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An electric field is given by (6 i+5 j+3 k) N/C. The electric flux through a surface area 30 i m2 lying in YZ-plane (in SI unit) is :
(3) 180
= E A=(6 i+5 j+3 k) (30 i)
=6 30=180 N m2 C-1
The area vector of a YZ-plane surface is along i, so only Ex gives flux.
Answer: (3)
A thin metallic wire having cross sectional area of 10-4 m2 is used to make a ring of radius 30 cm. A positive charge of 2 C is uniformly distributed over the ring, while another positive charge of 30 pC is kept at the centre of the ring. The tension in the ring is N; provided that the ring does not get deformed (neglect the influence of gravity). (given, 14 0 = 9 109 SI units)
Two charges of -4 and +4 are placed at the points A(1, 0, 4) m and B(2, -1, 5) m located in an electric field E=0.20 i V/cm. The magnitude of the torque acting on the dipole is 8 10-5 Nm, where = .