Potential Difference

1. 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




2. 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




3. 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

  
  
  
4. 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