Chapter 4, Problem 4.1P

Given E = E_xa_x + E_ya_y + E_z3_z V/m, where E_X, E_y, and E_z are constants, determine the incremental work required to move charge q through a distance 6: (a) along the positive x axis, (b) in a, direction at 45 degrees from the x axis in the first quadrant; (c) along a line in the first octant having equal x, y, and z components, and moving away from the origin.

To determine

(a)

The value of incremental work along positive $x$ axis.

Answer to Problem 4.1P

The value of incremental work along positive $x$ axis is $-q{E}_x\delta$.

Explanation of Solution

Given:

The expression for electric field is $E=\left(E_x{a}_x+E_y{a}_y+E_z{a}_z\right)\text{V/m}$.

Value of $E_x$ , $E_y$ and $E_z$ are constant.

Concept used:

Write the expression for incremental work.

   $w=-Q{\displaystyle {\int }_a^{b}E\cdot dl}$ ...... (1)

Here, $w$ is incremental work, $Q$ is the value of charge, $a$ is the initial value of length, $b$ is the final value of length and $E$ is electric field.

Calculation:

To calculate incremental work along $x$ axis, component of length along $y$ axis and $z$ axis is zero.

So value of $dl$ is $dx{a}_x$.

Substitute $q$ for $Q$ , $\left(E_x{a}_x+E_y{a}_y+E_z{a}_z\right)$ for $E$ , $dx{a}_x$ for $dl$ , $0$ for $a$ and $\delta$ for $b$ in equation (1).

   $\begin{array}{c}w=-q{\displaystyle {\int }_0^{\delta }\left(E_x{a}_x+E_y{a}_y+E_z{a}_z\right)\cdot dx{a}_x}\\ =-q{\displaystyle {\int }_0^{\delta }{E}_{x}dx}\\ =-q{E}_x{\left(x\right)}_0^{\delta }\\ =-q{E}_x\left(\delta -0\right)\end{array}$

Simplify further.

   $w=-q{E}_x\delta$

Conclusion:

Thus,value of incremental work along positive $x$ axis is $-q{E}_x\delta$.

To determine

(b)

The value of incremental work in direction at $45°$ from $x$ axis in first quadrant.

Answer to Problem 4.1P

The value of incremental work in direction at $45°$ from $x$ axis in first quadrant is

   $-q\frac{\delta }{\sqrt{2}}\left(E_x+E_y\right)$.

Explanation of Solution

Concept used:

Write the expression for incremental work in $xy$ plane.

   $w=-Q{\displaystyle {\int }_a^{b}E\cdot \left(dx{a}_x+dy{a}_y\right)}$ ...... (2)

Here, $w$ is incremental work, $Q$ is the value of charge, $a$ is the initial value of length, $b$ is the final value of length, $a_x$ is the unit vector in $x$ direction, $a_y$ is the unit vector in $y$ direction and $E$ is electric field.

Write the expression for component of length in $x$ axis.

   $x=l\cos \theta$ ...... (3)

Here, $x$ is the value of component of length in $x$ axis, $l$ is the value of length and $\theta$ is the value of angle.

Write the expression for component of length in $y$ axis.

   $y=l\sin \theta$ ...... (4)

Here, $y$ is the value of component of length in $y$ axis.

Calculation:

Substitute $\delta$ for $l$ and $45°$ for $\theta$ in equation (3).

   $\begin{array}{c}x=\delta \cos 45°\\ =\delta \left(\frac{1}{\sqrt{2}}\right)\\ =\frac{\delta }{\sqrt{2}}\end{array}$

So value of component of length in $x$ axis is $\frac{\delta }{\sqrt{2}}$.

Substitute $\delta$ for $l$ and $45°$ for $\theta$ in equation (4).

   $\begin{array}{c}y=\delta \sin 45°\\ =\delta \left(\frac{1}{\sqrt{2}}\right)\\ =\frac{\delta }{\sqrt{2}}\end{array}$

So value of component of length in $y$ axis is $\frac{\delta }{\sqrt{2}}$.

Substitute $q$ for $Q$ and $\left(E_x{a}_x+E_y{a}_y+E_z{a}_z\right)$ for $E$ in equation (2).

   $\begin{array}{c}w=-q{\displaystyle {\int }_a^b\left(E_x{a}_x+E_y{a}_y+E_z{a}_z\right)\cdot \left(dx{a}_x+dy{a}_y\right)}\\ =-q{\displaystyle {\int }_a^b{E}_{x}dx+E_{y}dy}\end{array}$

Simplify further.

   $w=-q{\displaystyle {\int }_a^b{E}_{x}dx}-q{\displaystyle {\int }_c^d{E}_{y}dy}$

Here, $a$ is the initial value of length in $x$ axis, $b$ is the final value of length in $x$ axis, $c$ is the initial value of length in $y$ axis and $d$ is the final value of length in $y$ axis.

Substitute $0$ for $a$ , $\frac{\delta }{\sqrt{2}}$ for $b$ , $0$ for $c$ and $\frac{\delta }{\sqrt{2}}$ for $d$ in above equation.

   $\begin{array}{c}w=-q{\displaystyle {\int }_0^{\frac{\delta }{\sqrt{2}}}{E}_{x}dx}-q{\displaystyle {\int }_0^{\frac{\delta }{\sqrt{2}}}{E}_{y}dy}\\ =-q{E}_x{\left(x\right)}_0^{\frac{\delta }{\sqrt{2}}}-q{E}_y{\left(y\right)}_0^{\frac{\delta }{\sqrt{2}}}\\ =-q{E}_x\left(\frac{\delta }{\sqrt{2}}-0\right)-q{E}_y\left(\frac{\delta }{\sqrt{2}}-0\right)\\ =-q{E}_x\frac{\delta }{\sqrt{2}}-q{E}_y\frac{\delta }{\sqrt{2}}\end{array}$

Simplify further.

   $w=-q\frac{\delta }{\sqrt{2}}\left(E_x+E_y\right)$

Conclusion:

Thus,the value of incremental work in direction at $45°$ from $x$ axis in first quadrant is

   $-q\frac{\delta }{\sqrt{2}}\left(E_x+E_y\right)$.

To determine

(c)

The value of incremental work along a line in the first octant having equal $x$ , $y$ and $z$ component.

Answer to Problem 4.1P

The value of incremental work along a line in the first octant having equal $x$ , $y$ and $z$ component is $-q\frac{\delta }{\sqrt{3}}\left(E_x+E_y+E_z\right)$.

Explanation of Solution

Concept used:

Write the expression for incremental work in $xyz$ coordinate system.

   $w=-Q{\displaystyle {\int }_a^{b}E\cdot \left(dx{a}_x+dy{a}_y+dz{a}_z\right)}$ ...... (5)

Here, $w$ is incremental work, $Q$ is the value of charge, $a$ is the initial value of length, $b$ is the final value of length, $a_x$ is the unit vector in $x$ direction, $a_y$ is the unit vector in $y$ direction, $a_z$ is the unit vector in $z$ direction and $E$ is electric field.

Write the expression for length in $xyz$ coordinate system.

   $l=\sqrt{a^2+b^2+c^2}$ ...... (6)

Here, $l$ is value of length in $xyz$ coordinate system, $a$ is length in $x$ axis, $b$ is length in $y$ axis and $c$ is length in $z$ axis.

Calculation:

Value of length is same in $x$ axis, $y$ axis and $z$ axis.

So $a=b=c$

Substitute $\delta$ for $l$ , $a$ for $b$ and $a$ for $c$ in equation (6).

   $\begin{array}{c}\delta =\sqrt{a^2+a^2+a^2}\\ =\sqrt{3a^2}\\ =\sqrt{3}a\end{array}$

Rearrange for $a$.

   $a=\frac{\delta }{\sqrt{3}}$

Substitute $q$ for $Q$ and $\left(E_x{a}_x+E_y{a}_y+E_z{a}_z\right)$ for $E$ in equation (5).

   $\begin{array}{c}w=-q{\displaystyle {\int }_a^b\left(E_x{a}_x+E_y{a}_y+E_z{a}_z\right)\cdot \left(dx{a}_x+dy{a}_y+dz{a}_z\right)}\\ =-q{\displaystyle {\int }_a^b{E}_{x}dx+E_{y}dy+E_{z}dz}\end{array}$

Simplify further.

   $w=-q{\displaystyle {\int }_a^b{E}_{x}dx}-q{\displaystyle {\int }_c^d{E}_{y}dy}-q{\displaystyle {\int }_e^f{E}_{z}dz}$

Here, $e$ is the initial value of length in $z$ axis and $f$ is the final value of length in $z$ axis.

Substitute $0$ for $a$ , $\frac{\delta }{\sqrt{3}}$ for $b$ , $0$ for $c$ , $\frac{\delta }{\sqrt{3}}$ for $d$ , $0$ for $e$ , $\frac{\delta }{\sqrt{3}}$ for $f$ in above equation.

   $\begin{array}{c}w=-q{\displaystyle {\int }_0^{\frac{\delta }{\sqrt{3}}}{E}_{x}dx}-q{\displaystyle {\int }_0^{\frac{\delta }{\sqrt{3}}}{E}_{y}dy}-q{\displaystyle {\int }_0^{\frac{\delta }{\sqrt{3}}}{E}_{z}dz}\\ =-q{E}_x{\left(x\right)}_0^{\frac{\delta }{\sqrt{3}}}-q{E}_y{\left(y\right)}_0^{\frac{\delta }{\sqrt{3}}}-q{E}_z{\left(z\right)}_0^{\frac{\delta }{\sqrt{3}}}\\ =-q{E}_x\left(\frac{\delta }{\sqrt{3}}-0\right)-q{E}_y\left(\frac{\delta }{\sqrt{3}}-0\right)-q{E}_z\left(\frac{\delta }{\sqrt{3}}-0\right)\\ =-q{E}_x\frac{\delta }{\sqrt{3}}-q{E}_y\frac{\delta }{\sqrt{3}}-q{E}_z\frac{\delta }{\sqrt{3}}\end{array}$

Simplify further.

   $w=-q\frac{\delta }{\sqrt{3}}\left(E_x+E_y+E_z\right)$

Conclusion:

Thus,the value of incremental work along a line in the first octant having equal $x$ , $y$ and $z$ component is $-q\frac{\delta }{\sqrt{3}}\left(E_x+E_y+E_z\right)$.