Calculating Electric Field of Spherical Conductor

In summary, for a spherical conductor with a radius of x, an inner cubic cavity with a side of b, a charge of y inside the cavity, and a charge density of z on the surface of the sphere, the electric field at a point g>r can be calculated using Gauss's Law. The charge enclosed by the Gaussian surface is Q = y, and the electric field at point g is equal to (1/4*pi*eo)*∫chargedensity(z) *radius(x)^2 * sin(theta) d(x) d(theta) d (phi). This takes into account the charge density on the surface of the sphere.
  • #1
jackxxny
39
0

Homework Statement


Let's say I have :
Spherical conductor of radius=x;
Spherical conductor has a inner cubic cavity of side = b;
inside the cubic cavity we have a charge = y;
the surface of the sphere has a charge density = z;
I need to calculate the electric field at some point g, where g>r;


Homework Equations



I believe we can say that the sphere is uniformly charged.

The Attempt at a Solution



I don't need all of these information right?

All I do is

E= y/4*pi*E0*g2

?
 
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  • #2
HINT: This is a Gauss's Law problem. How much charge is enclosed by your Gaussian surface?
 
  • #3
the charge enclose to the surface is just Q.
 
  • #4
What do you mean by Q? Express the charge enclosed using the variables and quantities defined by the problem.
 
  • #5
sorry i mean y.

so all i have to do for the outside electric field is:y/4*pi*eog2
 
  • #6
That's all i have to do right?
 
  • #7
What about the charge density on the surface of the sphere? You need to take that into account.
 
  • #8
i think this may work:



Electric field at point g=


(1/4*pi*eo)*∫ p d(tao)

= (1/4*pi*eo)*∫chargedensity(z) *radius(x)^2 * sin(theta) d(x) d(theta) d (phi)

will that work?
 

FAQ: Calculating Electric Field of Spherical Conductor

How do you calculate the electric field of a spherical conductor?

To calculate the electric field of a spherical conductor, you can use the formula E = kQ/r^2, where E is the electric field, k is Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the charge of the spherical conductor, and r is the distance from the center of the conductor. This formula assumes that the spherical conductor has a uniform charge distribution.

What is the direction of the electric field at a point outside of a charged spherical conductor?

The direction of the electric field at a point outside of a charged spherical conductor is radially outward, away from the center of the conductor. This is because the electric field lines always point away from positive charges and towards negative charges, and a charged conductor will have an excess of either positive or negative charge on its surface.

Can the electric field inside a charged spherical conductor be zero?

Yes, the electric field inside a charged spherical conductor can be zero. This is because, according to Gauss's law, the electric field inside a conductor is zero. This means that any charges placed inside the conductor will redistribute themselves on the surface in such a way that the electric field inside is cancelled out.

How does the electric field of a spherical conductor change with distance?

The electric field of a spherical conductor follows an inverse square law, meaning that it decreases with distance from the center of the conductor. The electric field is strongest closer to the surface of the conductor and weaker further away, following the equation E = kQ/r^2. This relationship is the same as the electric field of a point charge.

Can the electric field of a spherical conductor be negative?

Yes, the electric field of a spherical conductor can be negative. This would occur if the spherical conductor has a net negative charge, causing the electric field to point towards the center of the conductor instead of away from it. However, it is important to note that the magnitude of the electric field is always positive, as it is a measure of the strength of the electric field at a particular point.

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