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Electric Field Questions from 2004 and 2005

Electric Field Questions from 2004 and 2005. The diagram shows two particles at a distance d apart. One particle has charge + Q and the other -2 Q .

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Electric Field Questions from 2004 and 2005

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  1. Electric Field Questions from 2004 and 2005

  2. 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 x0.1m = 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. 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 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

  10. 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) 20/4 = 5 μN

  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

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