Chapter 24, Problem 48CP

An electric dipole is located along the y axis as shown in Figure P24.48. The magnitude of its electric dipole moment is defined as p = 2aq. (a) At a point P, which is far from the dipole (r ≫ a), show that the electric potential is

$V=\frac{k_{e}p\cos \theta }{r^2}$

(b) Calculate the radial component E_r and the perpendicular component E_θ of the associated electric field. Note that E_θ = −(1/r)(∂V/∂θ). Do these results seem reasonable for (c) θ = 90° and 0°? (d) For r = 0? (e) For the dipole arrangement shown in Figure P24.48, express V in terms of Cartesian coordinates using r = (x² + y²)^1/2 and

$\cos \theta =\frac{y}{{(x^2+y^2)}^{1/2}}$

(f) Using these results and again taking r ≫ a, calculate the field components E_x and E_y.

Figure P24.48

To determine

To show: The electric potential at a point P is $V=\frac{k_{\text{e}}p\cos \theta }{r^2}$.

Answer to Problem 48CP

The charge on the insulating sphere is $-4.01\times {10}^{-9}\text{C}$.

Explanation of Solution

Given info: The magnitude of the electric dipole moment is $p=2aq$, the radial component is $E_{\text{r}}$ and perpendicular component is $E_{\text{θ}}$. and $E_{\text{θ}}=-\left(\frac{1}{r}\right)\left(\frac{\partial V}{\partial \theta }\right)$.

The expression to calculate the total electric potential of the dipole is,

$\begin{array}{c}V=\frac{k_{\text{e}}q}{r_1}-\frac{k_{\text{e}}q}{r_2}\\ =\frac{k_{\text{e}}q}{r_1{r}_2}\left(r_2-r_1\right)\end{array}$

Since, $r>>a$, $r_2-r_1=2a\cos \theta$.

Substitute $2a\cos \theta$ for $r_2-r_1$ in above equation.

$V=\frac{k_{\text{e}}q}{r_1{r}_2}\left(2a\cos \theta \right)$

Substitute $p$ for $2aq$ in above equation.

$V=\frac{k_{\text{e}}p\cos \theta }{r^2}$

Conclusion:

Therefore, the electric potential at a point P is $V=\frac{k_{\text{e}}p\cos \theta }{r^2}$.

To determine

The radial component $E_{\text{r}}$ and perpendicular $E_{\text{θ}}$.

Answer to Problem 48CP

The radial component is $\frac{2k_{\text{e}}p\cos \theta }{r^3}$ and perpendicular component is $\frac{k_{\text{e}}p\sin \theta }{r^3}$.

Explanation of Solution

Given info: The magnitude of the electric dipole moment is $p=2aq$, the radial component is $E_{\text{r}}$ and perpendicular component is $E_{\text{θ}}$. and $E_{\text{θ}}=-\left(\frac{1}{r}\right)\left(\frac{\partial V}{\partial \theta }\right)$.

Write the expression of radial component $E_{\text{r}}$.

$E_{\text{r}}=-\frac{\partial V}{\partial r}$

Substitute $\frac{k_{\text{e}}p\cos \theta }{r^2}$ for $V$ in above equation.

$\begin{array}{c}{E}_{\text{r}}=-\frac{\partial }{\partial r}\left(\frac{k_{\text{e}}p\cos \theta }{r^2}\right)\\ =\frac{2k_{\text{e}}p\cos \theta }{r^3}\end{array}$

Write the expression to calculate the and perpendicular component $E_{\text{θ}}$.

$E_{\text{θ}}=-\frac{1}{r}\left(\frac{\partial V}{\partial r}\right)$

Substitute $\frac{k_{\text{e}}p\cos \theta }{r^2}$ for $V$ in above equation.

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

Thus, the radial component is $\frac{2k_{\text{e}}p\cos \theta }{r^3}$ and perpendicular component is $\frac{k_{\text{e}}p\sin \theta }{r^3}$.

Conclusion:

Therefore, radial component is $\frac{2k_{\text{e}}p\cos \theta }{r^3}$ and perpendicular component is $\frac{k_{\text{e}}p\sin \theta }{r^3}$.

To determine

Whether the values of $E_r$ and $E_{\text{θ}}$ are reasonable or not.

Answer to Problem 48CP

These values of radial component and perpendicular component of electric field seem reasonable.

Explanation of Solution

Given info: The magnitude of the electric dipole moment is $p=2aq$, the radial component is $E_{\text{r}}$ and perpendicular component is $E_{\text{θ}}$. and $E_{\text{θ}}=-\left(\frac{1}{r}\right)\left(\frac{\partial V}{\partial \theta }\right)$.

The radial component of electric filed at point P is,

$E_{\text{r}}=\frac{2k_{\text{e}}p\cos \theta }{r^3}$

Substitute $90°$ for $\theta$ in above equation.

$\begin{array}{c}{E}_{\text{r}}=\frac{2k_{\text{e}}p\cos 90°}{r^3}\\ =0\end{array}$

Substitute $0$ for $\theta$ in above equation.

$\begin{array}{c}{E}_{\text{r}}=\frac{2k_{\text{e}}p\cos 0°}{r^3}\\ =\frac{2k_{\text{e}}p}{r^3}\end{array}$

The perpendicular component of electric field at point P is,

$E_{\text{θ}}=\frac{k_{\text{e}}p\sin \theta }{r^3}$

Substitute $90°$ for $\theta$ in above equation.

$\begin{array}{c}{E}_{\text{θ}}=\frac{k_{\text{e}}p\sin 90°}{r^3}\\ =\frac{k_{\text{e}}p}{r^3}\end{array}$

Substitute $0$ for $\theta$ in above equation.

$\begin{array}{c}{E}_{\text{θ}}=\frac{k_{\text{e}}p\sin 0°}{r^3}\\ =0\end{array}$

Thus, these values of radial component and perpendicular component of electric field seem reasonable.

Conclusion:

Therefore, these values of radial component and perpendicular component of electric field seem reasonable.

To determine

Whether the values of $E_r$ and $E_{\text{θ}}$ are reasonable or not at $r\to 0$.

Answer to Problem 48CP

The value of radial component and perpendicular component of electric field at $r\to 0$ does not seem reasonable.

Explanation of Solution

Given info: The magnitude of the electric dipole moment is $p=2aq$, the radial component is $E_{\text{r}}$ and perpendicular component is $E_{\text{θ}}$. and $E_{\text{θ}}=-\left(\frac{1}{r}\right)\left(\frac{\partial V}{\partial \theta }\right)$.

The magnitude of the electric field between the charges of the dipole is not infinite.

The radial component of electric filed at point P is,

$E_{\text{r}}=\frac{2k_{\text{e}}p\cos \theta }{r^3}$ (1)

The perpendicular component of electric field at point P is,

$E_{\text{θ}}=\frac{k_{\text{e}}p\sin \theta }{r^3}$ (2)

The values of $E_{\text{r}}$ and $E_{\text{θ}}$ at $r\to 0$ is tends to infinite which is not possible.

Conclusion:

Therefore, the value of radial component and perpendicular component of electric field at $r\to 0$ does not seem reasonable.

To determine

The electric potential of dipole in terms of Cartesian coordinates.

Answer to Problem 48CP

The electric potential of dipole in terms of Cartesian coordinates is $\frac{k_{\text{e}}py}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}$.

Explanation of Solution

Given info: The magnitude of the electric dipole moment is $p=2aq$, the radial component is $E_{\text{r}}$ and perpendicular component is $E_{\text{θ}}$. and $E_{\text{θ}}=-\left(\frac{1}{r}\right)\left(\frac{\partial V}{\partial \theta }\right)$.

The electric potential at a point P is,

$V=\frac{k_{\text{e}}p\cos \theta }{r^2}$. (3)

If $r_1\approx r_2\approx r$ then,

$r={\left(x^2+y^2\right)}^{\frac{1}{2}}$ and $\cos \theta =\frac{y}{{\left(x^2+y^2\right)}^{\frac{1}{2}}}$

Substitute ${\left(x^2+y^2\right)}^{\frac{1}{2}}$ for $r$ and $\frac{y}{{\left(x^2+y^2\right)}^{\frac{1}{2}}}$ for $\cos \theta$ in equation (3).

$\begin{array}{c}V=\frac{k_{\text{e}}p\frac{y}{{\left(x^2+y^2\right)}^{\frac{1}{2}}}}{{\left({\left(x^2+y^2\right)}^{\frac{1}{2}}\right)}^2}\\ =\frac{k_{\text{e}}py}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}\end{array}$

Conclusion:

Therefore, the electric potential of dipole in terms of Cartesian coordinates is $\frac{k_{\text{e}}py}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}$.

To determine

The field component $E_{\text{x}}\text{and}{E}_{\text{y}}$.

Answer to Problem 48CP

The field component $E_{\text{x}}\text{and}{E}_{\text{y}}$ are $\frac{3k_{\text{e}}pxy}{{\left(x^2+y^2\right)}^{\frac{5}{2}}}$ and $\frac{k_{\text{e}}p\left(2y^2-x^2\right)}{{\left(x^2+y^2\right)}^{\frac{5}{2}}}$ respectively.

Explanation of Solution

Given info: The magnitude of the electric dipole moment is $p=2aq$, the radial component is $E_{\text{r}}$ and perpendicular component is $E_{\text{θ}}$. and $E_{\text{θ}}=-\left(\frac{1}{r}\right)\left(\frac{\partial V}{\partial \theta }\right)$.

The x component of the electric field is,

$E_{\text{x}}=-\frac{\partial V}{\partial x}$

Substitute $\frac{k_{\text{e}}py}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}$ for $V$ in above equation.

$\begin{array}{c}{E}_{\text{x}}=-\frac{\partial }{\partial x}\frac{k_{\text{e}}py}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}\\ =\frac{3k_{\text{e}}pxy}{{\left(x^2+y^2\right)}^{\frac{5}{2}}}\end{array}$\

The y component of the electric field is,

$E_{\text{y}}=-\frac{\partial V}{\partial y}$

Substitute $\frac{k_{\text{e}}py}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}$ for $V$ in above equation.

$\begin{array}{c}{E}_{\text{y}}=-\frac{\partial }{\partial y}\frac{k_{\text{e}}py}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}\\ =\frac{k_{\text{e}}p\left(2y^2-x^2\right)}{{\left(x^2+y^2\right)}^{\frac{5}{2}}}\end{array}$

Conclusion:

Therefore, the field component $E_{\text{x}}\text{and}{E}_{\text{y}}$ are $\frac{3k_{\text{e}}pxy}{{\left(x^2+y^2\right)}^{\frac{5}{2}}}$ and $\frac{k_{\text{e}}p\left(2y^2-x^2\right)}{{\left(x^2+y^2\right)}^{\frac{5}{2}}}$ respectively.