# Please help with image method

1. Apr 13, 2007

### brad sue

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 attach a picture.
I get lost at some point

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

to answer the question 1-:
electric field from charge Q:
EQ=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)

Here is my first problem: the direction of the two electric fields EQ and E-Q.
I am not sure about the electric field of the expressions.

(I suspect I get some mistakes because I tried a similar problem but with a more simple expression for the 2 fields and when I add them, I get zero)

I would really have some help with this problem.

please can someone help me?
Thank you

#### Attached Files:

• ###### method image.doc
File size:
30 KB
Views:
65
2. Apr 14, 2007

### Mentz114

Reverse the direction of the unit vector y for $$E_{-Q}$$ because the polarity is opposite to $$E_{Q}$$.

3. Apr 14, 2007

### brad sue

thanks.
I will try to continue from there.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook