The electric field intensity at all points in space is given by `vecE = sqrt(3)hati - hatj` `V//m`. The nature of equipotential lines is x-y plane is given by

Options

A. a.The magnitude of electric flux increases from initial value as `theta` is increased
B. b.The magnitude of electric flux increases from initial value as `theta` is decreased
C. c.The magnitude of electric flux may increase or decrease from initial value as `theta` is changed
D. d.The magnitude of electric flux will decrease from initial value as `theta` is changed.

Answer

D

Explanation

The direction of electric field is in x-y plane as shown in figure
The magnitude of electric field is E =
`sqrt(E_(x)^(2)+E_(y)^(2))=sqrt(3+1)=2V//m`
The direction of electric field is given by `theta =`
`"tan"^(-1)(E_(y))/(E_(x))"tan"^(-1)(1)/(sqrt(3))=30^(@) ="tan"^(-2)(1)/(sqrt(3))=30^(@)`
Hence electric field is normal to square frame LMNO as shown in figure
`:.` electric flux `vec(E) = vec(A)`
`= E A cos 0 = 2 xx 1 = 2 V//m`
`= E A cos 0 = 2 xx 1 = 2 V//m`
`:.` C is correct of Q.7
Since flux is maximum at `0 = 60^(@)`, rotation of `30^(@)` either way would lead to decrease in flux
`:.` D is correct option of Q.9
Lines ON and NM are both normal to uniform electric field `vec(E)`. Hence work done by electric field as a point charge `2 mu C` is taken from O to M is zero.
`:.` A is correct option of Q.8.