Chapter 21.1, Problem 4PP

To determine

To Complete:The given table.

Answer to Problem 4PP

The complete table is:

  $\begin{array}{ccc}\begin{array}{l}\text{Test Charge }\\ \text{Strength (C)}\end{array}& \begin{array}{l}\text{Force Exerted on }\\ \text{Test Charge (N) }\end{array}& \begin{array}{l}\text{Electric Field }\\ \text{Intensity (N/C)}\end{array}\\ 1.0\times {10}^{-6}& 0.30& \underset{\_}{3.0\times {10}^5}\\ 2.0\times {10}^{-6}& \underset{\_}{0.66}& 3.3\times {10}^5\\ \underset{\_}{3.0\times {10}^{-6}}& 0.45& 1.5\times {10}^5\end{array}$

Explanation of Solution

Given:

The table to complete is:

  $\begin{array}{ccc}\begin{array}{l}\text{Test Charge }\\ \text{Strength (C)}\end{array}& \begin{array}{l}\text{Force Exerted on }\\ \text{Test Charge (N) }\end{array}& \begin{array}{l}\text{Electric Field }\\ \text{Intensity (N/C)}\end{array}\\ 1.0\times {10}^{-6}& 0.30& \\ 2.0\times {10}^{-6}& & 3.3\times {10}^5\\ & 0.45& 1.5\times {10}^5\end{array}$

Formula used:

The electric field strength is given by:

  $E=\frac{F_{\text{on }q\text{'}}}{q\text{'}}$

Where,

  $F_{\text{on }q\text{'}}\text{'}$ is the force exerted on q’ .

Calculation:

Using the above formula, the electric field intensity in the 1^st row of the table will be;

  $\begin{array}{c}E=\frac{F_{\text{on }q\text{'}}}{q\text{'}}\\ =\frac{0.30}{1.0\times {10}^{-6}}\\ =3.0\times {10}^5\text{N/C}\end{array}$

And, the force in the 2^nd row of the table will be,

  $\begin{array}{c}E=\frac{F_{\text{on }q\text{'}}}{q\text{'}}\\ 3.3\times {10}^5=\frac{F_{\text{on }q\text{'}}}{2.0\times {10}^{-6}}\\ F_{\text{on }q\text{'}}=0.66\text{N}\end{array}$

And, the charge in the 3^rd row of the table will be,

  $\begin{array}{c}E=\frac{F_{\text{on }q\text{'}}}{q\text{'}}\\ 1.5\times {10}^5=\frac{0.45}{q\text{'}}\\ q\text{'}=3.0\times {10}^{-6}\text{C}\end{array}$

Conclusion:

Hence, using the formula of the electric field strength in terms of force and charge, the given table can be completed.