Point charge on a string in an electric field

[swede5670]

Homework Statement

I need some help solving this question, it is the last one I have to do and I'm not sure how to solve it.

A point charge (m = 1.0 g) at the end of an insulating string of length 53 cm is observed to be in equilibrium in a uniform horizontal electric field of 9000 N/C with the charge 1.0 cm above the lowest (vertical) position. If the field points to the right determine the magnitude and sign of the point charge.

Find Magnitude in C:

Homework Equations

T*cosTheta = mg
T*sinTheta = F(elec) = qE

The Attempt at a Solution

I found this question on a thread from a while ago but I'm not sure entirely what it is saying. Since I have those equations I need to solve for T in the first one to insert it in the second
T*cosTheta = mg
T=mg / Cos(theta)

When I plug that into the second equation
T*sinTheta = F(elec) = qE

(mg/cosineTheta) * sineTheta = qE = F(elec)

My question is, do I need all three parts of this last equation? Or can I simplify it to this:
(mg/cosineTheta) * sineTheta = qE

Even if I do this, how do I figure out the angle?
And in this case what does E represent?

 

[swede5670]

I tried this, but the answer was incorrect. Can you see what I did wrong?


| \
| \
| \
52 | \ 53
| \
| T \
*

So the point charge is basically the dobber of a simple pendulum. In this case you have two forces acting on it: gravity, and electrostatic force. Electrostatic force can be described as:

F = E * q

where F is the force vector, E is the electric field vector and q is the charge on the particle. This force vector is directed parallel with the electric field (to the right).

So now look at gravity:

F = m * g

where m = mass of the particle and g = force of gravity (9.8 kg/s^2). Give a 1 g (.001kg) particle, the gravitational force is .0098 N). This is directed straight down (toward the earth).

The string will counter-act both of these forces and can be described as:

Fx = F cos T
Fy = F sin T

Give the diagram above:

sin T = 52 / 53
(because the particle is 1 cm above the lowest point and the string is 53cm long)

So, Fy has to counteract gravity:

Fy = F sin T = m * g
F * (52/ 53) = 0.0098 N
F = 0.0098 * 53 / 52

So Fx has to counteract the electrostatic force:

Fx = F cos T = E * q

Solve for q as:

q = (F cos T) / E

Where F = 0.0098 * 53 / 52
T = arcsin (52/53)
E = 9000 N/C

Plugging in these values:

1.376 * [cosine{arcsin (52/53)}]
1.376 * .999999
q=1.375999/9000
q=1.5288e-4

But this is wrong for q, does anyone one know what went wrong?

 

[LowlyPion]

The first thing to do is figure the angle.

That's determined by the height of 1 cm.

That means that the Cos of the angle with the vertical must be

(53-1)/53 = 52/53

Cos^-1(52/53) = angle

The Tangent of that angle is the ratio of the 2 forces ... mg down and the Fe horizontally - then isn't it?

 

[swede5670]

Ok so cos-1 (52/53) = 11.14 degrees

So can I plug that into the equations I was given up above?

T*cos(11.14) = mg
T*sin(11.14) = F(elec) = qE

 

[swede5670]

I would need to convert g=9.81 m/s to g=0.00981 km/s to get coulombs correct?
If that does work,
T= (0.00981 * 1)/cos(11.14)
T=.009998389

T*sin(11.14) = F(elec) = qE
.009998389 * sin(11.14) =qE
[.009998389 * sin(11.14)]/E = q
.0019317587 / E = q
.0019317587 / 9000 = q
2.146398 e-7

Correct?

 

[LowlyPion]

Ok so cos-1 (52/53) = 11.14 degrees

So can I plug that into the equations I was given up above?

T*cos(11.14) = mg
T*sin(11.14) = F(elec) = qE

I'd go from here with T = mg/cos(11.14)
So ...

q*E = tan(11.14)*mg

 

[LowlyPion]

I would need to convert g=9.81 m/s to g=0.00981 km/s to get coulombs correct?
If that does work,
T= (0.00981 * 1)/cos(11.14)
T=.009998389

T*sin(11.14) = F(elec) = qE
.009998389 * sin(11.14) =qE
[.009998389 * sin(11.14)]/E = q
.0019317587 / E = q
.0019317587 / 9000 = q
2.146398 e-7

Correct?

I'd use the previous equation in SI units and then the answer is already in C.

q = .197*.001*9.81/9000

It yields your answer.

But I would express it as .215 μC

The answer they are looking for is in C though.