Calculating the electric field of a single charge

In summary: The electric field around a proton is caused by the charge on the proton. The electric field points away from a positive charge, toward a negative one. The electric field around a proton is weaker at the center of the proton than it is on the proton's surface.
  • #1
FilthyMove
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Homework Statement


Calculate the Electric Field vector at the surface of a proton.

Homework Equations


F = qE --> E = F/q

F = [k(q1)(q2)] / (r^2)

The Attempt at a Solution


Well basically I'm not sure... I know it's a really simple question (it's not really even a homework exercise, just something that was put up in a lecture) but I'm having trouble understanding how to use these equations since Coulomb's law describes the force caused on one charge by another (ie involves two charges).
But this question is asking to find an intrinsic electric field held by one isolated charge.

If you sub F = [k(q1)(q2)] / (r^2) into E = F/q I can see that you could end up with
E = (kq) / (r^2) but that's just by cancelling out the q's without really understanding what they represent.
I assume either (q1) or (q2) is going to have to be e = 1.6*(10E-19), but as for the other? And even the value 'r'... at the surface, shouldn't the distance be 0?
Eh, I'm sure it's really simple, but any help would be great. Thanks.
 
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  • #2
A charge causes an electric field around itself with magnitude
E = kq/r^2 where r is the distance from the center of the charge to the point where you want to know the strength of E. The E field points away from a positive charge, toward a negative one.

A proton does have a (very small) radius, so E is not infinite at its surface.

Historically everyone used F = kqQ/r^2 for the force of one charge Q on another charge q. Later physicists became uncomfortable with the idea of "action at a distance". They figured that there had to be something extending from Q that did the pushing on q. It couldn't happen instantaneously - that would violate the speed of light speed limit for information travel. When q is moving, it feels the force instantly when it arrives at a new point in space, so the invisible hand of Q must already be there. So the electric field was invented and, where it matters, we now split the force formula into two parts:
E = kQ/r^2 - a charge causes an electric field around itself
F = qE - an electric field causes a force on a charge q

It turns out the E field really does exist and in fact it works with the magnetic field to make electromagnetic waves (radio, TV, etc.).
 
  • #3
Right, so it's not so much a case of having to conceptualise the equation F = kqQ/r^2 for a system that only involves one particle, so much as initially understanding F = kQ/r^2 for one particle, and then understanding that F = kqQ/r^2 can be used to relate the force felt by q due to the field of Q...
Okay, but in this case 'r' is still a bit of a mystery, since the Efield really only exists at the surface? So r (the distance from the charge) would be zero? Unless you say that the point-charge actually originates from the centre of the proton, so the distance from the surface to the "point" is r = 1.2*(10^-15)... That gives the correct answer, I believe, but is my thinking correct? More than anything I want to understand what you're actually doing with these equations rather than just solving them for numerical values.
 

What is the formula for calculating the electric field of a single charge?

The formula for calculating the electric field of a single charge is E = kQ/r^2, where E is the electric field in Newtons per Coulomb, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), Q is the magnitude of the charge in Coulombs, and r is the distance from the charge in meters.

How do I determine the direction of the electric field?

The direction of the electric field is always away from a positive charge and towards a negative charge. This can also be determined by using the right hand rule, where you point your thumb in the direction of the charge and your fingers will curl in the direction of the electric field.

Can the electric field of a single charge be negative?

Yes, the electric field of a single charge can be negative. This occurs when the charge is negative, which means the electric field points towards the charge instead of away from it.

What unit is used to measure the electric field?

The electric field is measured in Newtons per Coulomb (N/C), which is equivalent to Volts per meter (V/m).

How does the distance from the charge affect the electric field?

The electric field is inversely proportional to the square of the distance from the charge. This means that as the distance increases, the electric field decreases. Therefore, the electric field is stronger closer to the charge and weaker farther away from the charge.

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