Chapter 16, Problem 56QAP

To determine

(a)

The electric field at a distance $x=0.06m$

Answer to Problem 56QAP

Net electric field at ( $x=0.06m$ ) is $6.87\times {10}^7\text{N/C}$

Explanation of Solution

Given:

Charge 1, $q_1=+5\times {10}^{-6}\text{C}$

Charge 2, $q_2=-10\times {10}^{-6}\text{C}$

Distance of the point from charge 1, $r_1=0.06m$

Distance of the point from charge 2, $r_2=0.04m$

  

Formula used:

Electric field at a distance x from the charge q.

  $\begin{array}{l}{E}_x=\frac{k\left|q\right|}{r^2}\\ \text{Where,}\\ r\text{=}\text{distance}\text{form}\text{the}\text{charge}\text{q}\\ \text{constant,}k\text{=}8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\end{array}$

Calculation:

  $\begin{array}{l}{E}_{1x}=\frac{k\left|q_1\right|}{r_1^2}=\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|5.00\times {10}^{-6}\text{C}\right|}{{(0.060\text{m})}^2}=1.249\times {10}^7\frac{\text{N}}{\text{C}}\\ E_{2x}=\frac{k\left|q_2\right|}{r_2^2}=\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|-10.0\times {10}^{-6}\text{C}\right|}{{(0.040)}^2}=5.619\times {10}^7\frac{\text{N}}{\text{C}}\end{array}$

Net electric field is the sum of the electric field due to both charges.

  $\begin{array}{l}{E}_x=E_{1x}+E_{2x}\\ =1.249\times {10}^7\frac{\text{N}}{\text{C}}+5.619\times {10}^7\frac{\text{N}}{\text{C}}\\ =6.87\times {10}^7\frac{\text{N}}{\text{C}}\end{array}$

Conclusion:

Net electric field at ( $x=0.06m$ ) is $6.87\times {10}^7\text{N/C}$

To determine

(b)

The point on the x -axis at which the electric field is zero.

Answer to Problem 56QAP

The point where E = 0 is at distance $x=-24.1\text{ cm}$.

Explanation of Solution

Given:

Charge 1, $q_1=+5\times {10}^{-6}\text{C}$

Charge 2, $q_2=-10\times {10}^{-6}\text{C}$

Distance of the point from charge (1) is x

Distance of the point from charge (2) is $\left(0.1-x\right)$

  

Formula used:

The electric field at a distance x from the charge q.

  $\begin{array}{l}{E}_x=\frac{k\left|q\right|}{r^2}\\ \text{Where,}\\ r\text{=}\text{distance}\text{form}\text{the}\text{charge}q\\ \text{constant,}k\text{=}8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\end{array}$

Calculation:

  $\begin{array}{l}E=E_1+E_2=\left(\frac{k{q}_1}{x^2}\right)+\left(\frac{k{q}_2}{{(x-0.100)}^2}\right)=0\\ \frac{q_1}{x^2}+\frac{q_2}{{(x-0.100)}^2}=0\\ \frac{-5.00\times {10}^{-6}}{x^2}+\frac{10.00\times {10}^{-6}}{{(x-0.100)}^2}=0\\ 5x^2+x-0.05=0\\ x=\frac{-1\pm \sqrt{1^2-4(5)(-0.05)}}{2(5)}\\ x=-0.241m,+0.0414m\end{array}$

The point where E = 0 must lie to the left of both charges, where it is closer to the smaller charge

Conclusion:

The point where E = 0 is at distance $x=-24.1\text{ cm}$.