2The 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?
3Distance doubles so force drops to a quarter of that before Product of charges before was+Q x -2Q = -2Q2Product of new charges is2Q x -Q = -2Q2So force doesn’t change in sign or size due to charge changeOverall change produces an attractive force of F/4
4The 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?
5V = Er100mm = 0.1mSo V = 3 × 106 Vm-1 x 0.1m= 3 × 105 V
6At 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?
7Field strength is an inverse square relationship therefore doubling the distance will reduce E to E/4Potential is an inverse relationship – doubling the distance will halve the potential V to V/2
8Two 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?
9At 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 PXV = constant x Q/rAt X: 0 = constant (4/r – 6/(100-r)4/r = 6/(100-r)r = 6r10r = 400r = 40mm
10Two 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?
11Distance is doubled therefore force drops to a quarter of what it was before (inverse square law)
12The 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?
13Moving 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 VWork done = QDV = 30 V x 4 μC= 120 μJ