Question

1) A small object of mass 3.8 g and charge 15 mC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field?

2)Two small identical conducting spheres are placed with their
centers 0.3 m apart. One is given a charge of
10 10^{-9} C, the other a charge of
–18 10^{-9} C.

(a) Find the electrostatic force exerted on one sphere by the other.

(b) The spheres are connected by a conducting wire. Find the electrostatic force between the two after equilibrium is reached, where both spheres have the same charge

3)

An electric field of magnitude 3.25 10^{5}
N/C points due south at a certain location.

(a) Find the magnitude of the force on a +6 mC charge at this
location.

(b) Find the direction of the force on a +6 mC charge at this location.

Answer #1

1. The weight of the object acts vertically downward. As the object is held motionless, there must be a force on the object acting vertically upward, which in this case, will be provided by the electric force. Now, as the object is positively charged, so, to have the direction of electric force vertically upward, the electric field must point vertically upward.

For equilibrium, we must have,

Weight of object = Electric force on the object

Or, mg = qE

Or, E = mg/q

Here, m = 3.8*10^{-3} kg, q = 15*10^{-3} C,
so,

Electric field, E = (3.8*10^{-3}*9.8/15*10^{-3})
N/C

= 2.482 N/C

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magnitude
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