A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m, then the value of Q is nC. ( 0=8.85 10-12 C2/Nm2 and =3.14)
- 2.14
- 2.44
- 3.25
- 0.7
Answer
(1) 2.14
Solution
[figure]
Linear charge density: = Q r
Components of dE parallel to the diameter cancel in pairs; the perpendicular components add:
E= 0 k r d r2 = 2k r
⇒ E= 24 0 Q r2= Q2 2 0r2
∴ Q=2 2 0r2E
=2 (3.14)2 8.85 10-12 (0.35)2 100
=2.14 10-9 C =2.14 nC
Answer: (1)