1. Electric Field Questions from 2004 and 2005

2. The diagram shows two particles at a distance d apart
The diagram shows two particles at a distance d apart. One particle has charge +Q and the other -2Q.
The two particles exert an electrostatic force of attraction, F, on each other. Each particle is then given an additional charge +Q and their separation is increased to a distance of 2d.
What is the force that now acts between the two particles?

3. Distance doubles so force drops to a quarter of that before
Product of charges before was
+Q x -2Q = -2Q2
Product of new charges is
2Q x -Q = -2Q2
So force doesn’t change in sign or size due to charge change
Overall change produces an attractive force of F/4

4. The electrical field strength, E, and the electrical potential, V, at the surface of a sphere of radius r carrying a charge Q are given by the same equations as that for a point charge on your data sheet.
A school van de Graaff generator has a dome of radius 100 mm. Charge begins to leak into the air from the dome when the electric field strength at its surface is approximately 3 × 106 Vm-1.
What, approximately, is the maximum potential to which the dome can be raised without leakage?

5. V = Er
100mm = 0.1m
So V = 3 × 106 Vm-1 x 0.1m
= 3 × 105 V

6. At a distance R from a fixed charge, the electric field strength is E and the electric potential is V.
What is the electric field strength and electric potential at a distance 2R from the charge?

7. Field strength is an inverse square relationship therefore doubling the distance will reduce E to E/4
Potential is an inverse relationship – doubling the distance will halve the potential V to V/2

8. Two charges, P and Q, are 100 mm apart
Two charges, P and Q, are 100 mm apart. X is a point on the line between P and Q.
If the potential at X is 0V, what is the distance from P to X?

9. At X: 0 = constant (4/r – 6/(100-r)
At X the positive electric potential from the +4mC charge is equal to the negative potential from the -6mC charge
– then when you add them they will cancel out!
Let r be the distance PX
V = constant x Q/r
At X: 0 = constant (4/r – 6/(100-r)
4/r = 6/(100-r)
r = 6r
10r = 400
r = 40mm

10. Two isolated point charges are separated by 0
Two isolated point charges are separated by 0.04 m and attract each other with a force of 20 μN. If the distance between them is increased by 0.04 m, what is the new force of attraction?

11. Distance is doubled therefore force drops to a quarter of what it was before (inverse square law)

12. The diagram shows a uniform electric field of strength 10Vm-1
A charge of 4 μC is moved from P to Q and then from Q to R. If the distance PQ is 2m and QR is 3m, what is the change in potential energy of the charge when it is moved from P to R?

13. Moving from P to Q no work is done (perp to line of action of the force!) so we ignore that!
Moving from Q to R work is done:
DV = 10 V/m x 3 m = 30 V
Work done = QDV = 30 V x 4 μC
= 120 μJ