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Electric fields

Question: 1. Calculate the electrostatic force between an alpha particle (charge = 2e) and a gold nucleus (charge = 79e) when they are separated by 3.6x10-14 m

2. Calculate the electric field intensity at the following places:
(i) between a thunder cloud and the ground if the p.d. between the cloud and the ground is 1000MV and the cloud is 1 km above it
(Relative permittivity for air = 1.0000536.)
(ii) 4.5m from the dome of a Van de Graaff generator if the charge on the dome is 2.5x10-5 C.

3. Calculate the electrostatic force between the proton and the electron in the hydrogen atom if the separation of the two particles is 1.0x10-10 m

4. Two small equally charged are 10 cm apart and the repulsive force between them is 3.6x10-6 N
(i) what is the charge on each sphere?
(ii) the spheres are negatively charged. How many excess electrons are there on each sphere?


The formulae you will need are:

F = [1/4peo]Q1Q2/d2 E (radial field) = [1/4peo]Q/d2 and E (uniform field) = V/d

I am assuming that you have been given the values of all the basic constants.
Remember that the charges must be in coulombs and the distances in metres because of the value taken for the constant.
A useful value that I will use is that 1/4peo = 9x109 using the correct units (see above) for distance and charge.

1. Use: F = [1/4peo]Q1Q2/d2
F = [1/4peo]Q1Q2/d2 = [9x109x2x1.6x10-19x79x1.6x10-19]/(3.6x10-14)2 = 28.1 N

2. (i) Use: E = V/d = 1.0000536x1000x106/1000 = 1.0000536x106 Vm-1 = 1.0000536 MVm-1
(ii) Use: E (radial field) = [1/4peo]Q/d2 = 9x109x2.5x10-5/4.52 = 1.11x104 Vm-1 = 11 000 Vm-1

3. Use: F = [1/4peo]Q1Q2/d2
F = [1/4peo]Q1Q2/d2 = [9x109x1.6x10-19x1.6x10-19]/(1.0x10-10)2 = 2.3x10-8 N

4. Use: F = [1/4peo]Q1Q2/d2 but Q1 = Q2 = Q
(i) Therefore: Q2 = Fd2/[1/4peo] = 3.6x10-6x0.12/9x109 = 4x10-18 C = 40x10-19 C
(ii) Number of electrons = Q/e = 40x10-19 /1.6x10-19 = 25 electrons
© Keith Gibbs 2007