 #1
skg94
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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 .40m2*10^6C.40m3*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.
[itex]\sqrt{.333^2 + .333^2}[/itex] =.4709331163
then used tan to find the angle, .333/.333= 1 tan1(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



BC