A semicircular ring of radius 0.5m is uniformly charged with a total charge of 1.4×10^-9 C. The electric field intensity at the centre of the ring is
Answers
Answer:
Explanation:
=> It is given that,
r=0.5m
q=1.4 x 10-19 C
electric field intensity = ?
=> Now think about a small element of length ( dl ) on the semicircular ring
suppose θ is the angle between the base and radius vector.
so, at centre angle subtended by this element = d θ
θ = (arc/radius)
or dl = Rd θ ...(1)
charge density = p
charge of this element = dq
dq = p(dl) = pRd θ ...(2)
=> Electric field dE = (k dq)/(R) 2 [due to this element at centre]
dE (magnitude) = (Kdq)/(R)2cos θ (i) + (kdq)/(R)2sin θ (-j)
dE = (Kdq)/(R)2[cos θ i - sin θ j]
dE = (kp)/(R )[ cos θ d θ (i) - sin θ d θ (j) ]
E = (kp)/R [ -2j ]
=> now we have p (charge per unit length) = (Q) /(πR)
Therefore, E = (2kQ)/(πR2) (-j)
placeing all the values,
E = 2 * 9 * 10^9 * 1.4 * 10^-9 / π * (0.5)^2
E = 32.08 V/m
Thus, the electric field intensity at the centre of the ring is 32.08 V/m