# Math Help - method of image

1. ## method of image

HI I dont understand the method of image.
I have this problem:

A uniform charged disk is suspended ( at a distance h) above an infinite conducting plane ( disk is parallel to the plane). The disk carries a charge Q and has a diameter d. Assume d>>h.
1- Calculate the electric field vector near the center and below the disk.
2- Calculate the capacitance of the system using the formula C=Q/V.
3- Calculate the capacitance of the system using the energy formulas.
I tried to draw a picture.

^Y(unit vector direction)
|
|
_____________ (disk)
*
h ( distance between disk and plane)
*
*
\\\\\\\\\\\\\\\\\\ (plane) .-----> x(unit vector)

I dont understand the procedure.
Thank you

HI I dont understand the method of image.
I have this problem:

A uniform charged disk is suspended ( at a distance h) above an infinite conducting plane ( disk is parallel to the plane). The disk carries a charge Q and has a diameter d. Assume d>>h.
1- Calculate the electric field vector near the center and below the disk.
2- Calculate the capacitance of the system using the formula C=Q/V.
3- Calculate the capacitance of the system using the energy formulas.
I tried to draw a picture.

^Y(unit vector direction)
|
|
_____________ (disk)
*
h ( distance between disk and plane)
*
*
\\\\\\\\\\\\\\\\\\ (plane) .-----> x(unit vector)

I dont understand the procedure.
Thank you
I'm not clear on what part you don't understand...

The method of images is used when we are finding the electric field due to a charge (density) in the presence of a conductor. We may assume that we can model this same electric field by replacing the conductor with an effective charge density, along with suitable boundary conditions (ie V = 0) where the surface of the conductor was. To be clear: the electric field of the original problem will be the same as the vector sum of the electric fields of the two charge densities.

The case of the infinite conducting plane is quite simple: replace the conductor with a charge density that is the "mirror" of the original charge density. For example, if we have a point charge q a perpendicular distance of "a" from the plane, we put a "mirror charge" -q at distance a from the other side of the plane. (An obvious extension of this gives the case of a charge distribution on one side of the plane.

If the problem is with part 2, you need to calculate the total charge induced on the surface of the conductor. (Hint: It will be -Q. This is a result of Gauss' Law since the interior of the conductor must have E = 0.) We may find this by observing that the normal component of the electric field at the surface of a conductor is given by:
E_n = (surface charge density)/(epsilon0) <-- Make sure you are using the correct units for your coursework!

-Dan

3. Thank you for helping me.
My first problem is about finding the correct expression (with the signs and direction)

the electric fiel for a disk ( in free space)is:
E=Qy/(2*pi*epsilon)*(1/y-1/(sqrt(y^2+R^2)) y (unit vector)

electric field from charge Q:
E[Q]=Q(h-y)/(2*pi*epsilon)*(1/(h-y)-1/(sqrt((h-y)^2+R^2)) (-y) (unit vector)

electric field from charge -Q ( the image):
E[-Q]=-Q(h+y)/(2*pi*epsilon)*(1/(h+y)-1/(sqrt((h+y)^2+R^2)) (-y) (unit vector)

The direction for both fields is downward.
are the expression for E correct?

Thank you for helping me.
My first problem is about finding the correct expression (with the signs and direction)

the electric fiel for a disk ( in free space)is:
E=Qy/(2*pi*epsilon)*(1/y-1/(sqrt(y^2+R^2)) y (unit vector)

electric field from charge Q:
E[Q]=Q(h-y)/(2*pi*epsilon)*(1/(h-y)-1/(sqrt((h-y)^2+R^2)) (-y) (unit vector)

electric field from charge -Q ( the image):
E[-Q]=-Q(h+y)/(2*pi*epsilon)*(1/(h+y)-1/(sqrt((h+y)^2+R^2)) (-y) (unit vector)

The direction for both fields is downward.
are the expression for E correct?
They look good to me.

-Dan