Finding the magnitude of electric field

In summary, a student attempted to solve a problem for electric field magnitude using equations they had not previously learned but failed. They found help online and managed to figure it out all on their own.
  • #1
jlmccart03
175
9

Homework Statement


A uniform electric field exists in the region between two oppositely charged parallel plates 1.59 cmapart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval 1.50×10−6 s .

Find the magnitude of the electric field.
Use 1.60×10−19 C for the magnitude of the charge on an electron and 1.67×10−27 kg for the mass of a proton.

Homework Equations


Coulomb: F = kQq/r2
F=ma
F=qE
E=kQ/r2
a=F/m

The Attempt at a Solution


First I have taken this course last spring and failed it. I simply do not understand how I set up these problems and get answers without being given an example. I literally bang my head against my desk and now I am doing the exact same thing just in a new semester. So I attempted to solve the problem this way.

I took E=kQ/r2 thinking that I could simly take the magnitude of Charge times Coloumbs constand divided by the distance squared. I get this E = 1.60*10-19 * 9*109 / 0.01592 to get 9*10-8 N/C. It's wrong. So I looked up a way to do this problem by Googling how to find electric field between two planes. I get E = V/d, but that is something we haven't learned yet so I think I am missing something and simply do not know where to go. How do you find electric field magnitude with the given equations? More so, how can I learn to tackle these problems since I just keep going one after the other not knowing what to do?

Thanks for any and all help given!
 
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  • #2
jlmccart03 said:
E=kQ/r2
This is the electric field around a point charge, which is not relevant here.
jlmccart03 said:
How do you find electric field magnitude with the given equations?
You should first find the equation for the force on a charge. Once you have this, you can work out the needed variables from F=ma and the one-dimensional motion equations.
 
Last edited:
  • #3
jlmccart03 said:
I literally bang my head against my desk
We've all been there...
jlmccart03 said:
More so, how can I learn to tackle these problems since I just keep going one after the other not knowing what to do?
There's no simple answer. My best advice is to try and understand the meaning of the formulas and their relation to reality as thoroughly as possible. Once you do that, it is just a matter of algebra. The biggest mistake you can make is to just try and memorize equations and hope they work.
 
  • #4
NFuller said:
This is the electric field around a point charge, which is not relevant here.

You should first find the equation for the force on a charge. Once you have this, you can work out the needed variables from F=ma and the one-dimensional motion equations.
I managed to figure it out all on my own! And yeah I struggled last semester. I going to see my professor and TA's to see what I can do to make sure I understand the material. It's possible, I know I can do it, it just takes time and persistance to fight for an A in the class, but better an understading of how to do physics. Thanks!
 
  • #5
jlmccart03 said:

Homework Statement


A uniform electric field exists in the region between two oppositely charged parallel plates 1.59 cmapart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval 1.50×10−6 s .

Find the magnitude of the electric field.
Use 1.60×10−19 C for the magnitude of the charge on an electron and 1.67×10−27 kg for the mass of a proton.

Homework Equations


Coulomb: F = kQq/r2
F=ma
F=qE
E=kQ/r2
a=F/m

The Attempt at a Solution


First I have taken this course last spring and failed it. I simply do not understand how I set up these problems and get answers without being given an example. I literally bang my head against my desk and now I am doing the exact same thing just in a new semester. So I attempted to solve the problem this way.

I took E=kQ/r2 thinking that I could simly take the magnitude of Charge times Coloumbs constand divided by the distance squared. I get this E = 1.60*10-19 * 9*109 / 0.01592 to get 9*10-8 N/C. It's wrong. So I looked up a way to do this problem by Googling how to find electric field between two planes. I get E = V/d, but that is something we haven't learned yet so I think I am missing something and simply do not know where to go. How do you find electric field magnitude with the given equations? More so, how can I learn to tackle these problems since I just keep going one after the other not knowing what to do?

Thanks for any and all help given!
1499915841528.jpeg

I guess this will help.
 

Related to Finding the magnitude of electric field

1. What is the definition of electric field magnitude?

The electric field magnitude is a measure of the strength of an electric field at a specific point in space. It is defined as the force per unit charge experienced by a test charge placed at that point.

2. How is the magnitude of electric field calculated?

The magnitude of electric field is calculated by dividing the force acting on a test charge by the magnitude of the test charge itself. Mathematically, it can be represented as E = F/q, where E is the electric field, F is the force, and q is the test charge.

3. What is the unit of measurement for electric field magnitude?

The unit of measurement for electric field magnitude is newtons per coulomb (N/C) in SI units. It can also be expressed in volts per meter (V/m) in the CGS system.

4. How does the distance from a charged object affect the electric field magnitude?

The electric field magnitude is inversely proportional to the square of the distance from a charged object. This means that as the distance increases, the electric field decreases, and vice versa. This relationship is known as the inverse square law.

5. Can the magnitude of electric field be negative?

Yes, the magnitude of electric field can be negative. This indicates that the direction of the electric field is opposite to the direction of the force acting on a positive test charge. In other words, the electric field is pointing towards the source of the negative charge.

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