Question

If I have an electric field around a charged object that is 2.95 x 10^6 N/C at a distance of 0.400m, how would I find the charge on the object?

If you could just direct me with a formula-then I'll do it and check back with my answer-Thank you for you help

Answer 1

Look up Gauss's law

Answer 2

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

Answer 3

Do it with a sphere with charge at the center and a radius of 0.400 m

Note - Surface area of sphere = (4/3) pi r^2

Answer 4

Oh wow, look at this:

http://academicearth.org/lectures/electric-flux-and-gauss-law

Answer 5

I took that course in that room in 1957 !

Answer 6

area = 4 pi r^2

Answer 7

Thanks for your assistance-I'm not sure still how to set it up but I'm going to go to those webiste and investigate it-

Answer 8

The idea is that for a sphere the field is constant over the surface if the charge is considered at the center.
Therefore E is proportional to charge/(4 pi r^2)

Answer 9

Electric field

To help visualize how a charge, or a collection of charges, influences the region around it, the concept of an electric field is used. The electric field E is analogous to g, which we called the acceleration due to gravity but which is really the gravitational field. Everything we learned about gravity, and how masses respond to gravitational forces, can help us understand how electric charges respond to electric forces.

The electric field a distance r away from a point charge Q is given by:

Electric field from a point charge : E = k Q / r2

The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. Like the electric force, the electric field E is a vector. If the electric field at a particular point is known, the force a charge q experiences when it is placed at that point is given by :

F = qE

If q is positive, the force is in the same direction as the field; if q is negative, the force is in the opposite direction as the field.

Answer 10

k = 1/(4 pi eo)
= 9 * 10^9 N m^2/C^2

Answer 11

Would I use this: E=kQ/r^2
and then I rearrange it and I have-Q=(2.95 *10^6N/C)(0.400m)^2/(8.988 * 10^9n*m^2/C^2)
=5.25E13
Is that even close?

Answer 12

3*10^6 * .16 / 10^10 = .5 * 10^-4

Answer 13

10^6/10^9 = 10^-3, I suspect you added

Answer 14

Note - I only carried one significant figure, doing in my head

Answer 15

Get it ?

Answer 16

Thanks-I think I'm beginning to get it-we just started this in class yesterday and I was trying to do the work due next week-thanks for your patience and explanations-I really appreciate it

Answer 17

Great, good luck !

Answer 18

Thank you very much!!