2fer: Infinite Line and Plane Energy Field

[Rettro]

Got some help with my first problem here, so let's give it a go with the other 2 that stump me. The chapter is on point, infinite line, infinite plane, and parallel plate capacitors, so the line is treated as infinite and the plates are treated as infinite planes.

Homework Statement

A 10 cm long thin glass rod is uniformly charged to +40nC. A small glass beat, charged to +6nC, is 4cm from the center of the rod. What is the force (magnitude and direction) on the bead?

Homework Equations

E = K2A/r
F = Eq
A = q/L
(Note, A represents the traditional symbol Lambda)

The Attempt at a Solution

A = 40E-9C / 0.1m
A = 4E-7 C/m

E = (9E9 Nm^2/C^2)(2)(4E-7 C/m) / 0.04m
E = (7,200 Nm/C) / 0.04m
E = 180,000 N/C

F = (180,000 N/C)(6E-9C)
F = 0.00108N away = 1.08E-3 N away

Book Answer = 8.4E-4N away

Obviously 1.08E-3 N away /= 8.4E-4N away



Second Problem:

Homework Statement

Two 10 cm diameter charged disks face each other, 20 cm apart. The left dist is charged to -50nC and the right dist is charged to +50nC.
a) What is the electric field E both magnitude and direction at the midpoint between the two disks?
b) What is the force F on a -1nC charge placed at the midpoint?

Homework Equations

E = n2K(pi)
n = q/A
A = Area

The Attempt at a Solution

20cm apart is far too far for a parallel plate capacitor, so I treated it as two planes.

n = 50E-9C / 0.0025(pi) m^2
n = 2E-5/(pi) C/m^2 =~ 6.37E-6 C/m <-- I deal with the pi version, habit

E1 = (2E-5/(pi) C/m^2)(2)(pi)(9E9 Nm^2/C^2)
E1 = 3.6E-13 N/C left <-- See how the pi goes away, cleaner that way

Logically, E2 = E1, E = E1+E2, E = 2E1
E = 7.2E-13 N/C left <-- answer to a)

F = (7.2E-13 N/C)(1E-9 C)
F = 7.2E-22 N right <-- answer to b)


Book answers are:
a) 7.6E4 N/C left
b) 7.6E-5 N right

I understand the left and right, and how the answer to b is equal to a E-9, but how they got that for a is beyond me.

 

[ideasrule]

E = K2A/r
F = Eq
A = q/L
(Note, A represents the traditional symbol Lambda)

This is the formula for an infinite line charge. 10 cm is far from infinite, so you'll need to use integration to get the answer.

20cm apart is far too far for a parallel plate capacitor, so I treated it as two planes.

As soon as you begin treating finite plates as planes, you *are* dealing with a parallel plate capacitor. The derivations of all equations relating to the parallel plate capacitor treats the plates as two planes, in exactly the way you have done.

n = 50E-9C / 0.0025(pi) m^2
n = 2E-5/(pi) C/m^2 =~ 6.37E-6 C/m <-- I deal with the pi version, habit

E1 = (2E-5/(pi) C/m^2)(2)(pi)(9E9 Nm^2/C^2)
E1 = 3.6E-13 N/C left <-- See how the pi goes away, cleaner that way

Check your algebra. You have the right formula, but messed up somewhere in the calculation.

 

[Rettro]

The problem I have with integrating is that this chapter is supposed to be about points and infinite features, in fact integration to find electric field doesn't show it's ugly head until chapter 30, this is chapter 27, and only chapters 27 and 29 are on the test. To be perfectly honest I haven't figured out how to derive for field yet or even tried because of this fact.


The second problem, I did have an error, but repairing and re-doing treating it as a parallel plate capacitor yielded the same answer, 7.2 x10^5 N/C for A, which is closer but still not correct.

n = 50E-9 C / 0.0025(pi) m^2
n = 6.3662E-6 C/m^2

E = n/eo
E = 6.3662E-6 C/m^2 / 8.85E-12 C^2/Nm^2
E = 719,344 N/C = 7.2E5 N/C
Correct = 7.6E4 N/C