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

1. May 14, 2016

### Kuzon

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
F= qV/d

3. 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.

2. May 14, 2016

### blue_leaf77

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

3. May 14, 2016

### Kuzon

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

4. May 14, 2016

### blue_leaf77

5. May 14, 2016

### Kuzon

I don't know sorry

6. May 14, 2016

### blue_leaf77

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

7. May 14, 2016

### Kuzon

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.

8. May 14, 2016

### Kuzon

9. May 14, 2016

### blue_leaf77

10. May 14, 2016

### Kuzon

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

11. May 14, 2016

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: May 14, 2016
12. May 14, 2016

### Staff: Mentor

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?

13. May 14, 2016