Electric field of charge distributions

As you can see, the field due to the dipole is not zero inside the sphere. Therefore the field due to the sphere must be such as to cancel the field due to the dipole.
  • #1
demonhunter19
10
0

Homework Statement


15-54.jpg

A small, thin, hollow spherical glass shell of radius R carries a uniformly distributed positive charge +Q, as shown in the diagram above. Below it is a horizontal permanent dipole with charges +q and -q separated by a distance s (s is shown greatly enlarged for clarity). The dipole is fixed in position and is not free to rotate. The distance from the center of the glass shell to the center of the dipole is L.

The charge on the thin glass shell is +6e-09 coulombs, the dipole consists of charges of 4e-11 and -4e-11 coulombs, the radius of the glass shell is 0.15 m, the distance L is 0.45 m, and the dipole separation is 2e-05 m. Calculate the net electric field at the center of the glass shell. The x-axis runs to the right, the y-axis runs toward the top of the page, and the z axis runs out of the page, toward you.

E^^->_(net) = < , , > N/C
15-55.jpg

If the upper sphere were a solid metal ball with a charge +6e-09 coulombs, what would be the net electric field at its center?
E^^->_(net) = < , , > N/C

Which of the diagrams below best shows the charge distribution in and/or on the metal sphere?
foil_a1.gif
A
foil_i4.gif
B
foil_i2.gif
C
foil_i3.gif
D
foil_i1.gif
E
foil_i5.gif
F
foil_i6.gif
G
foil_i7.gif
H
foil_i8.gif
J
foil_k1.gif
K

Homework Equations


Electric Force of Dipole: (1/4*pi*epsilon_0)(q*s)/r^3
Electric Force of Sphere: (1/4*pi*epsilon_0)(Q/r^2)

The Attempt at a Solution



The question is kind of confusing to me, as it asks to calculate the net electric field at the CENTER of the glass shell. Right now I'm assuming that the electric field AT THE CENTER of the glass shell is zero (is this assumption not correct? since r<R inside the sphere) and that to answer this first part of the question, it would just be the electric field of the DIPOLE, which is <.000079,0,0>N/C.
Also, if I do calculate the electric field of the glass shell, from the center of the glass shell to the center of the dipole, whose distance is .45m, then I would obtain <0,266.667,>N/C.

I am really confused about this first part.

As for the second part, I would assume it would be the same values as when I calculate the net electric field at the center of the glass shell.

For the last part (multiple choice) I'm not entirely sure, but I'm leaning towards C.

HELP WOULD BE MUCH APPRECIATED, and thank you in advance.
 
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  • #2
You solved the first part correctly. To find the total field, you must sum up the fields of the glass sphere and of the dipole (principle of superposition). Glass sphere gives zero because charge is distributed uniformly, dipole gives according to the formula.

In the second part, not that the field inside any conductor is always zero, i.e. the charge on the surface of the metal sphere distributes in such a way so as to cancel the field of the dipole inside the sphere.
 
  • #3
So, what you're saying for the second part is that the only electric field within this system is the metal sphere? Would the dipoles have any effect? If not, then the only net electric is to the negative y direction (as seen in the picture)? So then to calculate the electric field of just the metal sphere, it is 1/(4*pi*epsilon_0)*Q/A?
 
  • #4
demonhunter19 said:
So, what you're saying for the second part is that the only electric field within this system is the metal sphere? Would the dipoles have any effect? If not, then the only net electric is to the negative y direction (as seen in the picture)? So then to calculate the electric field of just the metal sphere, it is 1/(4*pi*epsilon_0)*Q/A?

I'm not saying that. All the charge in the surroundings contribute to the electric field, both the sphere and the dipole. However, the charge on the sphere distributes in such a way as to oppose the dipole and make the field zero inside the sphere. It's the nature of conductors. Under any circumstances the field inside the conductor is zero.
 

FAQ: Electric field of charge distributions

What is an electric field?

An electric field is a physical quantity that describes the influence of an electric charge on other charges in its vicinity. It is a vector field, meaning it has both magnitude and direction, and is created by the presence of electric charges.

How is the electric field of a charge distribution calculated?

The electric field of a charge distribution is calculated by using Coulomb's law, which states that the electric field at a point is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.

What is the difference between a point charge and a charge distribution?

A point charge is a single, isolated charge with no physical dimensions, while a charge distribution is a collection of charges that have a finite size and are distributed in space. The electric field of a point charge can be described by a single value, while the electric field of a charge distribution varies at different points in space.

How does the shape and arrangement of charges affect the electric field of a charge distribution?

The shape and arrangement of charges in a distribution can greatly affect the electric field. For example, a spherical charge distribution will have a more uniform electric field compared to a linear charge distribution. Similarly, the direction of the electric field can also be influenced by the arrangement of charges.

What are some real-life applications of understanding the electric field of charge distributions?

Understanding the electric field of charge distributions is crucial in many fields, including electronics, telecommunications, and energy production. For example, it is used in designing and optimizing circuits, determining the strength of signals in communication systems, and calculating the forces on charged particles in particle accelerators.

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