Electric Field Question (Very Difficult)

In summary: ThanksIn summary, the conversation is about a question involving a small plastic ball with a mass and charge suspended between the plates of a capacitor. The problem is to find the magnitude of the charge on each plate using the area of the plates. The solution involves analyzing the forces acting on the ball, using the conditions for equilibrium, and applying knowledge about capacitors. The use of area is crucial in finding the electric field and ultimately, the charge on each plate.
  • #1
cheechnchong
132
1

Homework Statement



While you read this question look at this diagram http://i3.tinypic.com/3zho01t.jpg...

A small plastic ball with a mass of 6.50 x 10^-3 kg and with a charge of +0.150 uC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). the ball is in equilibrium, with the thread making an angle of 30.0 degrees with respect to the vertical. the area of each plate is 0.0150 m^2. what is the magnitude of the charge on each plate?


The Attempt at a Solution



I tried to find q using Q = mg/E, but it ended up in a mess since I was looking for the magnitude. The area is throwing me off...I don't know how to utilize it. Honestly, this is a very tough question! would love all the help i can get. i give up after 2 days!

Answer from the textbook is 3.25 x 10^-8 C
 
Last edited:
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  • #2
upping! still waiting for some help.
 
  • #3
cheechnchong said:
I tried to find q using Q = mg/E, but it ended up in a mess since I was looking for the magnitude.
Start by analyzing all the forces acting on the ball. Hint: Three forces act on the ball. Then apply the conditions for equilibrium--net force equals zero--to solve for the electric field. Hint: Write equations for vertical and horizontal force components and then combine them.
The area is throwing me off...I don't know how to utilize it.
Once you've found the electric field, use what you know about capacitors to figure out the charge on each plate needed to produce that field. That's when the area will come in handy.
 
  • #4
^Awesome. i appreciate the help doc.

i used your advice and it worked out just fine...i got the answer i was lookin for! that hint sparked something in my mind...
 
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What is an electric field?

An electric field is a physical field that is created by electrically charged objects. It is a force that surrounds and interacts with charged particles, causing them to either attract or repel each other.

How is the strength of an electric field measured?

The strength of an electric field is measured in terms of electric potential, which is the amount of work needed to move a unit of charge from one point to another in the field. It is typically measured in volts per meter (V/m).

What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude of the charges creating the field, the distance between the charges, and the medium through which the field is passing. It is also affected by the presence of other charges in the vicinity.

How is the direction of an electric field determined?

The direction of an electric field is always directed away from positive charges and towards negative charges. It is also determined by the direction in which a positive test charge would move if placed in the field.

How do electric fields interact with conductors and insulators?

Electric fields can interact differently with conductors and insulators. Conductors allow electric charges to flow freely, so the electric field within a conductor is always zero. Insulators, on the other hand, do not allow electric charges to flow, so the electric field within an insulator is non-zero.

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