Chapter 21, Problem 60A

To determine

Magnitude of electric field intensity.

Answer to Problem 60A

  $E=3\times {10}^4\text{N/C}$ in the same direction of force.

Explanation of Solution

Given:

The given figure is,

  

Positive test charge, $q=1\times {10}^{-5}\text{C}$

Force experienced, $F=0.3\text{N}$.

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

  $\begin{array}{c}E=\frac{F}{q}\\ E=\frac{0.3}{1\times {10}^{-5}}\text{N/C}\\ =3\times {10}^4\text{N/C}\end{array}$

Since charge q is positive therefore direction of electric field is same as that of force.

Conclusion:

Therefore, the magnitude of electric field strength is

  $E=3\times {10}^4\text{N/C}$ in the same direction of force.