Electric Potaential of two spheres

In summary, on an airplane, electric charge can accumulate and cause discharge. To prevent this, needle-shaped metal extensions on the wing tips and tail allow charge to leak off. The electric field around the needle is larger than the body of the airplane, and to model this, assume two charged spherical conductors connected by a long wire. With a 9.00 µC charge, the potential at the surface of the spheres is 101 V, and the electric field is 1.69 x 10^5 V/m at the body and 5.06 x 10^5 V/m at the needle tip.
  • #1
crybllrd
120
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?
 
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  • #2
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
 
  • #3
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
 
  • #4
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
 
  • #5
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?
 
  • #6
Never mind, I figured it out.
Thanks a lot!
 

1. What is the electric potential of two spheres?

The electric potential of two spheres refers to the amount of electrical energy per unit charge at a specific point between the two spheres. It is a measure of how much work is needed to move a unit of charge from one sphere to another.

2. How is the electric potential of two spheres calculated?

The electric potential of two spheres can be calculated using the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of one sphere, and r is the distance between the two spheres.

3. Can the electric potential of two spheres be negative?

Yes, the electric potential of two spheres can be negative. This means that the electrical energy at that point is less than a reference point, typically at infinity. It can also be positive, meaning the electrical energy is greater than the reference point.

4. How does the electric potential change as the distance between the spheres increases?

As the distance between the two spheres increases, the electric potential decreases. This is because the electrical energy is spread out over a larger distance, resulting in a lower potential. Similarly, as the distance decreases, the electric potential increases.

5. What are some real-life applications of the electric potential of two spheres?

The electric potential of two spheres has many applications in daily life, such as in capacitors, batteries, and lightning rods. It is also used in industry for electrostatic painting, electroplating, and electrostatic separation of materials.

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