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# what is the graph of electric field due to  uniformly charged circular disk?

36 Points
10 years ago

Dear student,

Divide the disk into rings of radius r and thickness dr. If you cut the ring and pull it out into a rectangle, it has length 2pr and thickness dr. Its area dA = 2pr dr. If its charge per unit area = dq/dA = s, dq = sdA = s2pr dr. The field due to a small amount of charge dq at the top of the ring of radius r is dEtop = kdq/(x+ r2down and to the right as shown in Fig. 7.  The electric field for an equal amount of charge dq from the bottom is dEbottom = kdq/(x+ r2up and to the right as shown in Fig. 7.  The dE's from the entire ring form a cone about point P.  The components in the Y and Z direction cancel, leaving the resultant field in the +X - direction.

The component of the dE in the X-direction is:

kdq/(x+ r2) cos Q.

From Fig. 7, we see that cos Q = x/(x+ r2)1/2,    so

dEx = kdq/(x+ r2) cos Q = kdqx/(x+ r2)3/2.

For the ring of Fig. 7 with charge  s2pr dr,

dE = k(s2pr)xdr/(x+ r2)3/2