Explanation:

An electric field stronger than 3.0 x 106 V/m will cause SPARKS to occur in air. What is the maximum potential DIFFERENCE that can be applied across two metal plates 1.0 mm apart before sparking begins ?

V = Eh = 3.0 x 106 x 1.0 x 10-3 = 3000 Volts

A cloud is at a potential of 8.0 x 106 V relative to the ground. A charge of 40 C is transferred in a lightning stroke between the cloud and the ground. FIND the energy dissipated.

Energy dissipated = QV = 40 x 8.0 x 106 = 3.2 x 108 J

In charging a 20 kg storage battery, a total of 2.0 x 105 C is transferred from one set of electrodes to another. The potential difference between the electrodes is 14 V.

(a) How much energy is stored in the battery ?

Electrical PE = QV = 2.0 x 105 x 14 = 2.8 x 106 J

(b) If this energy were used to raise the battery above the ground, how high would it go ?

mgh = PE electrical

h = PE/mg

h = 2.8 x 106/ (20 x 9.8) = 1.43 x 104 m = 14.3 km

(c) If this energy were used to provide the battery with kinetic energy, what would its speed be ?

KE = PE electrical

½mv2 = 2.8 x 106

v2 = (2.8 x 106 x 2)/m = (2.8 x 106 x 2)/20

v = 530 m/s

A potential difference of 20 kV is used to accelerate electrons in a television picture TUBE.

(a) How much kinetic energy do the electrons acquire ?

Electrical PE lost = KE gained

QV = KE = ½mv2

KE = 1.6 x 10-19 x 20000

KE = 3.2 x 10-15 J

(b) What is their speed ?

KE = ½mv2

v2 = 2(KE)/m = (2 x 3.2 x 10-15 )/9.1 x 10-31

v = 0.84 x 108 m/s