The electric field `vecE_1` at one face of a parallelopiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field `vecE_2` is also uniform over the entire face and is directed into that face (as shown in figure). The two faces in question are inclined at `30^@` from the horizontal, `vecE_1 and vecE_2`(both horizontal) have magnitudes of `2.50 xx 10^(4) NC^(-1)` and `7.00 xx 10^(4)NC^(-1)`. respectively. Assuming that no other electric field lines cross the surfaces of the parallelopiped, the net charge contained with in is

Options

A. `-67.5epsilon_0C`
B. `37.5epsilon_0C`
C. `105 epsilon_0C`
D. `-105 epsilon_0C`

Answer

A

Explanation

a. To find the charge enclosed,
we need the flux through the
parallelepiped:
`Phi_1 = AE_1cos60^@`
`=(0.0500m)(0.0600m)(2.50 xx 10^(4)NC^(-1)) cos 60^@`
`= 37.5 Nm^2C^(-1)`
`Phi_2 = AE_2cos120^@`
`= (0.0500m)(0.0600m)(7.00 xx 10^(4)NC^(-1))cos120^@`
` =-105Nm^2C^(-1)`
So the flux is
`Phi = Phi_1 + Phi_2 = (37.5 - 105)Nm^2C^(-1) = -67.5Nm^2C^(-1)`
`q = Phiepsilon_0 = (-67.5Nm^2C^(-1))epsilon_0 = -5.97 xx 10^(-10)C`
There must be a net charge (negative) in the parallelepiped since
there is a net flux flowing into the surface. Also, there must be an
external field, otherwise all lines would point towards the slab.