Physics Homework -

Serendipity

Sorry to bother you guys, but I have this homework question that I'm stumped on:

"You're 1.5m from a charge distribution whose size is much less than 1m. You measure an electric field strength of 282N/C. You move to a distance of 2.0m, and the field strength becomes 119N/C. What is the net charge of the distribution?"

I was trying to find out a way to calculate the charge, but I'm not sure where to go with this. All I understand is that the electric field decreases with distance?? Any help would be greatly appreciated!

BaggedLancer

The answer is C. Statistically C is the correct answer 80% of the time.

tonyb

I hate physics....if you don't know the answer damn it...CHEAT!!!

jdhdiggs

42 -Its the answer to everything!

jakelm

i get 807.44n/c

A 42% increase ever .5m you move closer.

But I could be wrong

jakelm

282 @.1.5m
400.44 @1m
568.62 @.5m
807.44@0m

tonyb

jakelm wrote: »

282 @.1.5m
400.44 @1m
568.62 @.5m
807.44@0m

Looks more like Russ's booze level on a Friday night.:p

Serendipity

jakelm wrote: »

i get 807.44n/c

A 42% increase ever .5m you move closer.

But I could be wrong

How did you get that? I'm less interested in the answer, more interested in understanding the problem.

As I said before I was trying to calculate the charge, but there must be an easier way...

jakelm

appadv wrote: »

How did you get that? I'm less interested in the answer, more interested in understanding the problem.

As I said before I was trying to calculate the charge, but there must be an easier way...

Ok . You have the measurement between 2.0m (119) and 1.5m (282). 119 devided by 282 is .42

42% is the increase every .5m you get closer.

Now go from 1.5m to 1.0 and add 42%
Now go from 1.0m to .5. and add 42%
now go from .5m to 0m and again add 42%

a charge is constant, it does not change. so an increase between 2.0m and 1.5m is the same as the increase between .5m and 0m. So the % of increase or decrease never changes.

Demiurge

BaggedLancer wrote: »

The answer is C. Statistically C is the correct answer 80% of the time.

They've done studies, you know. 60% of the time it works, every time.

jakelm

Demiurge wrote: »

They've done studies, you know. 60% of the time it works, every time.

60% is an F.

But 100% of 60% of the time is pretty good.

Serendipity

jakelm wrote: »

Ok . You have the measurement between 2.0m (119) and 1.5m (282). 119 devided by 282 is .42

42% is the increase every .5m you get closer.

Now go from 1.5m to 1.0 and add 42%
Now go from 1.0m to .5. and add 42%
now go from .5m to 0m and again add 42%

a charge is constant, it does not change. so an increase between 2.0m and 1.5m is the same as the increase between .5m and 0m. So the % of increase or decrease never changes.

Okay, cool! I was getting nowhere when trying to calculate the charge, but then after you explained I realized the charge is constant and you can solve it your way.

Thanks!

Demiurge

jakelm wrote: »

60% is an F.

But 100% of 60% of the time is pretty good.

jakelm

Or I could be completely wrong and it could be something like this.


http://www.ux1.eiu.edu/~cfadd/1360/25ElPot/Continu.html

jakelm

I think I am competely wrong at this equation, I must look like an idiot...lol

Serendipity

Nah, I don't get it either...

jakelm

So what is it? Did you figure it out? What equation are you using?

Serendipity

jakelm wrote: »

So what is it? Did you figure it out? What equation are you using?

I thought I was close, but you can say that I didn't figure it out yet.

jakelm

I understand.

Is there something missing from your question? A charge distribution less than 1m? Thats it? Is it an atom? What formula are you using? I wish I could help.. Good luck

nms

This sounds like a Physics 208 problem (electromagnetism, charges, electricity, etc). Couldn't you use point charge formulas? I threw away all my notes from last semester so I can't really help you too much

jakelm

Inside the field is always zero.

Maybe this..

sigma=-qd/[r^2+_d^2]^{3/2}

Dont ask me....I read somewhere about using an image charge..

jakelm

http://www.physics247.com/physics-homework-help/electric-field.php

Serendipity

nms wrote: »

This sounds like a Physics 208 problem (electromagnetism, charges, electricity, etc). Couldn't you use point charge formulas? I threw away all my notes from last semester so I can't really help you too much

I'm not sure if I could use point charge formulas in this case???

Anyways, the question in the first post is the entire question.