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 $$300$$ from the horizontal, E→$$1$$ and E→$$2$$ both $$horizontal)$$ have magnitudes of $$2.50$$×$$104$$ NC−$$1$$ and $$7.00$$×$$104$$ NC−$$1$$,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 $$300$$ from the horizontal, E→$$1$$  and  E→$$2$$ both $$horizontal)$$ have magnitudes of  $$2.50$$×$$104$$  NC−$$1$$  and  $$7.00$$×$$104$$ NC−$$1$$,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$$1=AE1cos600=(0.0500$$  $$m)(0.0600$$  $$m)(2.50$$×$$104$$  NC−$$1)cos600=37.5$$  $$Nm2C$$−$$1$$Φ$$2=AE2cos1200=(0.0500$$  $$m)(0.0600$$  $$m)(7.00$$×$$104$$  NC−$$1)cos1200=$$−$$105$$  $$Nm2C$$−$$1So$$, the total flux isΦ$$=$$Φ$$1+$$Φ$$2=(37.5$$−$$105)Nm2C$$−$$1=$$−$$67.5$$  $$Nm2C$$−$$1q=$$Φε$$0=($$−$$67.5$$ $$Nm2C$$−$$1)$$ε$$0=$$−$$5.97$$×$$10$$−$$10$$ CThere 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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