Chapter 23, Problem 34P

(a)

To determine

The total electric potential at a large distance from the quadrupole.

Answer to Problem 34P

The total electric potential at a large distance from the quadrupole is $\frac{2kB{\cos }^2\theta }{r^3}$

Explanation of Solution

Formula used:

The expression for the electric potential due to quadrupole is given as,

  $V_{\text{quadrupole}}=V^{+}+V^{-}$

Here, $V$ is the electric potential at a large distance from dipole.

Calculation:

  

Figure (1)

The electric potential at a large distance from dipole is calculated as,

  $V=\frac{kqL\cos \theta }{r^2}$

The electric potential due to quadrupole is calculated as,

  $\begin{array}{c}{V}_{\text{quadrupole}}=V^{+}+V^{-}\\ =\frac{kqL\cos \theta }{r_1^2}-\frac{kqL\cos \theta }{r_2^2}\end{array}$

  $\therefore \left(\begin{array}{l}{r}_2-r_1=L\cos \theta \\ r_2+r_1=2r\\ r_1=r_2=r\end{array}\right)$

Substitute the values,

  $\begin{array}{c}{V}_{\text{quadrupole}}=kqL\cos \theta \left(\frac{\left(r_2^2-r_1^2\right)}{r_1^2{r}_2^2}\right)\\ =kqL\cos \theta \left(\frac{2r\left(L\cos \theta \right)}{r^4}\right)\\ =\frac{2kq{L}^2{\cos }^2\theta }{r^3}\therefore \left(B=q{L}^2\right)\\ =\frac{2kB{\cos }^2\theta }{r^3}\end{array}$

Conclusion:

Therefore, the total electric potential at a large distance from the quadrupole is $\frac{2kB{\cos }^2\theta }{r^3}$

(b)

To determine

The electric field for $Z>>L$ .

Answer to Problem 34P

The electric field for $Z>>L$ is $\overrightarrow{E}=\left(6kB/z^4\right)\widehat{k}$ .

Explanation of Solution

Formula used:

The expression for the electric potential due to quadrupole on the $z$ -axis is given as,

  $V_z=\frac{2kB{\cos }^2\theta }{r^3}$

Calculation:

On $z$ -axis the angle $\theta$ will be zero and the distance $r$ is equal to $z$ .

  $\begin{array}{c}{V}_z=\frac{2kB{\cos }^2\theta }{r^3}\\ =\frac{2kB{\cos }^2\left(0\right)}{z^3}\\ =\frac{2kB}{z^3}\end{array}$

The electric field on the positive $z$ -axis is calculated as,

  $\begin{array}{c}\overrightarrow{E_z}=-\frac{\partial }{\partial z}\left(\frac{2kB}{z^3}\right)\widehat{k}\\ =-\left(2kB\right)\frac{\partial }{\partial z}\left(z^{-3}\right)\widehat{k}\\ =\left(\frac{6kB}{z^4}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the electric field for $Z>>L$ is $\overrightarrow{E}=\left(6kB/z^4\right)\widehat{k}$ .

(c)

To determine

The result of part (b) will be obtained by the addition of three point charges.

Answer to Problem 34P

The result of part (b) will be obtained by the addition of three point charges and the magnitude of the electric field is $\overrightarrow{E}=\left(6kB/z^4\right)\widehat{k}$ .

Explanation of Solution

Formula used:

The expression for the net electric field on the positive $z$ -axis is given as,

  $\overrightarrow{E}=\left(\frac{kq}{{\left(z-L\right)}^2}\right)\widehat{k}+\left(\frac{k\left(-2q\right)}{z^2}\right)\widehat{k}+\left(\frac{kq}{\left(z+L^2\right)}\right)\widehat{k}$

Calculation:

The net electric field on the positive $z$ -axis is calculated as,

  $\begin{array}{c}\overrightarrow{E}=\left(\frac{kq}{{\left(z-L\right)}^2}\right)\widehat{k}+\left(\frac{k\left(-2q\right)}{z^2}\right)\widehat{k}+\left(\frac{kq}{\left(z+L^2\right)}\right)\widehat{k}\\ =kq\left[\frac{1}{{\left(z-L\right)}^2}-\frac{2}{z^2}+\frac{1}{{\left(z+L\right)}^2}\right]\widehat{k}\\ =kq\left[\frac{1}{z^2{\left(1-\frac{L}{z}\right)}^2}-\frac{2}{z^2}+\frac{1}{z^2{\left(1+\frac{L}{z}\right)}^2}\right]\widehat{k}\\ =\frac{kq}{z^2}\left[{\left(1-\frac{L}{z}\right)}^{-2}-2+{\left(1+\frac{L}{Z}\right)}^{-2}\right]\widehat{k}\end{array}$ ....... (1)

By using binomial theorem,

  $\begin{array}{c}{\left(1-\frac{L}{z}\right)}^{-2}=1+2\frac{L}{z}+3{\left(\frac{L}{z}\right)}^2+\mathrm{.............}\\ {\left(1+\frac{L}{z}\right)}^{-2}=1-2\frac{L}{z}+3{\left(\frac{L}{z}\right)}^2+\mathrm{.............}\end{array}$

Substitute the binomial equation in equation (1)

  $\begin{array}{c}\overrightarrow{E}=\frac{kq}{z^2}\left[\left(1+2\frac{L}{z}+3{\left(\frac{L}{z}\right)}^2\right)-2+\left(1-2\frac{L}{z}+3{\left(\frac{L}{z}\right)}^2\right)\right]\widehat{k}\\ =\frac{kq}{z^2}\left[\frac{3L^2}{z^2}+\frac{3L^2}{z^2}\right]\widehat{k}\\ =\left(\frac{6kB}{z^4}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the result of part (b) will be obtained by the addition of three point charges and the magnitude of the electric field is $\overrightarrow{E}=\left(6kB/z^4\right)\widehat{k}$ .