Home
Class 12
PHYSICS
A uniform electric field is given as [E ...

A uniform electric field is given as `[E = 100 hat(i) N//C]` for `x gt 0 and E- 100 hat(i) N//C` for `x lt 0`. A right circular cylinder of length 20 cm and radius 5cm has its centre at the origin and its axis along the X-axis, so that one face is at `x= +10` cm and other is at `x= -10`cm.
What is the net outward flux through each flat face?

Text Solution

Verified by Experts

On the left face, the outward flux is `phi_(L)= E.Delta S= -100hat(i) .Delta S= 100Delta S`
Since, `hat(i).Delta S= -Delta S= 100 xx pi (0.05)^(2)`
`=0.785 N-m^(2)C^(-1)`
Similarly, on the right face, E and `Delta S` are parallel and therefore
`phi_(R )= E.Delta S`
`=0.785 N-m^(2)C^(-1)`
Promotional Banner

Similar Questions

Explore conceptually related problems

A uniform electric field is given as [E = 100 hat(i) N//C] for x gt 0 and E- 100 hat(i) N//C for x lt 0 . A right circular cylinder of length 20 cm and radius 5cm has its centre at the origin and its axis along the X-axis, so that one face is at x= +10 cm and other is at x= -10 cm. What is the net outward flux through the cylinder?

A uniform electric field is given as [E = 100 hat(i) N//C] for x gt 0 and E- 100 hat(i) N//C for x lt 0 . A right circular cylinder of length 20 cm and radius 5cm has its centre at the origin and its axis along the X-axis, so that one face is at x= +10 cm and other is at x= -10 cm. What is the flux through the side of cylinder?

A point charge of 2.0 μC is at centre of cubical Gaussian surface 9.0 cm on edge. what is the net electric flux through the surface?

obtain the equation of hyperbola in each of the following cases: centre at (0,0) conjugate axis along x-axis of length 6 and eccentricity 2.