Electric Potaential of two spheres

  • Thread starter Thread starter crybllrd
  • Start date Start date
  • Tags Tags
    Electric Spheres
AI Thread Summary
The discussion revolves around calculating the electric potential and electric field at the surfaces of two charged spherical conductors representing an airplane and its needle extensions. The first sphere has a radius of 6.00 cm and the second has a radius of 2.00 cm, with a total charge of 9.00 µC distributed between them. The user initially miscalculated the electric potential, resulting in confusion over the units and magnitude of the answer. After clarification, it was confirmed that the potential should be expressed in volts, and the user realized their error in scientific notation calculations. The conversation highlights the importance of careful unit management and accuracy in calculations when dealing with electric potential.
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top