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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Forces between charges A2 Level Notes.

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Presentation on theme: "Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Forces between charges A2 Level Notes."— Presentation transcript:

1 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Forces between charges A2 Level Notes

2 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. This formula is on the data sheet It applies to point charges a distance ‘r’ apart but can be used for any charged objects as long as they are small in diameter compared to ‘r’ – ‘r’ is then the distance between their centres. The fact that F is proportional to 1/r 2 is called the inverse square law of electrostatics.

3 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Q1 Calculate the force between two small conducting spheres with charges +1.0 nC and +9.0 nC whose centres are 30 cm apart in air. Is the force attractive or repulsive? Permittivity of air  AIR = 8.9 pN m -1

4 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. A1 From the data sheet F = 1.0 x x 9.0 x  x 8.9 x x (0.30) 2 F = 8.9 x N (  ) The force is positive therefore it is a repulsive force.

5 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Electric Field In the vicinity of any charge Q there is a region within which other charges will ‘feel the force’ and be repelled by it or attracted to it. That region is called the electric field of charge Q. It is a radial field (spreads out from a point).

6 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Electric Field We can describe the strength of that field by considering the force that would be experienced by a tiny test charge ‘q’ placed at any point in that field. The nearer the charge is to the centre of the field the stronger the force it will experience.

7 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Electric Intensity E If we divide the force experienced by the test charge ‘q’ by ‘q’ we eliminate needing to specify the size of our charge. We call this ‘E’ the electric intensity or the electric field strength. F = Q q 4  r 2 On your data sheet E = F q On your data sheet E = Q 4  r 2 The unit for E could be NC -1 (but Vm -1 is preferred)

8 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Radial field Field lines have arrows showing the direction in which a positive point charge would move if placed at that position. Field lines get further apart as you leave the source – therefore the field gets weaker

9 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Uniform field Field lines are parallel – therefore the field is constant in intensity. Here you can see that the unit for E can also be Vm -1

10 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Q2 Point charges are located at points A and B as shown in the diagram. Calculate the electric field strength at point C.  AIR = 8.9 pF m -1

11 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. A2 Both charges will be having an effect of the field strength at point C. Both charges will be producing a radial field around themselves. We therefore have to consider the components of the intensity of the field at C from each of the charges. From the data book we have the equation:

12 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. A2 Let us find the magnitude of the field intensity at point C from each charge. lE CA l = 3.6 x /(4  x 8.9 x x (0.030) 2 ) lE CA l = 3.58 x 10 4 Vm -1 (  ) lE CB l = 3.2 x /(4  x 8.9 x x (0.040) 2 ) lE CB l = 1.79 x 10 4 Vm -1 (  ) (We are going to use these later so they need to be to one more sig fig than is given in the question)

13 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. A2 We can now add the vectors at point C. We know the size and direction they act in – we can then construct a diagram to add them

14 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. A2 By Pythagoras E 2 = E CA 2 + E CB 2 E 2 = ( 3.58 x 10 4 ) 2 + ( 1.79 x 10 4 ) 2 lEl = 4.0 x 10 4 Vm -1 (  ) (Final answer – therefore to 2sf) Tan  = E CB /E CA = 1.79 x 10 4 / 3.58 x 10 4 = 0.5  = 26.6 o (  ) E = 40 kVm -1 at 27 o to CA (or 63 o to CB) (  ) (Final answer needs direction as E is a vector)

15 Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved.


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