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**Electric Field Questions from 2004 and 2005**

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**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?

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**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

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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?

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

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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?

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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

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**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?

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**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

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**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?

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

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**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?

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**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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Wednesday, Sept. 21, 2005PHYS 1444-003, Fall 2005 Dr. Jaehoon Yu 1 PHYS 1444 – Section 003 Lecture #7 Wednesday, Sept. 21, 2005 Dr. Jaehoon Yu Electric.

Wednesday, Sept. 21, 2005PHYS 1444-003, Fall 2005 Dr. Jaehoon Yu 1 PHYS 1444 – Section 003 Lecture #7 Wednesday, Sept. 21, 2005 Dr. Jaehoon Yu Electric.

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