Chapter 20, Problem 82P

(a)

To determine

The expression for the electric potential.

Answer to Problem 82P

The expression for the electric potential is $V=\frac{kp\cos \theta }{r^2}$.

Explanation of Solution

The diagram for the system is given by figure 1.

Write the expression total potential.

$V=V_1-V_2$        (I)

Here, $V$ is the total charge and $V_1,V_2$ are the potential differences at two points.

Write the equation for potential difference at first point.

  $V_1=\frac{kq}{r_1}$        (II)

Here, $k$ is the Coulomb’s constant, $q$ is the charge and $r_1$ is the distance.

Write the equation for potential difference at second point.

  $V_2=\frac{kq}{r_2}$        (II)

Here, $k$ is the Coulomb’s constant, $q$ is the charge and $r_2$ is the distance.

For $r>>a$ write the expression for the difference between the distances.

$\left(r_2-r_1\right)\approx 2a\cos \theta$        (III)

Here. $a$ is the distance.

The distance between the points are almost same, $r_1\approx r_2$.

Write the expression for dipole.

$p=2aq$        (IV)

Conclusion:

Substitute, $\frac{kq}{r_1}$ for $V_1$, $\frac{kq}{r_2}$ for $V_2$, $2a\cos \theta$ for $\left(r_2-r_1\right)$, $r$ for $r_1$, $r$ for $r_2$, $p$ for $2aq$ in Equation (I) to find $V$.

  $\begin{array}{c}V=\left(\frac{kq}{r_1}\right)-\left(\frac{kq}{r_2}\right)\\ =\frac{kq}{r_1{r}_2}\left(r_2-r_1\right)\\ =\frac{k\left(p\cos \theta \right)}{r^2}\end{array}$

Thus, the expression for the electric potential is $V=\frac{kp\cos \theta }{r^2}$.

(b)

To determine

The radial and perpendicular component of electric field.

Answer to Problem 82P

The radial and perpendicular component of electric field is $\underset{\_}{\left(2kp\cos \theta /r^3\right)}$ and $\underset{\_}{\left(kp\sin \theta /r^3\right)}$ respectively.

Explanation of Solution

Write the expression for the radial component of electric field.

$E_r=-\frac{\partial V}{\partial r}$        (V)

Here, $E_r$ is the radial component of electric field and $\left(\partial V/\partial r\right)$ is the partial derivative of potential with respect to distance.

Write the equation for perpendicular component of electric field.

  $E_{\theta }=-\frac{1}{r}\frac{\partial V}{\partial \theta }$        (VI)

Here, $E_{\theta }$ is the perpendicular component of electric field, $\left(\partial V/\partial \theta \right)$ is the partial derivative of potential with respect to angle.

Conclusion:

Substitute, $\frac{kp\cos \theta }{r^2}$ for $V$ in Equation (V) to find $E_r$.

$\begin{array}{c}{E}_r=-\frac{\partial \left[\frac{kp\cos \theta }{r^2}\right]}{\partial r}\\ =\frac{2kp\cos \theta }{r^3}\end{array}$

Substitute, $\frac{kp\cos \theta }{r^2}$ for $V$ in Equation (VI) to find $E_{\theta }$.

$\begin{array}{c}{E}_{\theta }=-\frac{1}{r}\frac{\partial \left(\frac{kp\cos \theta }{r^2}\right)}{\partial \theta }\\ =\frac{kp\sin \theta }{r^3}\end{array}$

Thus, the radial and perpendicular component of electric field is $\underset{\_}{\left(2kp\cos \theta /r^3\right)}$ and $\underset{\_}{\left(kp\sin \theta /r^3\right)}$ respectively.

(c)

To determine

The electric field at $90°$ and $0°$.

Answer to Problem 82P

The electric field at $90°$ and $0°$ are $\underset{\_}{E_{\theta }\left(90\right)=\frac{kp}{r^3}}$.and $\underset{\_}{E_r\left(0\right)=\frac{2kp}{r^3}}$

Explanation of Solution

Write the expression for the radial component of electric field.

$E_r=\frac{2kp\cos \theta }{r^3}$        (VII)

Write the expression for the perpendicular component of electric field.

$E_{\theta }=\frac{kp\sin \theta }{r^3}$        (VIII)

Conclusion:

For $r>>a$

Substitute, $90°$ for $\theta$ in Equation (VII) and (VIII) to find $E_r,E_{\theta }$.

  $\begin{array}{c}{E}_r=\frac{2kp\cos \left(90\right)}{r^3}\\ =0\end{array}$

  $\begin{array}{c}{E}_{\theta }=\frac{kp\sin \left(90\right)}{r^3}\\ =\frac{kp}{r^3}\end{array}$

For $r>>a$

Substitute, $0°$ for $\theta$ in Equation (VII) and (VIII) to find $E_r,E_{\theta }$.

  $\begin{array}{c}{E}_r=\frac{2kp\cos \left(0\right)}{r^3}\\ =\frac{2kp}{r^3}\end{array}$

  $\begin{array}{c}{E}_{\theta }=\frac{kp\sin \left(0\right)}{r^3}\\ =0\end{array}$

The results are reasonable since the component of electric field is having finite value.

Thus, the electric field at $90°$ and $0°$ are $\underset{\_}{E_{\theta }\left(90\right)=\frac{kp}{r^3}}$.and $\underset{\_}{E_r\left(0\right)=\frac{2kp}{r^3}}$.

(d)

To determine

The electric field at $r=0$.

Answer to Problem 82P

The electric field at $r=0$ is $\underset{\_}{E_{\theta }\left(r=0\right)=\infty }$.and $\underset{\_}{E_r\left(r=0\right)=\infty }$

Explanation of Solution

Write the expression for the radial component of electric field.

$E_r=\frac{2kp\cos \theta }{r^3}$        (VII)

Write the expression for the perpendicular component of electric field.

$E_{\theta }=\frac{kp\sin \theta }{r^3}$        (VIII)

Conclusion:

Substitute, $0$ for $r$ in Equation (VII) and (VIII) to find $E_r,E_{\theta }$.

  $\begin{array}{c}{E}_r=\frac{2kp\cos \left(90\right)}{{\left(0\right)}^3}\\ =\infty \end{array}$

  $\begin{array}{c}{E}_{\theta }=\frac{kp\sin \left(90\right)}{{\left(0\right)}^3}\\ =\infty \end{array}$

The electric field at the centre of dipole is not infinite.

The results are not reasonable since the component of electric field is having infinite value.

Thus, The electric field at $r=0$ is $\underset{\_}{E_{\theta }\left(r=0\right)=\infty }$.and $\underset{\_}{E_r\left(r=0\right)=\infty }$.

(e)

To determine

The potential in Cartesian coordinate.

Answer to Problem 82P

The potential in Cartesian coordinate is $V=\frac{kpy}{{\left(x^2+y^2\right)}^{3/2}}$.

Explanation of Solution

Write the expression for the potential.

$V=\frac{kp\cos \theta }{r^2}$        (IX)

Conclusion:

Substitute, ${\left(x^2+y^2\right)}^{1/2}$ for $r$, $\frac{y}{{\left(x^2+y^2\right)}^{1/2}}$ for $\cos \theta$ in Equation (IX) to find $V$.

  $\begin{array}{c}V=\frac{kp\left[\frac{y}{{\left(x^2+y^2\right)}^{1/2}}\right]}{\left(x^2+y^2\right)}\\ =\frac{kpy}{{\left(x^2+y^2\right)}^{3/2}}\end{array}$

Thus, the potential in Cartesian coordinate is $V=\frac{kpy}{{\left(x^2+y^2\right)}^{3/2}}$.

(f)

To determine

The x and y component of electric field.

Answer to Problem 82P

The x and y component of electric field is $\underset{\_}{\left(3kpxy/{\left(x^2+y^2\right)}^{5/2}\right)}$ and $\underset{\_}{\left(kp\left(2y^2-x^2\right)/{\left(x^2+y^2\right)}^{5/2}\right)}$ respectively.

Explanation of Solution

Write the expression for the x component of electric field.

$E_x=-\frac{\partial V}{\partial x}$        (X)

Here, $E_x$ is the x component of electric field and $\left(\partial V/\partial x\right)$ is the partial derivative of potential with respect to x-distance.

Write the equation for y component of electric field.

  $E_y=-\frac{\partial V}{\partial y}$        (XI)

Here, $E_y$ is the y component of electric field, $\left(\partial V/\partial y\right)$ is the partial derivative of potential with respect to y-distance.

Conclusion:

Substitute, $\frac{kpy}{{\left(x^2+y^2\right)}^{3/2}}$ for $V$ in Equation (X) to find $E_x$.

$\begin{array}{c}{E}_x=-\frac{\partial \left(\frac{kpy}{{\left(x^2+y^2\right)}^{3/2}}\right)}{\partial x}\\ =\left(3kpxy/{\left(x^2+y^2\right)}^{5/2}\right)\end{array}$

Substitute, $\frac{kpy}{{\left(x^2+y^2\right)}^{3/2}}$ for $V$ in Equation (XI) to find $E_y$.

$\begin{array}{c}{E}_y=-\frac{\partial \left(\frac{kpy}{{\left(x^2+y^2\right)}^{3/2}}\right)}{\partial y}\\ =\left(kp\left(2y^2-x^2\right)/{\left(x^2+y^2\right)}^{5/2}\right)\end{array}$

Thus, the x and y component of electric field is $\underset{\_}{\left(3kpxy/{\left(x^2+y^2\right)}^{5/2}\right)}$ and $\underset{\_}{\left(kp\left(2y^2-x^2\right)/{\left(x^2+y^2\right)}^{5/2}\right)}$ respectively.