# Sheet with Hole

What is the electric field at a point P, a distance h = 29.9 cm above an infinite sheet of charge, with a charge distribution of 2.29 C/m^2 and a hole of radius r = 4.49 cm with P directly above the center of the hole, as shown in the figure?

Equations used:
Edisk= (-σ/2ε)[1-(z/sqrt(z^2 + r^2))]
Esheet= +σ/2ε
Enet= ƩE

I converted all the cm into m and plugged the numbers in. I got 3.64e10 N/C for the disk and 1.30e11 N/C for the sheet. I added the two together but my answer was wrong so i tried subtracting and still didn't get the right answer. I thought adding them would've been correct. Am I missing something?

CAF123
Gold Member
What is the electric field at a point P, a distance h = 29.9 cm above an infinite sheet of charge, with a charge distribution of 2.29 C/m^2 and a hole of radius r = 4.49 cm with P directly above the center of the hole, as shown in the figure?

Equations used:
Edisk= (-σ/2ε)[1-(z/sqrt(z^2 + r^2))]
Esheet= +σ/2ε
Enet= ƩE

I converted all the cm into m and plugged the numbers in. I got 3.64e10 N/C for the disk and 1.30e11 N/C for the sheet. I added the two together but my answer was wrong so i tried subtracting and still didn't get the right answer. I thought adding them would've been correct. Am I missing something?

Yes, adding your expressions together is correct. You are essentially considering the E field at the point P due to the sheet with surface charge density +σ and subtracting the E field due to the imaginary disk of negative surface charge density -σ.

Alternatively, just subtract the E field due to a disk from the E field due to the plane.
In your end expression, take the limit z>>R, and see if your result makes sense.

mfb
Mentor
Disk and sheet should have opposite signs (as removing stuff from a sheet reduces the electric field), so your formulas look right and one of your resulting values should have a minus sign. Add both, and you get a value smaller than for the sheet alone.

Did you check your values with WolframAlpha, Matlab or something similar?