# Basic electric force/field question I can't get

• Kuzon
In summary, a particle with a charge of +1.7 μC moves from point A to point B, experiencing a constant electric force along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +8.1 x 10-4 J. To find the magnitude of the electric force, the equation F=qV/d can be used, where V represents the change in voltage and d is the distance traveled. However, the equation F=V/d is not correct as the units on each side do not match. The correct equation is F=qV/d. To find the magnitude of the electric field, the equation E=kq/d^

## Homework Statement

A particle has a charge of +1.7 μC and moves from point A to point B, a distance of 0.26 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +8.1 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences.

F= qV/d

## The Attempt at a Solution

What's wrong with this reasoning:
F = V/d and V=Ed so E=F/q=V/d
Therefore F=qV/d
substitute all those values into that formula to get the electrostatic force and it's wrong?

I also tried b) using E=kq/d^2 and that didn't work either! (I got 226331.3609 as the answer)

By the way, there are no rounding requirements or anything, only a 2% margin of error so it's not that.

What you are given is ##\Delta E_p## which is equal to ##q\Delta V##. By the way, what's the correct answer?

blue_leaf77 said:
What you are given is ##\Delta E_p## which is equal to ##q\Delta V##. By the way, what's the correct answer?

Right so I tried going:
deltaE/q = deltaV
Then using F=(q*deltaV)/d to get the electrostatic force and I got the wrong answer again

blue_leaf77 said:
I don't know sorry

Then how did you know that your answer was wrong? Is it a homework given to you?

blue_leaf77 said:
Then how did you know that your answer was wrong? Is it a homework given to you?
It's this homework website and you get 5 attempts at getting the correct answer, it tells you when it is wrong but doesn't give you hints or the correct answer.

blue_leaf77 said:
I did it's in the OP. Do you know how to solve the question?

They give you the change in potential energy ## \Delta W=F \cdot d=q V ##. The change in voltage is often also referred to as the change in potential, but here they were quite specific that the particle experienced a change in its "potential energy", so that ## \Delta W=q V ##. (using ## W ## here for energy to save the letter ## E ## for the electric field. Notice also that ## V ## stands for voltage in electrical problems, and only becomes energy when multiplied by the charge ## q ##.)

Last edited:
Kuzon
Kuzon said:
What's wrong with this reasoning:
F = V/d and V=Ed so E=F/q=V/d
Therefore F=qV/d
substitute all those values into that formula to get the electrostatic force and it's wrong?
You start with F=V/d and show that F=qV/d? Something is wrong there. The units on each side of F=V/d don't match, so it can't be right.

What value did you use for V?

Kuzon said:
I did it's in the OP. Do you know how to solve the question?

They give you the change in potential energy ## \Delta W=F \cdot d=q V ##. The change in voltage is often also referred to as the change in potential, but here they were quite specific that the particle experienced a change in its "potential energy", so that ## \Delta W=q V ##. (using ## W ## here for energy to save the letter ## E ## for the electric field. Notice also that ## V ## stands for voltage in electrical problems, and only becomes energy when multiplied by the charge ## q ##.)
Thanks so much for this, I got it now.

## 1. What is the difference between electric force and electric field?

The electric force is the attraction or repulsion between two charged particles, while the electric field is the region around a charged particle where the force can be felt. Essentially, the electric field is the "messenger" of the electric force.

## 2. How is the strength of an electric field determined?

The strength of an electric field is determined by the distance between the charged particles and the magnitude of the charges. The closer the particles are, the stronger the electric field will be. Additionally, the larger the magnitude of the charges, the stronger the electric field will be.

## 3. What is the unit of measurement for electric force and electric field?

The unit of measurement for electric force is Newtons (N), which is the same unit used for measuring any type of force. The unit of measurement for electric field is Newtons per Coulomb (N/C).

## 4. How does the direction of an electric field relate to the direction of the electric force?

The direction of the electric field is always in the direction that a positive charge would move. The direction of the electric force, on the other hand, depends on the sign (positive or negative) of the charge that is experiencing the force. A positive charge will experience a force in the same direction as the electric field, while a negative charge will experience a force in the opposite direction.

## 5. Can an electric field exist without an electric force?

Yes, an electric field can exist without an electric force. This is because the electric field is a property of the space around a charged particle, while the electric force only exists when there is another charged particle present to interact with. Therefore, an electric field can exist without any other charged particles present, but an electric force cannot.

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