How Do You Calculate Voltage from Electric Field and Charge?

In summary, the homework statement states that a positively charged ball will be placed on an isolating tripod in order to measure the voltage between 2 parallel metal discs. After that, the balls charge will be turned off and the scale measurements will be read again. If the balls charge is 12*10^-9 nC and the difference in scale measurements is .25g, calculate the voltage between the discs.
  • #1
Zamze
11
0

Homework Statement


a positivly spheric charged ball is placed on an isolating tripod. its going to be used to measure the voltage between 2 parallel metaldiscs. trough a tiny hole in the lower metaldisc the ball reaches into the homogen field. the ball and the isolated tripod are on top of a scale. the values on the scale are registered. after that the charge on the metal discs are turned off and the scale numbers are read again.

calculate the voltage between the 2 metaldiscs if the balls charge is 12nC (12*10^-9) and the difference in the scale measurements are 0.25g. the distance between the discs are 5cm.


Homework Equations



E(N/C)=F/Q
E=(k*Q)/r^2
E=F/Q



The Attempt at a Solution



i found out that E=(mg)/Q but this still didnt help me. (E here is for electric field strenghts)
so i tried k which is 8.99*10^9.
(k*Q)/r^2 and got 43142 N/C for E.
then my E*d = 2157 volt,
so then i involved E=(mg)/Q
(E*Q)/9.82 and got m = 5.25*10^-5.
and since delta m is 0.25g i did my m + delta m and got 3.027*10^-4.

this i then used as U/d=F/Q with my new m value.
so (mg)/Q = 247709
this i multibplied with d, 0.05 and got 12385. so the closest thing i can come up to, to reach the answer 10kV is 12385-2157. but i wouldent exactly understand why :/.
thanks in advance!
 
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  • #2
Zamze said:

The Attempt at a Solution



i tried plugging in (k*Q)/r^2 for E, and then E*d=U which gives me the wrong answer. i should get 10kV.
I realized it has to do with the difference in mass which i don't know how to use at all. Use it with F i guess but i don't know how to.

Can you show your calculations? Then can see where your method is going wrong :smile:
 
  • #3
i found out that E=(mg)/Q but this still didnt help me. (E here is for electric field strenghts)
so i tried k which is 8.99*10^9.
(k*Q)/r^2 and got 43142 N/C for E.
then my E*d = 2157 volt,
so then i involved E=(mg)/Q
(E*Q)/9.82 and got m = 5.25*10^-5.
and since delta m is 0.25g i did my m + delta m and got 3.027*10^-4.

this i then used as U/d=F/Q with my new m value.
so (mg)/Q = 247709
this i multibplied with d, 0.05 and got 12385. so the closest thing i can come up to, to reach the answer 10kV is 12385-2157. but i wouldent exactly understand why :/.
thanks in advance!
 
  • #4
Zamze said:
i found out that E=(mg)/Q but this still didnt help me. (E here is for electric field strenghts)
[Edit: previous comment here removed.]
so i tried k which is 8.99*10^9.
(k*Q)/r^2 and got 43142 N/C for E.

Be careful with your equation for E. Does E = (k*Q)/r^2 apply to a system of parallel plates (or is that the equation for a point charge)? Perhaps you should check that. You can use Gauss' law to derive the formula if you don't have it written down somewhere.

(Hint: The electric field between the plates will be in terms of some other variables/constants such as Q, A (area), [itex] \epsilon _0 [/itex], and/or σ. Leave your intermediate answer (for the electric field between the plates) in terms of these additional variables/constants for now. They may end up canceling out later before you reach your final answer. :wink:)
 
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  • #5
collinsmark said:
[Edit: previous comment here removed.]


Be careful with your equation for E. Does E = (k*Q)/r^2 apply to a system of parallel plates (or is that the equation for a point charge)? Perhaps you should check that. You can use Gauss' law to derive the formula if you don't have it written down somewhere.

(Hint: The electric field between the plates will be in terms of some other variables/constants such as Q, A (area), [itex] \epsilon _0 [/itex], and/or σ. Leave your intermediate answer (for the electric field between the plates) in terms of these additional variables/constants for now. They may end up canceling out later before you reach your final answer. :wink:)

Youre right that was for point charge and not for parallel plates. that makes it harder for me since i can't calculate out the electric field strenght now.
 
  • #6
Zamze said:
that makes it harder for me since i can't calculate out the electric field strenght now.

Again, you can use Gauss' law to derive the electric field on your own. Show your work, if you're stuck. Alternately, you might be able to find the electric field formula for parallel plates in your text or notes (I recommend Guass' law though given the choice -- I hate remembering specific formulae, and Gauss' law is useful enough that one can derive many electric field equations on the fly, specific to the given configuration.) You can use Guass' law to get the electric field of a single plate. You can use superposition if you need to for parallel plates.

But like I said before, your electric field equation might end up a function of other unknown variables or constants, such as (a) area of the plates, (b) surface charge density of the plates, (c) total charge on the plates, and/or (d) electric constant [itex] \epsilon_0 [/itex]. But don't worry about that. Leave your electric field equation in terms of any of these variables/constants for now. They will go away later. You don't need to look for a specific 'number' yet.

Once you find an equation for the electric field, you can find an equation for the voltage between the plates, in terms of the same variables/constants.

Also, once you find the electric field equation, you can find an expression for the force on the ball, in terms of the same variables/constants.

Using the data from the scale, you'll have enough information such that with just a small bit of substitution, you can find the voltage between the plates as an actual number (any unknown variables/constants vanish in this step).
 
  • #7
I haven't learned Gauss law yet.
but what it comes down to i guess is E=F/Q and E=U/d
where E is the electric field strenght, F is the force and can be mg, Q is the charge of the ball. U is the voltage between the fields, and d is the distance between the fields.
ive tried with F/Q=u/d but no progress.
 
  • #8
Zamze said:
I haven't learned Gauss law yet.
but what it comes down to i guess is E=F/Q and E=U/d
where E is the electric field strenght, F is the force and can be mg, Q is the charge of the ball. U is the voltage between the fields, and d is the distance between the fields.
ive tried with F/Q=u/d but no progress.

Okay, I understand now. Those equations were given beforehand. It turns out that those equations will work too, since they are valid for this particular problem with parallel plates. And thus it turns out that this problem is much simpler than I originally suspected, since you don't need to derive those equations.

So you have F/Q = U/d. Solve for the voltage, U. Substitute in the other values given in the problem.

(Cautious hint: The scale's measurement difference was given to you already as a difference. So there is no need to calculate any absolutes [such as the mass of the ball or tripod]. It is this difference [as given] that is associated with the electric force, thus the voltage. But be careful, the scale's measurement was given in units of kilograms [as many scales do], which is a measusre of mass. You'll need to convert this measurement to units of Newtons, which is a measure of force.)

[Edit: By the way, I looked through original post and I think you made a miscalculation somewhere. Start over with your F/Q = U/d equation. All the other information you need is given in the problem statement -- just make sure you convert kg to N.]
 
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  • #9
thx a lot for the help, managed to solve it now =).
cheers!
 
  • #10
Great! :smile:
 

Related to How Do You Calculate Voltage from Electric Field and Charge?

1. What is electric field strength?

Electric field strength is a physical quantity that measures the force per unit charge on a charged particle in an electric field. It is represented by the symbol E and is measured in newtons per coulomb (N/C).

2. How is electric field strength calculated?

Electric field strength can be calculated by dividing the force exerted by an electric field on a charged particle by the magnitude of the charge. Mathematically, it is represented as E = F/q, where E is the electric field strength, F is the force, and q is the magnitude of the charge.

3. What factors affect electric field strength?

The electric field strength is affected by the magnitude and direction of the electric field, as well as the magnitude and sign of the charge. It also depends on the distance between the charged particles and the medium they are in.

4. How does electric field strength relate to electric potential?

Electric potential is the amount of work required to move a unit charge from one point to another in an electric field. Electric field strength is directly proportional to electric potential, meaning that as the electric field strength increases, the electric potential also increases.

5. What are some real-world applications of electric field strength?

Electric field strength has many practical applications, such as in electronic devices, power distribution systems, and medical equipment. It is also used in particle accelerators, plasma physics, and electrostatic precipitators for air pollution control.

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