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

## 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.

Thanks a lot for your help in advance!

## Answers and Replies

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

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
Science Advisor
Homework Helper
What's the correct answer according to your source?

What's the correct answer according to your source?

I don't know sorry

blue_leaf77
Science Advisor
Homework Helper
Then how did you know that your answer was wrong? Is it a homework given to you?

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.

Could someone please help with this question? Really struggling

blue_leaf77
Science Advisor
Homework Helper
Can you post your own answer?

Can you post your own answer?

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

Charles Link
Homework Helper
Gold Member
2020 Award
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
NascentOxygen
Staff Emeritus
Science Advisor
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?

blue_leaf77
Science Advisor
Homework Helper
I did it's in the OP. Do you know how to solve the question?
You mean this: 226331.3609 ? Isn't that your answer for (b). I asked for your answer for part (a).

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.

• Charles Link