Electric Potential / Electric Energy

[VanKwisH]

Homework Statement

A 0.50kg ball is suspended in a uniform electric field ( E = 1500 N/C )
as shown in the diagram ...

http://img201.imageshack.us/img201/9771/23vl0.th.png

Calculate the charge on the ball

Homework Equations

The Attempt at a Solution

I found fg = -4.9 N ...
but what next?

 

[Mentz114]

What is Fg ?

You need to find the forces on the ball due to gravity and the electric field and set them equal. That will give an equation for Q, the charge on the ball.

 

[VanKwisH]

so basically what ur saying is Fg = E ?

 

[Mentz114]

If the ball is not moving, then the electric force (which is horizontal) is balanced by the horizontal component of the gravitational force. Write down expressions for the forces.

 

[VanKwisH]

Fex + Fgx = 0
Fex = - Fgx
Fey + Fgy = 0
Fey = - Fgy ?

 

[Mentz114]

If you have a charge Q in an electric field E ( pointing in the x-direction), what is the force on the charge ?

 

[VanKwisH]

hmmmmmmm the answer says 5.76 x 10^-4
but i get 0.00331 ......
what i did was
i found Fex = 0.864
and Fey = 4.9
i used pythagoras to find the whole part of the Fe and
E = F / Q
so i re-arranged it so that Q = F / E
and that's how i got my answer ... what am i doing wrong?

 

[VanKwisH]

If you have a charge Q in an electric field E ( pointing in the x-direction), what is the force on the charge ?

i don't know ... my physics teacher didn't teach me that

 

[Mentz114]

I dunno. I would write

Q.E = -m.g.sin(10 deg)

 

[Mentz114]

i don't know ... my physics teacher didn't teach me that

You need to know that to solve the problem.

 

[VanKwisH]

hmmmmmm when i do that ... I get the right answer ...
but one question ... what happened to the Y part of the
formula ? isn't that relevant to the solution as well ?
because if u have an Fgy ... wouldn't u need an Fey
in order for it so not move??

 

[Mentz114]

Have a look here -
http://en.wikipedia.org/wiki/Electric_field

what happened to the Y part of the
formula ? isn't that relevant to the solution as well ?
because if u have an Fgy ... wouldn't u need an Fey
in order for it so not move??

There is a y-component in the tension of the string that offsets the vertical force.

 

[alphysicist]

Hi Mentz114,

The formula you have

Q.E = -m.g.sin(10 deg)

does not look right to me. There are three forces acting on the ball: gravity, tension from the string, and the electric force. Gravity is vertical, the electric force is horizontal, and the tension is at an angle.

So the vertical component of the tension (T cos(theta)) must be equal in magnitude to the gravitational force mg, and the horizontal component of the tension (T sin(theta) )must be equal in magnitude to the electric force qE. (This is shown by writing the x and y equations from your force diagram.)

At that point you'll have two equation with two unknowns which you can solve, and you'll end up with a tangent function instead of a sine.

 

[Mentz114]

alphysicist,
you could be right. OP says he got the right answer, but I didn't see the string until he raised the y-component issue.

 

[VanKwisH]

Hi Mentz114,

The formula you have

Q.E = -m.g.sin(10 deg)

does not look right to me. There are three forces acting on the ball: gravity, tension from the string, and the electric force. Gravity is vertical, the electric force is horizontal, and the tension is at an angle.

So the vertical component of the tension (T cos(theta)) must be equal in magnitude to the gravitational force mg, and the horizontal component of the tension (T sin(theta) )must be equal in magnitude to the electric force qE. (This is shown by writing the x and y equations from your force diagram.)

At that point you'll have two equation with two unknowns which you can solve, and you'll end up with a tangent function instead of a sine.

AHHHH ... that made so much more sense thank you ...and i got it :D

 

[Mentz114]

alphysicist,
Something bothers me about the above. To solve 2 equations in 2 unknowns you need 2 degrees of freedom. But if the string is always taut, x and y are not independent. Any change in x requires a change in y, so there's only one degree of freedom, the angle.

However I can see that a tan() is more intuitive. Can you display the solution ?

M

 

[alphysicist]

Mentz114,

From the force diagram I get:

x equation: T sin(10 degrees) = q E
y equation: T cos(10 degrees) = m g

Either solving one of these for T and plugging into the other, or dividing the x-equation over the y-equation gives:

(T sin(10 degrees) )/(T cos(10 degrees) ) = (qE)/(mg)

which gives qE = mg tan(10 degrees)


I might be misunderstanding your statement about the degrees of freedom; in the vertical direction there is no change in the forces no matter what the charge is since the vertical component of the tension is always equal to the weight. Varying the charge would change just the horizontal component of the tension. Therefore the real central part of the problem varying with charge would just be

(T_x) = q E

but since we don't want to use tension in the answer, we want the angle, we use extra expressions (the vertical equation) that relate tension, theta, and mg.

 

[Mentz114]

alphysicist,
thanks, all resolved. I missed the cos resolution. It is one equation in one unknown as required by the df.

M