Chapter 16, Problem 28P

To determine

To Find: Electric field midway point

Answer to Problem 28P

  $8.44\times {10}^7\text{N}/\text{C}$

Explanation of Solution

Given:

Electric field midway point between $-8.0\mu \text{C}$ and a $+7.0\mu \text{C}$ charges which are $8.0\text{cm}$

Formula Used:

Below sketch shows direction of electric field at a point mid away between the charges

  

The vector sum of the fields due to by $q_1$ and $q_2$ is the value of the electric field at the midpoint between $q_1$ and $q_2$ . Thus, both fields point to the left as shown in above the sketch.

The electric at a point mid-way between the two charges is given by a relation as,

  $\begin{array}{l}E=\frac{k\left|q_1\right|}{{\left(\frac{d}{2}\right)}^2}+\frac{k\left|q_2\right|}{{\left(\frac{d}{2}\right)}^2}\\ =\frac{k}{{\left(\frac{d}{2}\right)}^2}\left[\left|q_1\right|+\left|q_2\right|\right]\end{array}$

Here, $d$ is distance between the charges, $k$ is electro-static constant $\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)$

Calculations:

Conversion of units of charges from micro coulomb to coulomb as

  $\begin{array}{l}{q}_1=-8.0\mu \text{C}\\ =(-8.0\mu \text{C})\left(\frac{{10}^{-6}\text{C}}{\mu \text{C}}\right)\\ =-8.0\times {10}^{-6}\text{C}\\ q_2=+7.0\mu \text{C}\\ =(+7.0\mu \text{C})\left(\frac{{10}^{-6}\text{C}}{\mu \text{C}}\right)\\ =+7.0\times {10}^{-6}\text{C}\end{array}$

Conversion of units of distance between the charges from $\text{cm}$ to $\text{m}$ as, $d=8\text{cm}$

  $=(8\text{cm})\left(\frac{0.01\text{m}}{\text{cm}}\right)$

  $=0.08\text{m}$

To obtain electric field mid-way between the two charges, substitute $+7.0\times {10}^{-6}\text{C}$ for $q_1$

  $-8.0\times {10}^{-6}\text{C}$ for $q_2$ ,and $0.08\text{m}$ for $d,\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)$ for $k$ in the equation

  $\begin{array}{l}E=\frac{k}{{\left(\frac{d}{2}\right)}^2}\left[\left|q_1\right|+\left|q_2\right|\right]\\ =\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)}{{\left(\frac{0.08\text{m}}{2}\right)}^2}\left[7\times {10}^{-6}\text{C}+8\times {10}^{-6}\text{C}\right]\\ =8.44\times {10}^7\text{N}/\text{C}\end{array}$

Conclusion:

Electric field midway between the charges is $8.44\times {10}^7\text{N}/\text{C}$ towards $q_2$ .