# How you I determine the charge on the child's fingertip?

• Josh123
In summary, Thomas was crawling around on the rug. When he reached for a metal truck, a prominent spark lasting 5mmsec appeared between his fingertip and the object. His fingertip was about 2 mm from his toy. His finger burnt (the area of the burnt region was of 10^-4 m^2) Thomas' finger had a capacitance of 4.43*10^-13 Farads, and the electric field at the point of discharge was 3*10^6 Volts/meter. Using C=Q/V, Thomas found that the charge on his fingertip was 6000 volts. The resistance between the toy and the finger was found to be 4.43*10^-13
Josh123
There's a problem that I am currently working on. I know the theory, but I'm not quite sure how to start this particular problem:

"Thomas was crawling around on the rug. When he reached for a metal truck, a prominent spark lasting 5mmsec appeared between his fingertip and the object. His fingertip was about 2 mm from his toy. His finger burnt (the area of the burned region was of 10^-4 m^2)

On that day, the air was cold and dry causing it to become conducting when the electric field reached 3*10^6 N/C."

My question is, how you I determine the charge on the child's fingertip?
How do I estimate the resistance of the dry air between the toy
truck and the child's fingertip? (I just would like to know how to start this problem.. you don't have to do the entire thing)

Treat the child's finger as a capacitor. You are given the Area of the 'capacitor', and the separation. You also know the electric field at the point of discharge is exactly 3*10^6 N/C.

This should be all the information you need to solve the problem.

Claude.

I found the capacitance:

Dialectric constant of air: 1.00058=k
C=KEA/d = (1.00058)(8.85*10^-12 C^2/Nm^2) (10^-4m^2) / 0.002m
=4.43*10^-13 F

From here, I'm not quite sure what to do.

I can try using C=Q/V to find the charge... but I don't know "V". Should I use RC circuit principle to find "V"?

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josh,

"...I don't know 'V'..."

Do you know how to find the voltage between two points when you know their separation and the field between them?

Sure,to make that ratio,u need that electric field to be constant.A finger tip is not a plane surface and the gradient of the electric field is quite significant...

But to solve this problem,u need to make simplifying assumptions,even if those have nothing to do with the reality...

Daniel.

Ok thanks, I think I've figured out how to calculate "V":

Since E=3*10^6 N/C = 3*10^6 V/m
V=E*distance= (3*10^6 V/m)(0.002m)=6000V

Now that I know the voltage, I can find the charge using C=Q/V.

Once I have the charge, I can find the resistance between the toy and the finger using discharging principle: Q(t)=Qo*e^(t/RC)

Am I on the right track so far?

That is not an equation for discharging...It lacks a minus.

Daniel.

Oh, I forgot. I meant: Q(t)=Qo*e^-(t/RC). where Qo is initial charge, unknown: R=resistance, C=capacitance. I'm not sure about the value of Q(t). Is it 0 (since it is discharged?)?

for some t it is.

If I have the capacitance, the voltage and the charge... how do I find the resistance (resistance is the air)?

To find the resistance, must I assume that Q(t) is zero? Q(t)=Qo*e^-(t/RC)
When I do this the resistance gives zero.

The resistance you're looking for is the R in the discharge eq. You know the charges, time and the capacitance. There really isn't much to do but to take the logarithms and solve the eq for R.

Actually I only know one of the charges.

In the equation: q(t) = Qe^(-t/RC).. I found Q... but I don't know what q(t) represents

You can't assume Q(t) is zero, since there is no value for R that will satisfy the equation. You need to estimate a value for Q(t) that one would consider negligible (or equivalently, how many time constants one would consider to be negligible).

Claude.

Thank you for help

"You can't assume Q(t) is zero, since there is no value for R that will satisfy the equation. You need to estimate a value for Q(t) that one would consider negligible (or equivalently, how many time constants one would consider to be negligible)."

I too am working on a discharging problem. However, figuring out Q(t) is confusing. You say that I should assume a "negligeable" value for it. Is 0.00000001 a negligeable value (If we take the example mentioned in the first post where we know "t", "C" and "Q")?

since physics teachers are so creative, i got the exact same problem as JOSH123, but what I'm wondering is how you can figure out the current from all this. CAn you just use the formula I=deltaQ/t assuming that the fingertip completely dishcarges?

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## 1. How do I measure the charge on a child's fingertip?

To measure the charge on a child's fingertip, you will need a device called an electrostatic voltmeter. This device can detect the presence and magnitude of electrical charges on an object, such as a fingertip.

## 2. What is the unit of measurement for the charge on a fingertip?

The unit of measurement for electrical charge is called the Coulomb. A child's fingertip typically has a very small charge, measured in microcoulombs (μC). One microcoulomb is equal to one millionth of a Coulomb.

## 3. How do I ensure that my measurement of the charge on the fingertip is accurate?

To ensure accuracy, it is important to minimize any external factors that could affect the measurement, such as static electricity in the surrounding environment. It is also important to ground yourself and the electrostatic voltmeter to eliminate any potential interference.

## 4. Can the charge on a child's fingertip change?

Yes, the charge on a child's fingertip can change based on various factors, such as rubbing against different materials or coming into contact with other charged objects. Additionally, the charge on a fingertip can also be affected by the humidity and temperature of the surrounding environment.

## 5. Why is it important to measure the charge on a child's fingertip?

Measuring the charge on a child's fingertip can help us understand more about the behavior of electrical charges and how they interact with different objects and materials. It can also be useful in experiments and investigations related to electricity and static electricity.

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