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# Electric Field due to a Charged Disk: Using Symmetry

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The figure below shows a disk of radius $R$ with a uniform surface charge density $\sigma$. We are interested in the electric field at a distance $d$ along a perpendicular line from the center of the disk of charge. In order to set up the problem, and to ultimately find the vector electric field at $d$, consider the ring at radius $r$ with width $dr$.

The charge in this circular region will result in an electric field, $\vec E_r$ at the distance $d$ along the center line perpendicular to the disk.

Which of the expressions below describes the vector field due to the ring of charge? Be sure to use symmetry to determine which components of the electric field add and which cancel.

A

$\vec E_r = \cfrac {r \sigma d \:dr}{2 \epsilon_0 (r^2 + d^2)^{3/2}}$

B

$\vec E_r = \cfrac {r \sigma d \:dr}{2 \epsilon_0 (r^2 + d^2)^{1/2}} \widehat z$

C

$\vec E_r = \cfrac {r^2 \sigma d \:dr}{2 \pi \epsilon_0 (r^2 + d^2)^{3/2}} \widehat z$

D

$\vec E_r = \cfrac {r \sigma d \:dr}{2 \epsilon_0 (r^2 + d^2)} \widehat z$

E

$\vec E_r = \cfrac {r \sigma d \:dr}{2 \epsilon_0 (r^2 + d^2)^{3/2}} \widehat z$