ =3.497\times {10}^{19}\text{N/C}\end{array}$

Conclusion:

Therefore, the magnitude of electric field strength is $E=3.497\times {10}^{19}\text{N/C}$ .