Report

Electric Field Questions from 2004 and 2005 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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? Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 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? Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. V = Er 100mm = 0.1m So V = 3 × 106 Vm-1 x 0.1m = 3 × 105 V Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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? Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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? Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 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) 400 - 4r = 6r 10r = 400 r = 40mm Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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? Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • Distance is doubled therefore force drops to a quarter of what it was before (inverse square law) 20/4 = 5 μN Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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? Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. • 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 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved.