(a) A 2.00-µF capacitor is connected to a 18.0-V battery. How much energy is stored in the capacitor?

1.8 x 10⁻⁵J.

Explanation:

E = 1/2 x C x V²

Where,

E = energy stored in a capacitor

C = capaciitance

V = Voltage

From the question, given:

C = 2.00μF

= 2.00 x 10⁻⁶F

V = 18.0V

E = 1/2 x 2.00 x 10⁻⁶ x 18.0

= 1.8 x 10⁻⁵J.

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

ice to water 1 water to vapour 2 latent heat