Equipotential Surfaces and Potential Gradient Problems and Solutions

Q#23.44

A very large plastic sheet carries a uniform charge density $-6.00 \ nC/m^2$ of on one face. 

(a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? 

(b) Find the spacing between equipotential surfaces that differ from each other by 1.00 V. What type of surfaces are these? 

Answer:

By the definition of electric potential, if a positive charge gains potential along a path, then the potential along that path must have increased. The electric field produced by a very large sheet of charge is uniform and is independent of the distance from the sheet. 

(a) No matter what the reference point, we must do work on a positive charge to move it away from the negative sheet.

Since we must do work on the positive charge, it gains potential energy, so the potential increases.

(b) Since the electric field is uniform and is equal to $\sigma/2\epsilon_0$, we have

$\Delta V=Ed=\frac{\sigma}{2\epsilon_0}d$

Solving for d gives

$d = \frac{2\epsilon_0 \Delta V}{\sigma}$

$d = \frac{2(8.85 \times 10^{-12} \ C^2.N^{-1}.m^{-2})(1.00 \ V)}{6.00 \times 10^{-9} \ C/m^2}$

$d=0.00295 \ m$ = 2.95 mm

Since the spacing of the equipotential surfaces (d = 2.95 mm) is independent of the distance from the sheet, the equipotential surfaces are planes parallel to the sheet and spaced 2.95 mm apart. 

Q#23.45

In a certain region of space, the electric potential is V(x, y, z) = $Axy -Bx^2+Cy$ where A, B and C are positive constants. 

(a) Calculate the x-, y-, and z-components of the electric field. 

(b) At which points is the electric field equal to zero? 

Answer:

Use Eq. $E_x=-\frac{\partial V }{\partial x}$, $E_y=-\frac{\partial V }{\partial y}$ and $E_z=-\frac{\partial V }{\partial z}$ to calculate the components of $\vec{E}$.

V(x, y, z) = $Axy -Bx^2+Cy$

(a) $E_x=-\frac{\partial (Axy-Bx^2+Cy) }{\partial x}$

$E_x=-Ay+2Bx$

 $E_y=-\frac{\partial (Axy-Bx^2+Cy) }{\partial y}$

$E_y=-Ax-C$

 $E_z=-\frac{\partial (Axy-Bx^2+Cy) }{\partial z}$

$E_z=0$

(b) E = 0 requires that $E_x=E_y=E_z=0$

$E_z$ = everywhere. 

$E_y$ = 0 at $x=-C/A$

And $E_x$ is also equal to zero for this x, any value of z and

$y=2Bx/A=(2B/A)(-C/A)=-2BC/A^2$

V doesn’t depend on z so $E_z$ = 0 everywhere. 

Q#23.46

In a certain region of space the electric potential is given by V(x, y, z) = $+Ax^2y-Bxy^2$ where A = 5.00 $V/m^3$ and B = 8.00 $V/m^3$. 

Calculate the magnitude and direction of the electric field at the point in the region that has coordinates x = 2.00 m, y = 0.400 m, and z = 0.

Answer:

Apply Eq. $E_x=-\frac{\partial V }{\partial x}$ and eq. $\vec{E}=\frac{1}{4\pi \epsilon_0}\frac{|q|}{r^2}\hat{r}$ is the electric field due to a point charge q.

(a) $E_x=-\frac{\partial V }{\partial x}$

$E_x=-\frac{\partial}{\partial x}\left(\frac{kQ}{\sqrt{x^2+y^2+z^2}}\right)$

$E_x=\frac{kQx}{(x^2+y^2+y^2)^{3/2}}$

because $r^2=x^2+y^2+z^2$ so

$E_x=\frac{kQx}{r^3}$

For y-axis:

$E_y=-\frac{\partial}{\partial y}\left(\frac{kQ}{\sqrt{x^2+y^2+z^2}}\right)$

$E_y=\frac{kQy}{(x^2+y^2+y^2)^{3/2}}$

$E_y=\frac{kQy}{r^3}$

For z-axis:

$E_z=-\frac{\partial}{\partial z}\left(\frac{kQ}{\sqrt{x^2+y^2+z^2}}\right)$

$E_z=\frac{kQz}{(x^2+y^2+y^2)^{3/2}}$

$E_z=\frac{kQz}{r^3}$

(b) From part (a), 

$\vec{E}=\frac{kQ}{r}\left(\frac{x\hat{i}}{r}+\frac{y\hat{j}}{r}+\frac{z\hat{k}}{r}\right)$ 

$\vec{E}=\frac{kQ}{r^2}\hat{r}$ which agrees with Eq. $\vec{E}=\frac{1}{4\pi \epsilon_0}\frac{|q|}{r^2}\hat{r}$

V is a scalar $\vec{E}$ is a vector and has components. 

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