# 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.

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blue_leaf77
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
Homework Helper

I don't know sorry

blue_leaf77
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.

blue_leaf77
Homework Helper

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

Homework Helper
Gold Member
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:
NascentOxygen
Staff Emeritus
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
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 $q$.)