Understanding E=F/q: Solving for E at a Specific Point q_p

  • Thread starter Thickmax
  • Start date
In summary: The electric potential at a point is the difference in electric potentials of the point and... two infinitely far away points with the same charge.So the EPE of a charge is its charge multiplied by the electric potential at its... spot.
  • #1
Thickmax
31
8
Homework Statement
Calculate Electric field in 2 situations
Relevant Equations
See below
1625400101261.png
I have the above question and I think I am now understanding it.

We know that

E=F/q

F= K * ((q1*q2)/r^2)

So E = K * ((q1*q2)/r^2*q)But as we need to find the E at point q_p

I'm thinking for solution 1,

E = (K*(q/(r^2*q)))*2 as the forces are equal,

And for solution 2

E = K*(q/(r^2*q) - K*(q/(r^2*q)

Am I barking up the wrong tree?
 
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
You mixed up things a bit ...
To put things in some order:
Assume a positive charge ##q'## is in the middle. Then for each of the two cases do:
  • Calculate force ##F_1## from ##q_1## to ##q'## both in magnitude and direction
  • Calculate force ##F_2## from ##q_2## to ##q'## both in magnitude and direction
  • Calculate the electric field ##E_1=F_1/q'## from ##q_1## both in magnitude and direction
  • Calculate the electric field ##E_2=F_2/q'## from ##q_2## both in magnitude and direction
  • Calculate the total electric field which will be ##E=E_1+E_2## if ##E_1,E_2## have the same direction, or ##E=|E_1-E_2|## if ##E_1,E_2## have opposite directions (the bars stand for the absolute value).
 
  • Like
Likes Thickmax
  • #3
The question refers to a diagram "… shown on the right in the figure above". This is missing - it is not possible to tell if it is important or not.

This is a badly worded question because the diagrams for Situation 1 and Situation 2 show ##q_p=0## (I assume ##q_p## is the charge on the airplane). This means the airplane experiences no electric field even if an electric field is present (assuming we can ignore induced charges).

Presumably you are being asked for the electric field’s magnitude at the position of the the airplane.

For Situation 1, consider the symmetry. Any charge at the mid-point will experience two equal magnitude forces in opposite directions. What does that tell you?
 
  • #4
I got the answers nearly correct, but for the wrong situations :(

for situation 1 answer is 0,

For situation 2, the answer is 2*k*q/r^2

Thanks for your help :)
 
  • Like
Likes Steve4Physics and Delta2
  • #5
1625410893151.png


So I Know

E=F/q

F = k*q_1q_2/r^2

So k*q_1*q_2/r^2*q

so to rewrite I have put the sum of the 2, making sure q_2 acts at r_1+r_2^2

((k*q_p)/r_1^2)+((k*q_2)/(r_2+r_1)^2)Is this looking correct?
 
  • #6
Thickmax said:
I got the answers nearly correct, but for the wrong situations :(

for situation 1 answer is 0,

For situation 2, the answer is 2*k*q/r^2

Thanks for your help :)
Judging from the likes, @Steve4Physics and @Delta2 interpret the above post as saying you have now seen how to get the right answers. I read it as saying you are still getting them swapped over.
What you are overlooking is that q1 and q2 are on opposite sides, so if they are like charges their fields oppose and if they are opposite charges their fields add. Draw arrows for the field directions.
 
  • Like
Likes Delta2
  • #7
Thickmax said:
View attachment 285450

So I Know

E=F/q

F = k*q_1q_2/r^2

So k*q_1*q_2/r^2*q

so to rewrite I have put the sum of the 2, making sure q_2 acts at r_1+r_2^2

((k*q_p)/r_1^2)+((k*q_2)/(r_2+r_1)^2)Is this looking correct?

I'm confused both by the question and your attempt.
What does it mean by the "electric energy experienced" by a body? Does it mean the force, the field, or the potential energy?
In your attempt, you write the expression for the force cloud 2 exerts on cloud 1, but end up with the expression for the field cloud 2 and the plane produce at cloud 1.
 
  • #8
Thickmax said:
View attachment 285450

So I Know

E=F/q

F = k*q_1q_2/r^2

So k*q_1*q_2/r^2*q

so to rewrite I have put the sum of the 2, making sure q_2 acts at r_1+r_2^2

((k*q_p)/r_1^2)+((k*q_2)/(r_2+r_1)^2)Is this looking correct?
No. It is completely wrong and seems to have nothing to do with the question.

I agree with @haruspex that the question is badly written. Can you tell us what course you are on and where the question is from?

If you have 2 point charges, Q1 and Q2, separated by a distance ##d##, what is the standard formula for the electrical potential energy stored? If you don't know, refer to your notes and/or textbook. Then post your reply.

Also, note this is really a new question and ideally should be in a separate thread.
 
  • #9
the answer is apparently
k*(q_1*q_p/r_1 + q_1*q_2/(r_1+r_2))

It is a physics course on edx called pre university physics.

Looking at the difference, my answer is no a million miles away, but you make one mistake in physics and it doesn't really matter.

The question started in a new thread, but got moved into this pre-existing thread, but an administrator.
 
  • #10
Thickmax said:
the answer is apparently
k*(q_1*q_p/r_1 + q_1*q_2/(r_1+r_2))
Ok, that shows the question wants the net electric potential energy of the plane.
The EPE of a charge is its charge multiplied by the electric potential at its location.
 
  • #11
Thickmax said:
Looking at the difference, my answer is no a million miles away, but you make one mistake in physics and it doesn't really matter.t
That's not really true. A mistake (in physics or elsewhere) can be small or big. The mistake you made was fundamental, indicating you didn't really undertsand some basic physics.

It's a bit like confusing velocity and acceleration. It may appear to be a small mistake (e.g. units are m/s and m/s²) but the meanings are very different.

You used a formula giving force instead of the formula giving energy. Though the formulae look similar and are related, they mean very different things.

You never answered the question I asked in Post#8. Here is the question, slightly revised, if you would like to answer it:

You have 2 point charges, Q1 and Q2, separated by a distance d.
What is the formula for the force exerted on each charge?
What is the formula for the electrical potential energy stored?
Refer to your notes and/or textbook if required. And post your reply.

If we can't learn from our mistakes, we are doomed to repeat them!
 
  • Like
Likes Thickmax and Delta2
  • #12
Steve4Physics said:
That's not really true. A mistake (in physics or elsewhere) can be small or big. The mistake you made was fundamental, indicating you didn't really undertsand some basic physics.

It's a bit like confusing velocity and acceleration. It may appear to be a small mistake (e.g. units are m/s and m/s²) but the meanings are very different.

You used a formula giving force instead of the formula giving energy. Though the formulae look similar and are related, they mean very different things.

You never answered the question I asked in Post#8. Here is the question, slightly revised, if you would like to answer it:

You have 2 point charges, Q1 and Q2, separated by a distance d.
What is the formula for the force exerted on each charge?
What is the formula for the electrical potential energy stored?
Refer to your notes and/or textbook if required. And post your reply.

If we can't learn from our mistakes, we are doomed to repeat them!
Yes exactly, but when do you know you've made your mistake if everything you are doing is an assumption made on best case knowledge, you're adding tomatoes to a fruit salad...

Charge exerted on each charge can be found using Columbs law F=K*((Q1*Q1)/d)

Electrical potential energy = Total Energy - Kinetic Energy
 
  • #13
Thickmax said:
Electrical potential energy = Total Energy - Kinetic Energy
That's not useful here because you don't know the total energy (and besides, that could included gravitational PE, elastic PE, ...).
@Steve4Physics was looking for an equation relating electric potential, charge and electric potential energy.
 
  • #14
Thickmax said:
Yes exactly, but when do you know you've made your mistake if everything you are doing is an assumption made on best case knowledge, you're adding tomatoes to a fruit salad...

Charge exerted on each charge can be found using Columbs law F=K*((Q1*Q1)/d)

Electrical potential energy = Total Energy - Kinetic Energy
In fact, to a botanist, tomatoes are fruit - so adding them to a botanist's fruit salad should not be a problem!

You said:
"Charge [you meant 'Force'] exerted on each charge can be found using Columbs law F=K*((Q1*Q1)/d)"
Your mistakes are:
- not squaring the 'd'
- not using lower-case 'k'
- misspelling "Coulomb's"
(Also you don't need all those brackets.)

I would write F = kQ₁Q₂/d²

That sort of mistake can be avoided simply by learning what Coulomb's law is and how it is expressed mathematically. And if you don't remember, then you look it up. Also, attention to detail is important!

The formula for the energy stored by 2 (point) charges looks similar to the above formula for force. But it is not the same. Don't you have notes or a textbook you can refer to?
 
  • Like
Likes Thickmax and berkeman

FAQ: Understanding E=F/q: Solving for E at a Specific Point q_p

What is the equation E=F/q and how is it related to electric fields?

The equation E=F/q is known as Coulomb's law and it describes the relationship between electric force (F) and electric charge (q) in a system. It states that the electric field (E) at a specific point is equal to the force (F) exerted on a charge (q) at that point divided by the magnitude of the charge (q).

How is E=F/q used to solve for electric fields at a specific point?

To solve for the electric field (E) at a specific point, you need to know the value of the force (F) exerted on a charge (q) at that point and the magnitude of the charge (q). Once you have these values, you can simply plug them into the equation E=F/q and solve for E.

Can E=F/q be used to determine the direction of the electric field at a specific point?

Yes, E=F/q can be used to determine the direction of the electric field at a specific point. The direction of the electric field is in the same direction as the force exerted on a positive charge (q) at that point. If the charge is negative, the direction of the electric field will be opposite to the direction of the force.

How does the value of q affect the electric field at a specific point?

The value of q has a direct impact on the magnitude of the electric field at a specific point. As the value of q increases, the electric field also increases. This means that a larger charge will experience a stronger electric field compared to a smaller charge at the same point.

Are there any limitations to using E=F/q to solve for electric fields at a specific point?

Yes, there are some limitations to using E=F/q to solve for electric fields at a specific point. This equation only applies to point charges and does not take into account the effects of distance or the presence of other charges in the system. It also assumes that the electric field is constant at that point, which may not always be the case in real-world situations.

Back
Top