# Charges mass on the end of a light string

• MissPenguins
In summary, a charged mass attached to a point on a uniformly charged vertical sheet is subject to an unknown angle and acceleration due to gravity. Using the equation tan(θ) = (q*E)/(m*g), the angle was found to be 21.254189 degrees. For part 2, the formula E = σ/2ε0 was used, and after rearranging and substituting known values, the charge density was found to be 7.51633 x 10^-7 μC/m^2.
MissPenguins

## Homework Statement

A charged mass on the end of a light string is
attached to a point on a uniformly charged
vertical sheet (with areal charge density
0.25 μC/m2) of infinite extent.

Look at attachment for diagram
Find the angle $\theta$ the thread makes with the
vertically charged sheet. The acceleration of
gravity is 9.8 m/s2 and the permittivity of free
space is 8.854 × 10−12 C2/N · m2 .
I got the first part right to be 21.254189 using $\theta$=tan-1(qE/mg)
I need help with part 2
Find $\sigma$ for an angle of 77o .

## Homework Equations

$\theta$=tan-1(qE/mg)
qE = $\sigma$/2Eo

## The Attempt at a Solution

Since $\theta$=tan-1(qE/mg), so I did tan(77o) = (qE)(mg)
qE = 0.0038118 (calculated from part 1)
mg = 0.0098
$\sigma$=(qE)(mg) = 7.51677X10-13
Converted back to uC = 7.51677X10-7

What did I do wrong? Thank you!

#### Attachments

• physicsforum.jpg
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E will be different for part 2, since the charge density is different.

Also, did you mean (qE)/(mg) instead of (qE)(mg)?

Redbelly98 said:
E will be different for part 2, since the charge density is different.

Also, did you mean (qE)/(mg) instead of (qE)(mg)?

Yea. tan(77) = (qE/mg) which is qE=mgtan(77) = 0.04244
I am not sure what other charge density do I have to use. E0is a constant, that's given.

Same question as this one https://www.physicsforums.com/showthread.php?t=109643
Except part 2 was not explained.

Okay, found another mistake:
MissPenguins said:
qE = $\sigma$/2Eo
Actually, it's E = $\sigma$/2eo (no q here).

Does that help?

Redbelly98 said:
Okay, found another mistake:

Actually, it's E = $\sigma$/2eo (no q here).

Does that help?

The previous one wasn't a mistake.
It's qE = mgtan(77) = 0.042448
I already did this $\sigma$ = (E)(2Eo).
So (0.042448)(2(8.85X10-12) = 7.5133X10-13. I converted back to uC = 7.51633X10-7. And it is wrong! :(

MissPenguins said:
The previous one wasn't a mistake.
It's qE = mgtan(77) = 0.042448
I agree, qE = 0.042448 N.

So based on that, what is E equal to?

I already did this $\sigma$ = (E)(2Eo).
So (0.042448)(2(8.85X10-12) = 7.5133X10-13.
Okay. But E is not 0.042448 -- see my comment above.

Redbelly98 said:
I agree, qE = 0.042448 N.

So based on that, what is E equal to?

Okay. But E is not 0.042448 -- see my comment above.

You said E is different for part 2 since the charge density changed. So how do I find E? I attempted to do (0.25E-6)/(2*8.854E-12) = 14117.91281 for E, is this right for E?
Then use (14117.91281)(2*8.854E-12) = 2.5E-7?
It doesn't make sense. Can you give me more hints? Thanks.

You've already written an equation of the form

tan(θ) = (q*E)/(m*g)

Can you not rearrange to find E? With E can you find σ ?

I figured the problem out. Thank you very much everyone!

## 1. What is the purpose of having a charge on the end of a light string?

The purpose of having a charge on the end of a light string is to create an electric field that can be used for various experiments and demonstrations. The charged object on the end of the string can interact with other charged objects, allowing for the observation and study of electric forces.

## 2. How is the mass of the charge on the end of the light string determined?

The mass of the charge on the end of the light string is typically determined by measuring the period of oscillation of the string and using the equation T = 2π√(m/k), where m is the mass of the charge and k is the string's spring constant. By rearranging the equation to solve for m, the mass can be calculated.

## 3. Can the charge on the end of the light string be changed?

Yes, the charge on the end of the light string can be changed by using different materials with different electrical properties. For example, a metal object will have a different charge than a non-conductive object. Additionally, the charge can be changed by using a different power source or by manipulating the charge with other objects.

## 4. What factors can affect the motion of the charge on the light string?

Several factors can affect the motion of the charge on the light string, including the strength of the electric field, the mass of the charge, and the length and tension of the string. Other external forces such as air resistance or friction can also impact the motion of the charge.

## 5. Is the charge on the end of the light string affected by gravity?

Yes, the charge on the end of the light string is affected by gravity, just like any other object. However, the effect of gravity may be overshadowed by the electric forces acting on the charge. In some cases, the charge may even appear to defy gravity due to the strength of the electric field.

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