The electric field ${\vec{E}}_{1}$ 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 ${\vec{E}}_{2}$ 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^o from the horizontal, ${\vec{E}}_{1}$ and ${\vec{E}}_{2}$ (both horizontal) have magnitudes of $2.50 \times 10^{4} {NC}^{- 1} and 7.00 \times 10^{4} {NC}^{- 1}$ , respectively. Assuming that no other electric field lines cross the surfaces of the parallelopiped, the net charge contained within is

Difficult

Solution

To find the charge enclosed, we need the flux through the parallelepiped:
$\begin{matrix} Φ_{1} = A E_{1} \cos 60^{\circ} \\ = (0.0500 m) (0.0600 m) (2.50 \times 10^{4} {NC}^{- 1}) \cos 60^{\circ} \\ = 37.5 {Nm}^{2} C^{- 1} \\ Φ_{2} = A E_{2} \cos 120^{\circ} \\ = (0.0500 m) (0.0600 m) (7.00 \times 10^{4} {NC}^{- 1}) \cos 120^{\circ} \\ = - 105 {Nm}^{2} C^{- 1} \end{matrix}$
So, the total flux is
$\begin{array}{l} Φ = Φ_{1} + Φ_{2} = (37.5 - 105) {Nm}^{2} C^{- 1} = - 67.5 {Nm}^{2} C^{- 1} \\ q = Φ ε_{0} = (- 67.5 {Nm}^{2} C^{- 1}) ε_{0} = - 5.97 \times 10^{- 10} C or - 67.5 ε_{0} C \end{array}$
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 toward the slab.

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