Problem #1
In Figure 1, two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge densities of opposite signs and magnitude 7.00 x 10─22 C/m2. In unit-vector notation, what is the electric field at points (a) to the left of the plates, (b) to the right of them, and (c) between them? |
| Fig.01 |
Known:
the plates have excess surface charge densities of opposite signs and magnitude σ = 7.00 x 10─22 C/m2
Sheet of charge:
E = σ/2ϵ0
the electric field on the side of the plate is shown in the figure 02.
 |
| Fig.02 |
ETotal = E1 + E2 = (σ/2ϵ0)i + (σ/2ϵ0)(─i) = 0
(b) electric field at points to the right of them,
ETotal = E1 + E2 = (σ/2ϵ0)(─
(c) electric field at points to between them
ETotal = E1 + E2 = (σ/2ϵ0)(─
Problem #2
In Figure 3, a small circular hole of radius R = 1.80 cm has been cut in the middle of an infinite, flat, nonconducting surface that has uniform charge density σ = 4.50 pC/m2. A z axis, with its origin at the hole’s center, is perpendicular to the surface. In unitvector notation, what is the electric field at point P at z = 2.56 cm?
 |
| Fig.03 |
Known:
circular hole of radius R = 1.80 cm = 1.8 x 10─2 m
uniform charge density σ = 4.50 pC/m2 = 4.50 x 10─12 C/m2
point P at z = 2.56 cm = 2.56 x 10─2 m
 |
| Fig.04 |
Esheet = (σ/2ϵ0)(k)
and the field due to the disk is
Edisk = (σ/2ϵ0)[1 ─ z(z2 + R2)─1/2](k)
We can assert that
E(P) = Esheet ─ Edisk
E(P) = (σz/2ϵ0)[(z2 + R2)─1/2](k)
= {(4.50 x 10─12 C/m2)(2.56 x 10─2 m)/(2 x 8.85 x 10─12 C2/N.m2)}{[(2.56 x 10─2 m)2 + (1.8 x 10─2 m)2]─1/2}
E(P) = (0.208 N/C)k
Problem #3
Figure 05a shows three plastic sheets that are large, parallel, and uniformly charged. Figure 05b gives the component of the net electric field along an x axis through the sheets. The scale of the vertical axis is set by Es = 6.0 x 105 N/C. What is the ratio of the charge density on sheet 3 to that on sheet 2?
 |
| Fig.05 |
Figure (a) and (b), with three large, parallel, uniformly charged plastic sheets. From the figure, since the electric field to the left of 1 and right of 3 is zero, the total charge on the three sheets is zero.
This follows directly from Gauss’ law. That, we draw our Gaussian surface to enclose corresponding section of all three plate. Working with the fields from the individual sheets as E1, E2 and E3 respectively (with positive values directed away from the plate producing it), we may write equations for the net field in each of the 4 regions as follows:
Left of (1) or right of (3)
─(E1 + E2 + E3) = 0 or
(E1 + E2 + E3) = 0
Between (1) and (2)
E1 – E2 – E3 = 2 x 105 N/C.
Between (2) and (3)
E1 + E2 – E3 = 6 x 105 N/C.
The last three linearly independent equations can be solved any number of ways. e.g., Adding the 1st equation to the 2nd equation yields
(E1 + E2 + E3) + (E1 – E2 – E3) = 2E1 = 2 x 105 N/C
E1 = 105N/C
Subtracting the 2nd equation from the 3rd equation gives
2E2 = 4 x 105 N/C
E2 = 2 x 105 N/C
Substituting these into the 1st equation gives
E3 = –(E1 + E2) = –3 x 105 N/C
The minus sign indicates the field is pointing toward the plate, and its charge is negative
E3/E2 = –1.5
Since E = σ/ε0 for the corresponding plates,
σ3/σ2 = –1.5 as well
Problem #4
In Fig. 06, a small, nonconducting ball of mass m = 1.0 mg and charge q = 2.0 x 10─8 C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle θ = 30° with a vertical, uniformly charged nonconducting sheet (shown in cross section). Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density s of the sheet.
 |
| Fig.06 |
Draw a free–body diagram for the sphere! This is done in Fig. 6. The forces on the ball are gravity, mg downward, the tension in the string T and the force of electrostatic repulsion (Felec, straight out from the sheet), arising from the sheet of positive charges. We know that the electrostatic force must point straight out from the sheet because the electric field arising from the charge points straight out, so the force exerted on the ball must point straight out as well. (We can assume the ball acts like a point charge with the charge concentrated at its center.)
 |
| Fig.07 |
T cos θ = mg and T sin θ = Felec
Divide the second of these by the first to cancel out T and give:
sin θ/cos θ = tan θ = Felec/mg
Felec = mg tan θ
Plug in the numbers (note: 1.0mg = 1.0 × 10−6 kg) and get
Felec = (1.0 × 10−6 kg)(9.80 m.s─2) tan 300 = 5.7 × 10−6 N
From Felec = |q|E we can get the magnitude of the electric field:
E = Felec/|q| = (5.7 × 10−6 N)/(2.0 × 10−8 C) = 2.8 × 102 N/C
This is the magnitude of an E field on one side of an infinite sheet of charge so that from Eq. 3.5 we can find the charge density of the sheet:
σ = 2ϵ0E
= 2(8.85 × 10−12 C2/N·m2)(2.8 × 102 N/C)
σ = 5.0 × 10−9 C/m2 = 5.0 nC/m2
Since the E field points away from the sheet, this is the correct sign for the charge density; the charge density of the sheet is +5.0 nC/m2.
Problem #5
An electron is shot directly toward the center of a large metal plate that has surface charge density ─2.0 x 10─6 C/m2. If the initial kinetic energy of the electron is 1.60 x 10─17 J and if the electron is to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?
Answer:
Known:
surface charge density, σ = ─2.0 x 10─6 C/m2
initial kinetic energy of the electron is W = 1.60 x 10─17 J
To find the distance between electron and plate we need to recall our dynamic knoowledge
W = F.Δx
in this situation
F = σe/ϵ0
F = (2.0 x 10─6 C/m2)( 1.6 x 10─19 C)/(8.85 x 10─12 C2/N.m2) = 3.6 x 10─14 N
We know that work don must be equal to change is the energy so W = 1.60 x 10─17 J,
Now we can calculate the distance by using these equations
W = F.Δx
1.60 x 10─17 J = (3.6 x 10─14 N)Δx
Δx = 0.44 x 10─3 m = 0.44 mm
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