Question

A thundercloud has the charge distribution illustrated above left. Treat this distribution as two point charges, a negative charge of –30 C at a height of 2 km above ground and a positive charge of +30 C at a height of 3 km. The presence of these charges induces charges on the ground. Assuming the ground is a conductor, it can be shown that the induced charges can be treated as a charge of +30 C at a depth of 2 km below ground and a charge of –30 C at a depth of 3 km, as shown above right. Consider point P₁, which is just above the ground directly below the thundercloud, and point P₂, which is 1 km horizontally away from P₁.

a. Determine the direction and magnitude of the electric field at point P₁.

b. i. On the diagram, clearly indicate the direction of the electric field at point P₂.

ii. How does the magnitude of the field at this point compare with the magnitude at point P₁? Justify your answer:

____ Greater ____Equal ____ Less

c. Letting the zero of potential be at infinity, determine the potential at these points.

i. Point P₁

ii. Point P₂

d. Determine the electric potential at an altitude of 1 km directly above point P₁.

e. Determine the total electric potential energy of this arrangement of charges.

Answer 1

a. E is the vector sum of kQ/r². Let fields directed upward be positive and fields directed downward be negative.

This gives E = k[– 30 C/(3000 m)² + 30 C/(2000 m)² + 30 C/(2000 m)² – 30 C/(3000 m)²] = 75,000 N/C upward

b. i

ii. Because it is a larger distance from the charges, the magnitude is less.

c. i. By symmetry, the potentials cancel and V = 0

ii. By symmetry, the potentials cancel and V = 0

d. V = ΣkQ/r = k[30 C/(2000 m) – 30 C/(1000 m) + 30 C/(3000 m) – 30 C/(4000 m)] = –1.12 × 10⁸ V

e. U = kq₁q₂/r for each pair of charges

= k[(30)(–30)/1000 + (30)(30)/5000 + (30)(–30)/6000 + –30(30)/4000 + –30(–30)/5000 + 30(–30)/1000] = –1.6 × 10¹⁰ J