Chapter 13.2, Problem 13.82P

To determine

(a)

$x$, $y$ and $z$ components of $P$.

Answer to Problem 13.82P

$x$ component:

$P_x=x{\left(x^2+y^2+z^2\right)}^{-1/2}$

$y$ component:

$P_y=y{\left(x^2+y^2+z^2\right)}^{-1/2}$

$z$ component:

$P_z=z{\left(x^2+y^2+z^2\right)}^{-1/2}$

Explanation of Solution

Given:

$V\left(x,y,z\right)=-{\left(x^2+y^2+z^2\right)}^{1/2}$

$x$ component:

$P_x=-\frac{\partial V}{\partial x}$

$P_x=-\frac{\partial \left[-{\left(x^2+y^2+z^2\right)}^{1/2}\right]}{\partial x}$

$P_x=x{\left(x^2+y^2+z^2\right)}^{-1/2}$

$y$ component:

$P_y=-\frac{\partial V}{\partial y}$

$P_y=-\frac{\partial \left[-{\left(x^2+y^2+z^2\right)}^{1/2}\right]}{\partial y}$

$P_y=y{\left(x^2+y^2+z^2\right)}^{-1/2}$

$z$ component:

$P_z=-\frac{\partial V}{\partial z}$

$P_z=-\frac{\partial \left[-{\left(x^2+y^2+z^2\right)}^{1/2}\right]}{\partial z}$

$P_z=z{\left(x^2+y^2+z^2\right)}^{-1/2}$

To determine

(b)

That the work done by $P$ is equal to the negative of change in potential from O to D.

Answer to Problem 13.82P

Work done by $P$ is $U_{OABD}=-\Delta V_{OD}.$

Explanation of Solution

Given:

$V\left(x,y,z\right)=-{\left(x^2+y^2+z^2\right)}^{1/2}$

$U_{OABD}=U_{OA}+U_{AB}+U_{BD}$

O − A : $P_y$ and $P_x$ are perpendicular to O − A and do no work.

Also, on O − A

$x=y=0$

And

$P_z=1$

Thus,

$U_{O-A}=\underset{0}{\overset{a}{\int }}{P}_{z}dz=\underset{0}{\overset{a}{\int }}dz=a$

A − B : $P_z$ and $P_y$ are perpendicular to A − B and do no work.

Also, on A − B

$y=0,z=a$

And,

$P_x=\frac{x}{{\left(x^2+a^2\right)}^{1/2}}$

Thus,

$U_{A-B}=\underset{0}{\overset{a}{\int }}\frac{xdx}{{\left(x^2+a^2\right)}^{1/2}}=a\left(\sqrt{2}-1\right)$

B − D : $P_x$ and $P_z$ are perpendicular to B − D and do no work.

Also, on B − D

$x=a,z=a$

And

$P_y=\frac{y}{{\left(y^2+2a^2\right)}^{1/2}}$

Thus,

$U_{B-D}=\underset{0}{\overset{a}{\int }}\frac{y}{{\left(y^2+2a^2\right)}^{1/2}}dy$

$U_{B-D}={{\left(y^2+2a^2\right)}^{1/2}|}_0^a$

$U_{B-D}={\left(a^2+2a^2\right)}^{1/2}-{\left(2a^2\right)}^{1/2}$

$U_{B-D}=a\left(\sqrt{3}-\sqrt{2}\right)$

$U_{OABD}=U_{OA}+U_{AB}+U_{BD}$

$U_{OABD}=a+a\left(\sqrt{2}-1\right)+a\left(\sqrt{3}-\sqrt{2}\right)$

$U_{OABD}=a\sqrt{3}$

$\Delta V_{OD}=V\left(a,a,a\right)-V\left(0,0,0\right)$

$\Delta V_{OD}=-{\left(a^2+a^2+a^2\right)}^{1/2}-0$

$\Delta V_{OD}=-a\sqrt{3}$

Thus,

$U_{OABD}=-\Delta V_{OD}$