invadingdingo

I don't understand how it doesn't change

But wouldn't we need to take the derivative of λ as well?

how does $-dq$ = lambda($dx$)?

So I'd be integrating the E equation, kQ/r^2?

Ok, so integrate and "add" all the pieces together?

I don't know how to go about finding the magnitude

I'm a little confused on part b.

So I worked part a of the problem and am getting -5.38e-14C/M. Does that look right? I want to make sure before going forward with the problem.

Alright, I don't think we covered that in class, but it's good to know.

And anytime it isn't the entire charge, we have to use calculus.

I got you. So we can think of formulas that don't use calculus as an "easier" method that only works sometimes, i.e, when it is talking about the entirety of a charge?

Ooh, I get it now

@Semiclassical's

Wait, I'm a little confused on what that statement means.

I thought integration wasn't involved until working with surfaces and volumes

It does say uniformly distributed

which really confused me

Chegg has it listed as -q/L for this equation

Ok, so that's what I was going to ask

And we only get 5 different attempts at a submission, which defeats the purpose of physics problem solving altogether. I shouldn't be scared to try new methods to get an answer..

Yeah, I hate the freaking software that this school uses.

Charge density along a line is lambda = q/L, right?

Ah, so just associate it with a new variable altogether?

fC *

not +4.47mC?

So use the quantity -4.47fC

HAHA

That's a really dumb way to go about that haha. It totally looks like the negatives should cancel

Ahh..

Seems to work on my end

Try that

edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c22/… There's the image

It reoccurs in part e.

(d) the rod and (e) a particle of charge -q = -4.47 fC that replaces the rod?

In the figure a nonconducting rod of length L = 8.34 cm has charge -q = -4.47 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (positive angle relative to the positive direction of the x axis) of the electric field produced at point P, at distance a = 14.7 cm from the rod? What is the electric field magnitude produced at distance a = 83 m by

That's the beginning to my problem. Do you think there's a typo with -q? I just don't think it makes sense to say -q = -charge..

In the figure a nonconducting rod of length L = 8.34 cm has charge -q = -4.47 fC uniformly distributed along its length.

I had a quick question

Ahh, I wanted to see what the general consensus on sites like Chegg is.

Are any of you guys physics professors?

I'm a super noob to all of this physics stuff, haha

Ahh, I guess it is much more accurate that way

It was to do with the notation. They wanted the raw answer, 189736.6596, not 1.8e5

I tried entering the number without scientific notation and it worked. Nowhere did it specify whether or not to use scientific notation or not....

Wait, nevermind.

which isn't considered right..

The answer to that is 1.8e5

Velocity should then be v = sqrt(2ax), where a is 1.8e12 and x is .01m(gotten from 1cm)

That's my problem. I solved part a. It's correct and has an answer of 1.8e12.