Electric Potaential of two spheres

  • Thread starter Thread starter crybllrd
  • Start date Start date
  • Tags Tags
    Electric Spheres
Click For Summary

Homework Help Overview

The problem involves calculating the electric potential and electric field at the surfaces of two charged spherical conductors, representing an airplane and its needle extensions, connected by a conducting wire. The spheres have different radii and a total charge is provided.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the charges on the spheres and their radii, attempting to derive the electric potential and field. There are questions about the units of electric potential and the calculations involved in determining the correct values.

Discussion Status

Some participants have provided feedback on the calculations, noting discrepancies in the units and magnitudes. There is ongoing exploration of the correct application of formulas, with one participant expressing confusion about their calculator's output. Another participant has indicated they resolved their confusion.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can use or share. There is an emphasis on ensuring the calculations are correct and consistent with the physical principles involved.

crybllrd
Messages
120
Reaction score
0

Homework Statement



Electric charge can accumulate on an airplane in flight. You may have observed needle-shaped metal extensions on the wing tips and tail of an airplane. Their purpose is to allow charge to leak off before much of it accumulates. The electric field around the needle is much larger than the field around the body of the airplane and, can become large enough to produce dielectric breakdown of the air, discharging the airplane. To model this process, assume that two charged spherical conductors are connected by a long conducting wire and a charge of 9.00 µC is placed on the combination. One sphere, representing the body of the airplane, has a radius of 6.00 cm, and the other, representing the tip of the needle, has a radius of 2.00 cm.
(a) What is the electric potential (V) of each sphere?
r = 6.00 cm
r = 2.00cm

(b) What is the electric field (V/m, direction) at the surface of each sphere?
r = 6.00 cm
r = 2.00cm

Homework Equations



r1 = 6e-2
r2 = 2e-2


The Attempt at a Solution


q1/q2=r1/r2
q1=3q2

q1+q2=9e-6

3q2+q2=9e-6
4q2=9e-6
q2=2.25e-6

q1+2.25e-6=9e-6
q1=6.75e-6

v1=kq1/r1

v1=101

When I submit it, it tells me that "Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully."

Any help?
 
Physics news on Phys.org
The potential is not a number, it has some unit. In what units is the potential 101? Check the magnitude of the data and give the potential in volts. ehild
 
Thanks for the quick reply.

v1=kq1/r1

v1=8.99e9 Nm2/C2*1.95e-5C/6e-2m

v1=101 Nm/C

Nm/C is a V
 
crybllrd said:
Thanks for the quick reply.

v1=kq1/r1

v1=8.99e9 Nm2/C2*1.95e-5C/6e-2m

v1=101 Nm/C

Nm/C is a V

q1=6.75 e-6C, is not it?

Check the magnitudes: You multiply 109 with 10-6; it is 103, and divide by 10-2: you get 105.

ehild
 
101e5 tells me that I am off by a multiple of ten.

I am still confused though- in my calculator I put in
(K*6.75*10e-6)/6e-2
(I have K stored as the ke constant)
and get 101. It sounds like you're telling me that my calculator cannot handle scientific notation, and I need to tag on the " e5 " to my calculation of 101. Why is it not giving me the right answer? Or am I misunderstanding this?
 
Never mind, I figured it out.
Thanks a lot!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
4K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 36 ·
2
Replies
36
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K