# Magnitude of charge with common Transparent tape

• FirstLanguageMath
In summary: Thank You for your...help?In summary, the homework statement says that if electrostatic force is great enough to support the weight of a 10.0 mg piece of tape held 1.00 cm above another, the charge on the tape is 0.000098 N.
FirstLanguageMath

## Homework Statement

a) Common transparent tape becomes charged when pulled from a
dispenser. If one piece is placed above another, the repulsive force can
be great enough to support the top piece’s weight. Assuming equal point
charges (only an approximation), calculate the magnitude of the charge if
electrostatic force is great enough to support the weight of a 10.0 mg
piece of tape held 1.00 cm above another.

## Homework Equations

F = k |q1 q2| / r2

## The Attempt at a Solution

I'm getting confused: I know r=1cm and the two pieces of tape have the same charge.
I am certain I need to add the weight in, but how?
I even tried using Newton's Law of gravitation with 10mg being the mass, but that brought no insight.

Draw a free body diagram for the floating piece of tape. What forces are acting?

For the piece of tape to be suspended in air, the sum of the forces has to equal zero - the same situation as if that piece of tape was lying on a table. Did you draw a free body diagram for the tape and sum the forces acting on the tape?

Welcome to Physics Forums.

Edit: @gneill, you are too fast for me. :)

gneill said:
Draw a free body diagram for the floating piece of tape. What forces are acting?
I did. You have gravity pulling down the top piece of tape and because of the pieces being equally charged the bottom tape is pushing the top one away.

FirstLanguageMath said:
I did. You have gravity pulling down the top piece of tape and because of the pieces being equally charged the bottom tape is pushing the top one away.
Okay, so you should be in a position to write an equation. What are the expressions for the two forces?

Sorry, I lost internet yesterday.
So would I use the equation of weight (w=mg) for the tape on the top and then use Coulomb's Law for the static force between the two?

FirstLanguageMath said:
Sorry, I lost internet yesterday.
So would I use the equation of weight (w=mg) for the tape on the top and then use Coulomb's Law for the static force between the two?
Yup.

So m=0.00001kg and g=9.8m/s2
which makes 0.000098N

Should I then place that answer as F in F = k|q1 q2|/r2 ?

What do you think? Try it. What do you get?

So if I continue with r=0.01m and k=8.988x109

then my equation becomes |q1 q2|=215.4895
if this path is right, I would then square root the answer since both pieces are to have equal point charges.
Right?

FirstLanguageMath said:
So if I continue with r=0.01m and k=8.988x109

then my equation becomes |q1 q2|=215.4895
That doesn't look right. It's far, far, far too large. Show your calculation in detail.

First we change it equation to find what I don't have:
F
= k|q1 q2|/r2
F/k = |q1 q2|/r2
(F/k)r2 = |q1 q2|
So we take F=0.000098N from the weight equation
k is the constant =8.988x109 Nm2/C2
r=0.01m
Oh! I messed up it's times r2
Which would make the solution=2.667x10-8

FirstLanguageMath said:
First we change it equation to find what I don't have:
F
= k|q1 q2|/r2
F/k = |q1 q2|/r2
(F/k)r2 = |q1 q2|
So we take F=0.000098N from the weight equation
k is the constant =8.988x109 Nm2/C2
r=0.01m
Oh! I messed up it's times r2
Okay so far.
Which would make the solution=2.667x10-8
Solution to what? What are the units? Also, it doesn't look like it would be a result coming from what you've shown so far.

Maybe try doing the calculation one step at a time. What do you get for F/k?

0.000098N/8.988x109 Nm2/C2 =1.09x10-14 m2/C2
Times that with r2=0.0001m2 = 1.09x10-18 C2
Then because the two pieces are equal we square root the answer to get 1.044x10-9 C

So I don't know how the numbers changed.

That's much better Your result looks good.

gneill said:
That's much better Your result looks good.
But now I'm lost how do I figure out if this magnitude from the electrostatic force is great enough to support the top piece of tape?

FirstLanguageMath said:
But now I'm lost how do I figure out if this magnitude from the electrostatic force is great enough to support the top piece of tape?
There's nothing to figure out; You found the charge using an equation that balanced the two forces. So the forces are exactly equal in magnitude.

gneill said:
There's nothing to figure out; You found the charge using an equation that balanced the two forces. So the forces are exactly equal in magnitude.
Oh, I see. Thank You very much now I understand what I did.
Thank You for your help.

You're welcome!

## 1. What is the magnitude of charge on common transparent tape?

The magnitude of charge on common transparent tape is around 10^-5 to 10^-6 Coulombs per meter.

## 2. How is the magnitude of charge on transparent tape determined?

The magnitude of charge on transparent tape is determined by measuring the tape's surface potential with an electrostatic voltmeter.

## 3. Can the magnitude of charge on transparent tape vary?

Yes, the magnitude of charge on transparent tape can vary depending on factors such as humidity, temperature, and surface conditions.

## 4. Is the magnitude of charge on transparent tape positive or negative?

The magnitude of charge on transparent tape can be either positive or negative, depending on the materials used in its production.

## 5. How does the magnitude of charge on transparent tape affect its adhesive properties?

The magnitude of charge on transparent tape can affect its adhesive properties by either increasing or decreasing its ability to stick to surfaces. This is because like charges repel each other, while opposite charges attract each other.

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