Chapter 16, Problem 90QAP

To determine

(a)

An expression for the electric field in the region x>0.

Answer to Problem 90QAP

An expression for the electric field in the region x>0 is, $E=kq\left[\frac{2}{x^2}-\frac{1}{{\left(x+a\right)}^2}\right]$

Explanation of Solution

Given:

Charge, $q_1=+2q$

Charge, $q_2=-q$

Distance, r

Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=Distance

Calculation:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

As charge 1 is placed at origin and charge 2 is placed at a distance a from origin. So, charge 1 is at distance x from reference point and charge 2 is at distance x+a from reference point.
  $\begin{array}{l}{E}_1=\frac{q_1}{4\pi {\epsilon }_0{r}_1^2}\\ E_2=\frac{q_2}{4\pi {\epsilon }_0{r}_2^2}\\ E=E_1+E_2\\ E=\frac{+2q}{4\pi {\epsilon }_0{x}^2}+\frac{-q}{4\pi {\epsilon }_0{\left(x+a\right)}^2}\\ E=kq\left[\frac{2}{x^2}-\frac{1}{{\left(x+a\right)}^2}\right]\end{array}$

To determine

(b)

An expression for the electric field in the region 0

Answer to Problem 90QAP

An expression for the electric field in the region 0E=kq[2x2−1( a−x)2]

Explanation of Solution

Given:

Charge, $q_1=+2q$

Charge, $q_2=-q$

Distance, r
Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=DistanceCalculation:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Thus,
  $\begin{array}{l}{E}_1=\frac{q_1}{4\pi {\epsilon }_0{r}_1^2}\\ E_2=\frac{q_2}{4\pi {\epsilon }_0{r}_2^2}\\ E=E_1+E_2\\ E=\frac{+2q}{4\pi {\epsilon }_0{x}^2}+\frac{-q}{4\pi {\epsilon }_0{\left(a-x\right)}^2}\\ E=kq\left[\frac{2}{x^2}-\frac{1}{{\left(a-x\right)}^2}\right]\end{array}$

To determine

(c)

An expression for the electric field in the region x>a.

Answer to Problem 90QAP

An expression for the electric field in the region x>a is, $E=kq\left[\frac{2}{x^2}-\frac{1}{{\left(x-a\right)}^2}\right]$

Explanation of Solution

Given:

Charge, $q_1=+2q$

Charge, $q_2=-q$

Distance, r

Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=Distance

Calculation:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Thus,
  $\begin{array}{l}{E}_1=\frac{q_1}{4\pi {\epsilon }_0{r}_1^2}\\ E_2=\frac{q_2}{4\pi {\epsilon }_0{r}_2^2}\\ E=E_1+E_2\\ E=\frac{+2q}{4\pi {\epsilon }_0{x}^2}+\frac{-q}{4\pi {\epsilon }_0{\left(x-a\right)}^2}\\ E=kq\left[\frac{2}{x^2}-\frac{1}{{\left(x-a\right)}^2}\right]\end{array}$

To determine

(d)

The point at which the electric field is zero.

Answer to Problem 90QAP

At $r_1=\sqrt{2}{r}_2$ the electric field is zero.

Explanation of Solution

Given:

Charge, $q_1=+2q$

Charge, $q_2=-q$

Distance, r

Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=Distance

Calculation:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Thus,
  $\begin{array}{l}{E}_1=\frac{q_1}{4\pi {\epsilon }_0{r}_1^2}\\ E_2=\frac{q_2}{4\pi {\epsilon }_0{r}_2^2}\\ E=E_1+E_2\\ 0=\frac{+2q}{4\pi {\epsilon }_0{r}_1^2}+\frac{-q}{4\pi {\epsilon }_0{r}_2^2}\\ \frac{2q}{4\pi {\epsilon }_0{r}_1^2}=\frac{q}{4\pi {\epsilon }_0{r}_2^2}\\ \frac{2}{r_1^2}=\frac{1}{r_2^2}\\ r_1^2=2r_2^2\\ r_1=\sqrt{2}{r}_2\end{array}$.

To determine

(e)

To plot:

The graph of E_xverses x.

Explanation of Solution

Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=Distance

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

  

To determine

(f)

An expression for the electric field in the region -8

Answer to Problem 90QAP

An expression for the electric field in the region -8E=0N/C

Explanation of Solution

Given:

Charge, $q_1=+2q$

Charge, $q_2=-q$

Distance, r

Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=Distance

Calculation:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Thus,
  $\begin{array}{l}{E}_1=\frac{q_1}{4\pi {\epsilon }_0{r}_1^2}\\ E_2=\frac{q_2}{4\pi {\epsilon }_0{r}_2^2}\\ E=E_1+E_2\\ E=\frac{+2q}{4\pi {\epsilon }_0{\left(-\infty \right)}^2}+\frac{-q}{4\pi {\epsilon }_0{\left(\infty -a\right)}^2}\\ E=0\text{N/C}\end{array}$.