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what is the graph of electric field due to uniformly charged circular disk?
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/(x2 + r2) down 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/(x2 + r2) up 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/(x2 + r2) cos Q.
From Fig. 7, we see that cos Q = x/(x2 + r2)1/2, so
dEx = kdq/(x2 + r2) cos Q = kdqx/(x2 + r2)3/2.
For the ring of Fig. 7 with charge s2pr dr,
dE = k(s2pr)xdr/(x2 + r2)3/2
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