Chapter 6, Problem 10P

(a)

To determine

State whether the given potential satisfies Laplace’s equation or not.

Answer to Problem 10P

No, the given potential does not satisfy Laplace’s equation.

Explanation of Solution

Calculation:

Consider the given potential expression.

$V_1=3xyz+y-z^2$        (1)

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

$\nabla V^2=\frac{\partial V_1}{\partial x^2}+\frac{\partial V_1}{\partial y^2}+\frac{\partial V_1}{\partial z^2}$        (2)

Substitute Equation (1) in (2).

$\begin{array}{c}\nabla V^2=\frac{\partial \left(3xyz+y-z^2\right)}{\partial x^2}+\frac{\partial \left(3xyz+y-z^2\right)}{\partial y^2}+\frac{\partial \left(3xyz+y-z^2\right)}{\partial z^2}\\ =\frac{\partial \left(3yz\right)}{\partial x}+\frac{\partial \left(3xz+1\right)}{\partial y}+\frac{\partial \left(3xy-2z\right)}{\partial z}\\ =0+0-2\end{array}$

$\nabla V^2=-2$        (3)

Write the expression for the Laplace’s equation.

$\nabla V^2=0$        (4)

Equation (2) is not equal to Laplace’s equation in Equation (4). Hence, the given potential equation does not satisfy the Laplace’s equation.

Conclusion:

No, the given potential does not satisfy Laplace’s equation.

(b)

To determine

State whether the given potential satisfies Laplace’s equation or not.

Answer to Problem 10P

Yes, the given potential satisfies Laplace’s equation.

Explanation of Solution

Calculation:

Consider the given potential expression.

$V_2=\frac{10\sin \varphi }{\rho }$        (5)

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

$\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial V_2}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2{V}_2}{\partial {\varphi }^2}+\frac{{\partial }^2{V}_2}{\partial z^2}=0$        (6)

Substitute Equation (5) in (6).

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

Simplify the equation as follows:

$\begin{array}{l}\frac{10\sin \varphi }{{\rho }^3}-\frac{10\sin \varphi }{{\rho }^3}=0\\ 0=0\end{array}$

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

Conclusion:

Yes, the given potential satisfies Laplace’s equation.

(c)

To determine

State whether the given potential satisfies Laplace’s equation or not.

Answer to Problem 10P

No, the given potential does not satisfy Laplace’s equation.

Explanation of Solution

Calculation:

Consider the given potential expression.

$V_3=\frac{5\sin \varphi }{r}$        (7)

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

$\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial V_3}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial V_3}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2{V}_3}{\partial {\varphi }^2}=0$        (8)

Substitute Equation (7) in (8).

$\begin{array}{l}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial \left(\frac{5\sin \varphi }{r}\right)}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \left(\frac{5\sin \varphi }{r}\right)}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2\left(\frac{5\sin \varphi }{r}\right)}{\partial {\varphi }^2}=0\\ \frac{1}{r^2}\frac{\partial }{\partial r}\left(-5\sin \varphi \right)+0+\frac{1}{r^2{\sin }^2\theta }\frac{\partial \left(\frac{5\cos \varphi }{r}\right)}{\partial \varphi }=0\\ 0+0+\frac{1}{r^2{\sin }^2\theta }\left(-\frac{5\sin \varphi }{r}\right)=0\\ -\frac{5\sin \varphi }{r^3{\sin }^2\theta }\ne 0\end{array}$

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

Conclusion:

No, the given potential does not satisfy Laplace’s equation.