The electric field 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 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 $$30o$$ from the horizontal, E→$$1$$ and E→$$2$$ (both$$ $$horizontal) have magnitudes of $$2.50$$×$$104NC$$−$$1$$ and $$7.00$$×$$104NC$$−$$1$$, respectively. Assuming that no other electric field lines cross the surfaces of the parallelopiped, the net charge contained within is

Question

The electric field 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 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 $$30o$$ from the horizontal, E→$$1$$ and E→$$2$$ (both$$ $$horizontal) have magnitudes of $$2.50$$×$$104NC$$−$$1$$ and $$7.00$$×$$104NC$$−$$1$$, respectively. Assuming that no other electric field lines cross the surfaces of the parallelopiped, the net charge contained within is

Infinity Learn · Accepted Answer

To find the charge enclosed, we need the flux through the parallelepiped:Φ$$1=AE1cos$$⁡$$60$$∘$$=(0.0500m)(0.0600m)2.50$$×$$104NC$$−$$1cos$$⁡$$60$$∘$$=37.5Nm2C$$−$$1$$Φ$$2=AE2cos$$⁡$$120$$∘$$=(0.0500m)(0.0600m)7.00$$×$$104NC$$−$$1cos$$⁡$$120$$∘$$=$$−$$105Nm2C$$−$$1So$$, the total flux isΦ$$=$$Φ$$1+$$Φ$$2=(37.5$$−$$105)Nm2C$$−$$1=$$−$$67.5Nm2C$$−$$1q=$$Φε$$0=$$−$$67.5Nm2C$$−$$1$$ε$$0=$$−$$5.97$$×$$10$$−$$10C$$ or −$$67.5$$ε$$0CThere$$ 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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