Chapter 6, Problem 4RQ

To determine

Choose the correct option that provides which potential does not satisfy Laplace’s equation.

Answer to Problem 4RQ

The correct option is $\underset{\_}{\left(\text{c}\right)r\cos \varphi }$.

Explanation of Solution

Calculation:

Consider the given potential expression in option (a).

$V=2x+5$        (1)

Write the expression for $\nabla V^2$.

$\nabla V^2=\frac{\partial V}{\partial x^2}+\frac{\partial V}{\partial y^2}+\frac{\partial V}{\partial z^2}$        (2)

Substitute Equation (1) in (2).

$\begin{array}{c}\nabla V^2=\frac{\partial \left(2x+5\right)}{\partial x^2}+\frac{\partial \left(2x+5\right)}{\partial y^2}+\frac{\partial \left(2x+5\right)}{\partial z^2}\\ =\frac{\partial \left(2\right)}{\partial x}+\frac{\partial \left(0\right)}{\partial y}+\frac{\partial \left(0\right)}{\partial z}\\ =0+0+0\end{array}$

$\nabla V^2=0$        (3)

Write the expression for the Laplace’s equation.

$\nabla V^2=0$        (4)

Equation (3) is equal to Laplace’s equation in Equation (4). Hence, the given potential equation satisfies the Laplace’s equation.

Consider the given potential expression in option (b).

$V=10xy$        (5)

Substitute Equation (1) in (2).

$\begin{array}{c}\nabla V^2=\frac{\partial \left(10xy\right)}{\partial x^2}+\frac{\partial \left(10xy\right)}{\partial y^2}+\frac{\partial \left(10xy\right)}{\partial z^2}\\ =\frac{\partial \left(10y\right)}{\partial x}+\frac{\partial \left(10x\right)}{\partial y}+\frac{\partial \left(0\right)}{\partial z}\\ =0+0+0\end{array}$

$\nabla V^2=0$        (6)

Equation (6) is equal to Laplace’s equation in Equation (4). Hence, the given potential equation satisfies the Laplace’s equation.

Consider the given potential expression in part (c).

$V=r\cos \varphi$        (7)

Write the expression for the Laplace’s equation in spherical coordinates.

$\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial V}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial V}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}V}{\partial {\varphi }^2}=0$        (8)

Substitute Equation (7) in (8).

$\begin{array}{l}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial \left(r\cos \varphi \right)}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \left(r\cos \varphi \right)}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2\left(r\cos \varphi \right)}{\partial {\varphi }^2}=0\\ \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\cos \varphi \right)+0+\frac{1}{r^2{\sin }^2\theta }\frac{\partial \left(-r\sin \varphi \right)}{\partial \varphi }=0\\ \frac{1}{r^2}\left(2r\cos \varphi \right)+\frac{1}{r^2{\sin }^2\theta }\left(-r\cos \varphi \right)=0\\ \frac{2}{r}\cos \varphi -\frac{\cos \varphi }{r{\sin }^2\theta }\ne 0\end{array}$

Hence, the given potential equation does not satisfy the Laplace’s equation.

Consider the given potential expression in part (d).

$V=\frac{10}{r}$        (9)

Substitute Equation (9) in (8).

$\begin{array}{l}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial \left(\frac{10}{r}\right)}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \left(\frac{10}{r}\right)}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2\left(\frac{10}{r}\right)}{\partial {\varphi }^2}=0\\ \frac{1}{r^2}\frac{\partial }{\partial r}\left(-\frac{10r^2}{r^2}\right)+0+0=0\\ 0=0\end{array}$

Hence, the given potential equation satisfies the Laplace’s equation.

Consider the given potential expression in option (e).

$V=\rho \cos \varphi +10$        (10)

Write the expression for the Laplace’s equation in cylindrical coordinates.

$\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial V}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^{2}V}{\partial {\varphi }^2}+\frac{{\partial }^{2}V}{\partial z^2}=0$        (11)

Substitute Equation (10) in (11).

$\begin{array}{l}\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial \left(\rho \cos \varphi +10\right)}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2\left(\rho \cos \varphi +10\right)}{\partial {\varphi }^2}+\frac{{\partial }^2\left(\rho \cos \varphi +10\right)}{\partial z^2}=0\\ \frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \cos \varphi \right)+\frac{1}{{\rho }^2}\frac{\partial \left(-\rho \sin \varphi \right)}{\partial \varphi }+0=0\\ \frac{1}{\rho }\left(\cos \varphi \right)-\frac{1}{{\rho }^2}\left(-\rho \cos \varphi \right)=0\\ 0=0\end{array}$

Hence, the given potential equation satisfies the Laplace’s equation.

Conclusion:

Thus, the correct option is $\underset{\_}{\left(\text{c}\right)r\cos \varphi }$.