Question

$x=-1.0cm$$A-2.0nC$charge and a $+2.0nC$charge are located on the x-axis at and $x=+1.0cm$, respectively.

a. Other than at infinity, is there a position or positions on the x-axis where the electric field is zero? If so, where?

b. Other than at infinity, at what position or positions on the x-axis is the electric potential zero?

c. Sketch graphs of the electric field strength and the electric potential along the x-axis.

Short answer

(a). There are no position or positions on the x-axis

(b). At the midpoint the net potential is zero.

(c). The graphs have been subemitted below.

Step-by-step solution

 Step1: Theory of  elctric potential field 

V is the electric potential, q is the charge and r is the distance

$$\begin{gathered} V=\frac{1/\left(4\pi {\epsilon }_0\right)q}{r} \\ {\epsilon }_0=8.85\times {10}^{-12}{C}^2/N\times m^2 \end{gathered}$$

On the basis of the figure

the two charges are -2.0nC and 2.0nC on the X axis position is x= -0.1 cm and +0.1cm

and q value from nano columb to coulomb is $2.0\times {10}^{-9}C$

(a)

When x is greater than 1 then the potential charge is larger and direction is positive but negative charge is smaller. both positive and negative direction is $\overline{x}$. due to this reason, there is no region and E = 0 at this position.

When x is belonged into $-1<x>+1$then the both direction is same and summation of the two points is zero.

Due to this reason, the electric field is zero and position is zero.

Defination the charge power on the midpoints.

The charge value at the midpoint for the charge $q_1$and V is the potential energy.

The charge value at the midpoint for the charge $q_1$

$V_1=\frac{k{q}_1}{r}$

The charge value at the midpoint for the charge $q_2$

$V_2=\frac{-k{q}_2}{r}$

The net charge is equal to the summation of the two charge

$\begin{array}{rcl}V& =& V_1+V_2\\ & =& \frac{k{q}_1}{r}+(-\frac{k{q}_2}{r})\\ & =& 0V\end{array}$

From the equation, there is a region but the electric field is zero. So the net potential at the midpoint is zero.

Defination the charge power of two points. 

As when the x > +1.0cm

Applying the given data

$$\begin{gathered} V=\sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{r_{x=+1.0cm}}+\frac{q_2}{r_{x=-1.0cm}}\right) \\ =2.0\times {10}^{-9}C/\left(4\pi {\epsilon }_0\right)\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \end{gathered}$$

As when the x > -1.0cm

$\begin{array}{rcl}V& =& \sum _{i}1/\left(4\pi {\epsilon }_0\right)q_i/r_i\\ V& =& \sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{r_{x=+1.0cm}}+\frac{q_2}{r_{x=-1.0cm}}\right)\\ q_1& =& 2.0\times {10}^{-9}C\\ q_2& =& -2.0\times {10}^{-9}{C}^{}\\ & & r_{x=+1.0cm}\\ & =& x+1\\ & & r_{x=-1.0cm}\\ & =& x-1\end{array}$

Applying the given data

$\begin{array}{rcl}V& =& \sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{r_{x=+1.0cm}}+\frac{q_2}{r_{x=-1.0cm}}\right)\\ & =& 2.0\times {10}^{-9}C/\left(4\pi {\epsilon }_0\right)\left(\frac{1}{x+1}+\frac{1}{1-x}\right)\\ & & \\ & & \end{array}$

As when the -1.0cm< x < +1.0cm

$$\begin{gathered} V=\sum _{i}1/\left(4\pi {\epsilon }_0\right)q_i/r_i \\ V=\sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{r_{x=+1.0cm}}+\frac{q_2}{r_{x=-1.0cm}}\right) \\ {q}_1=2.0\times {10}^{-9}C \\ {q}_2=-2.0\times {10}^{-9}C \\ {r}_{x=+1.0cm}=1-x \\ {r}_{x=-1.0cm}=x+1 \end{gathered}$$

Applying the given data

$$\begin{gathered} V=\sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{r_{x=+1.0cm}}+\frac{q_2}{r_{x=-1.0cm}}\right) \\ =2.0\times {10}^{-9}C/\left(4\pi {\epsilon }_0\right)\left(\frac{1}{1-x}+\frac{1}{x+1}\right) \end{gathered}$$

As when the x > +1 cm. Assume a point that has distance from origin at x distance

$\begin{array}{rcl}E& =& \sum _{i}1/\left(4\pi {\epsilon }_0\right)q_i/{r^2}_i\\ & & \\ E& =& \sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{{r_{x=+1.0cm}}^2}+\frac{q_2}{{r_{x=-1.0cm}}^2}\right)\\ q_1& =& 2.0\times {10}^{-9}C\\ q_2& =& -2.0\times {10}^{-9}C\\ r_{x=+1.0cm}& =& x-1\\ r_{x=-1.0cm}& =& x+1\end{array}$

Applying the given data

$E=2.0\times {10}^{-9}C/\left(4\pi {\epsilon }_0\right)\left(\frac{1}{{\left(x-1\right)}^2}-\frac{1}{{\left(x+1\right)}^2}\right)$

As when the x < -1 cm. Assume a point that has from origin at x distance

$\begin{array}{rcl}E& =& \sum _{i}1/\left(4\pi {\epsilon }_0\right)q_i/{r^2}_i\\ E& =& \sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{{r_{x=+1.0cm}}^2}+\frac{q_2}{{r_{x=-1.0cm}}^2}\right)\\ q_1& =& 2.0\times {10}^{-9}C\\ q_2& =& -2.0\times {10}^{-9}C\\ r_{x=+1.0cm}& =& x+1\\ r_{x=-1.0cm}& =& x-1\end{array}$

Applying the given data

$E=2.0\times {10}^{-9}C/\left(4\pi {\epsilon }_0\right)\left(\frac{1}{{\left(x+1\right)}^2}+\frac{1}{{\left(1-x\right)}^2}\right)$

As when the -1<x < +1 cm. Assume a point that has distance from origin at x distance

$$\begin{gathered} E=\sum _{i}1/\left(4\pi {\epsilon }_0\right)q_i/{r^2}_i \\ E=\sum _{i}1/\left(4\pi {\epsilon }_0\right)q\left(\frac{q_1}{{r_{x=+1.0cm}}^2}+\frac{q_2}{{r_{x=-1.0cm}}^2}\right) \\ {q}_1=2.0\times {10}^{-9}C \\ {q}_2=-2.0\times {10}^{-9}C \\ {r}_{x=+1.0cm}=1-x \\ {r}_{x=-1.0cm}=x+1 \end{gathered}$$

Applying the given data

$E=2.0\times {10}^{-9}C/\left(4\pi {\epsilon }_0\right)\times \left(\frac{1}{{\left(1-x\right)}^2}+\frac{1}{{\left(1+x\right)}^2}\right)$