- #1
Rettro
- 4
- 0
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.
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?
E = K2A/r
F = Eq
A = q/L
(Note, A represents the traditional symbol Lambda)
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:
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?
E = n2K(pi)
n = q/A
A = Area
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.
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.