Question

The electric field in a region of space has the components \(E_y = E_z = 0\) and \(E_x = (4.30 N/(C·m))x\). Point A is on the y axis at y = 3.10 m, and point B is on the x axis at x = 2.00 m. What is the potential difference (in V) \(V_B – V_A\)?

Answer 1

@TuringTest
Mi amigo, are you any good with this sort of stuff? It seems like it should be easy, but then I start thinking... then things go downhill after that.

Answer 2

draw the pic first

Answer 3

\[\Delta V=\int\vec E\cdot d\vec s=\vec E\cdot \vec s\]

Answer 4

so what's the equation for \(\vec s\) is the real question here

Answer 5

*expression

Answer 6

Stupid question, but, what is \(\vec s\)

And you just kinda beat me to the punch lol

Answer 7

well, it's the vector representation of the path from A to B
so how might you represent the path from A to B in terms of \(\hat i\) and \(\hat j\) ?

Answer 8

\(\vec s=-3.1\hat i + 2 \hat j\)

Does that seem right?

Answer 9

exactly
now just do the dot product

Answer 10

\(\vec E=4.30\hat i\)

Is this correct, as it is denoted x?

Answer 11

argh sorry bad connection :P
yes, your notation is better than theirs haha, using x for a direction confuses things eventually

Answer 12

If so, it would be \(\Delta V=-13.33\hat i\)

Answer 13

how did you get that?

Answer 14

first of all, you should *never* have a vector left over after taking the dot product, since \(\hat i\cdot\hat i=1\)

Answer 15

Voltage is a scalar quantity like energy. It can be positive or negative between two points, but has no direction; only magnitude

Answer 16

your expression for \(\vec s\) is right, and your expression for \(\vec E\) is right, so you just didn't take the dot product right

Answer 17

Sorry, this connection is killing me.

I think I just forgot to drop the \(\hat i\)

Answer 18

No, the magnitude it wrong, too

Answer 19

\(4.3\times -3.1~?\)

Answer 20

write out \(\vec E\cdot\vec s\)

Answer 21

which direction is 3.1 in?

Answer 22

\(\large{\Delta V = \vec E \cdot \vec s=4.3\hat i \cdot [2\hat i -3.1\hat j]=8.6}\)

Stupid error haha

Answer 23

oh I see, this is my fault, I let you get away with writing the equation for s wrong the first time
yes you got it :)

Answer 24

I just plugged it into WileyPLUS and it said it was wrong....

Answer 25

let's see...

Answer 26

hm... if it's not a sig fig issue I'm pretty puzzled. This is a pretty straight-forward question

Answer 27

unless by them saying \[\vec E=4.3x\]they mean\[\vec E=4.3x\hat i\]

Answer 28

in which case you have to integrate, as that would mean that E changes along x...

Answer 29

That could be... how would I go about doing that then?

Answer 30

Here is what it looks like exactly.

Answer 31

they probably mean the fiels is proportional to x I guess, as they did in that cube problem.
To get the answer you do basically the same thing, but you have to integrate at the end\[\int\vec E\cdot d\vec s=\int(4.30x\hat i)\cdot(\hat j+\hat i)ds=\int_0^24.3xdx\]then integrate from x=0 to x=2...

Answer 32

field*

Answer 33

I literally get the same answer.

Answer 34

the details of the math are out of scope of the class, so I would use a symmetry argument instead of the "line integral" that I did
oh yeah?

Answer 35

oh yeah, you would, by halving and then squaring x

Answer 36

@TuringTest are you any good at circuits?

Answer 37

@austinL how's it going?

Answer 38

Wait... are we doing \(V_A-V_B\)
Don't we need to do \(V_B-V_A\)?

Answer 39

um... right... so we are going from higher potential to lower potential, so could the problem be a minus sign?
do you have unlimited chances?

Answer 40

AHA!!!! :D

That was the problem, we were doing it backwards. The answer is -8.6

Answer 41

And no, 4 tries total on all of my problems

Answer 42

I see, I always forget that detail :P
The idea there is that because we integrated in the direction of the E-field, we are basically moving away from a positive charge, which means we are going from a higher to lower potential.
Oh damn you, Benjamin Franklin, for making the electron negative and not positive.

Answer 43

I have always wondered how they picked these things... make North, North when referring to magnetism and negative, negative when referring to electricity.