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# Semicircular ring of radius R=0.5 m is uniformly charged with q=1.4C.Find electric field at CENTRE OF CURVATURE??

10 years ago

consider a small element of length dl on the ring ..let @ is the angle be radious vector with base then angle substended

by this element at center is d@ ...

dl = Rd@     ..............1

let charge dencity = p then charge of this element is dq

dq = p(dl) = pRd@ ..............2

electric field due to this element at center dE (magnitude)= kdq/R2

dE = Kdq/R2cos@ (i) + kdq/R2sin@ (-j)          with direction

dE = kdq/R2[cos@i - sin@j]

dE = kp/R [ cos@d@ (i) - sin@d@ (j) ]            lim from 0 to pi

E = kp/R [ -2j ]

now we have p (charge per unit length) = Q/piR so

E = 2kQ/piR2 (-j)

putting all the values we get

E = 3.2x1012 N/C

10 years ago

As E=q/4 pi epsilon knot R

R=0.5m  q=1.4C    1/4 pi epsilon knot =9*10 raise to power 9

therefore, E = 9*10 raise to power 9*1.4/0.5

=25.2*10 raise to power 9 C per m Kushagra Madhukar
one year ago
Dear student,

Consider a small element of length dl on the ring ..let @ is the angle be radious vector with base then angle substended by this element at center is d@ ...arc/radius = @
dl = Rd@     ..............1
let charge dencity = p then charge of this element is dq
dq = p(dl) = pRd@ ..............2
electric field due to this element at center dE (magnitude)= kdq/R2
dE = Kdq/R2cos@ (i) + kdq/R2sin@ (-j)          with direction
dE = kdq/R2[cos@i - sin@j]
dE = kp/R [ cos@d@ (i) - sin@d@ (j) ]            lim from 0 to pi
E = kp/R [ -2j ]
now we have p (charge per unit length) = Q/piR
so E = 2kQ/piR2 (-j)
putting all the values we get
E = 3.2x1012 N/C

Hope it helps.
Thanks and regards,
Kushagra