Electrostatics questions, parallel plates, force, field

• skg94
In summary, the charges A and C are repelled and the centre charge is also repelled. The net electric force on the centre charge due to the other two charges is .561875N.
skg94

Homework Statement

1.
A 2*10^-6C charged object is .40m left of an another 2*10^-6C charged object, with a 3*10^-6C charged object .40m to the right of the centre object, creating three charged objects in a line. Calculate the magnitude of the net electric force on the centre charge due to the other two charges.

2*10^-6C ------.40m-------2*10^-6C------.40m----3*10^-6C

2. An electron traveling horizontally at a speed of 8.70 * 10^6m/s enters into an electric field of 1.32 *10^3 n/c between two horizontal, parallel plates. The plates are 14.0cm long and are separated by a distance of 8.00cm. Calculate the vertical displacement of the electron through the plates. [3*10^-2m]

3. A 3uC (A) charge is .333m above charged B (3uC), with charge C (3uC) 0.333m left to charge b, creating a triangle. Find the net electric force on charge C.

4. Two point charges are initially 6.0cm apart, and are then moved so that they are 2.0cm apart. If the initial force between these two point charges was F, what is the new force [9F]

[number inside these are the answers]

Homework Equations

E=fe/q
E=kq1q2/r^2
E=kq/r^2
V=E(energy)/q
E=v/d
k=8.99*10^9
charge of an electron= 1.6*10^-19
mass of an electron = 9.11*10^-31kg

The Attempt at a Solution

1.
E1= (8.99^10^9)(2*10^-6)(3*10^-6)/.40^2 = .337125
E2= (8.99*10^9)(2*10^-6)(2*10^-6)/.40^2= .22475
E1+E2=.561875N

I added them together based on coulombs law, charged A and C are repelled and the centre charge is also repelled. I don't have the answer so i don't know if I am right

2. First i found acceleration

a=Fe/m=Eq/m=(1.32*10^3 * 1.6*10^-19)/9.11*10^-31 = 2.318331504*10^14

I actually don't know where to go from here

using projectile motion i listed the x and y variables

x:
Vi= 8.7*10^6
d=.14m

y:
a=2.31...*10^14
d=.08m

how do i find time? then the displacement? I am a litte confused any help please

3.
F(BonC) = (8.99*10^9)(3*10^-6)(3*10^-6)/.333^2 = .7296485675

then for AonC

First i use pytha to find the length of the hypoteneuse connecting A and C.

$\sqrt{.333^2 + .333^2}$ =.4709331163

then used tan to find the angle, .333/.333= 1 tan-1(1) = 45, which makes the opposite triangle 45 degrees as well.

Then found the force on the hyp.

F= (8.99*10^9)(3*10^-6)(3*10^-6)/.4709..^2 = .3648242837

Now based on the diagram i have, , the x and y value triangle will use sin for x and cos for y based on opposite and adjacent, i just need some confirmation on this.
Also when i found the x and y values, would i subtract or add the x value of the force (BonC)?

A
|
|
|
B---------C

skg94 said:
how do i find time? then the displacement? I am a litte confused any help please

How long does it take the electron to travel along the plates?

voko said:
How long does it take the electron to travel along the plates?

it doesn't tell you you have to find it, and i do not know how well its usually t=$\sqrt{2d/a}$ however i can never come up with the right answer

based off d=vit+1/2at^2

The electron is traveling horizontally at a given speed along horizontal plates of a given length. Can't be too hard.

yes but i need to find the vertical displacement using projectile motion techniques try it

You said you cannot find time. What I said in #4 is enough to find time.

Well i can't get it so try helping?

1. What is electrostatics?

Electrostatics is a branch of physics that deals with the study of stationary electric charges and the forces they exert on each other.

2. How do parallel plates create an electric field?

Parallel plates create an electric field by having a positive and negative charge on each plate. The electric field lines are directed from the positive plate to the negative plate, creating a uniform electric field between the plates.

3. What is the force between two charged parallel plates?

The force between two charged parallel plates is directly proportional to the product of the charges on the plates and inversely proportional to the distance between them. This relationship is described by Coulomb's Law.

4. How does the distance between parallel plates affect the strength of the electric field?

The strength of the electric field between parallel plates is inversely proportional to the distance between them. This means that as the distance between the plates decreases, the electric field becomes stronger.

5. What is the relationship between the electric field and the force on a charged particle?

The force on a charged particle is directly proportional to the electric field at that point. This relationship is described by the equation F=qE, where F is the force, q is the charge of the particle, and E is the electric field.

Similar threads

• Introductory Physics Homework Help
Replies
3
Views
216
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
928
• Introductory Physics Homework Help
Replies
2
Views
924
• Introductory Physics Homework Help
Replies
4
Views
333
• Introductory Physics Homework Help
Replies
15
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
780
• Introductory Physics Homework Help
Replies
2
Views
880