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Forces between charges

Forces between charges. A2 Level Notes. 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.

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Forces between charges

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  1. Forces between charges A2 Level Notes

  2. 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/r2 is called the inverse square law of electrostatics.

  3. 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 eAIR = 8.9 pN m-1

  4. A1 From the data sheet F = 1.0 x 10-9 x 9.0 x 10-9 4p x 8.9 x 10-12 x (0.30)2 F = 8.9 x 10-7 N (P) The force is positive therefore it is a repulsive force.

  5. 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. 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. 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 4per2 On your data sheet E = F q On your data sheet E = Q 4per2 The unit for E could be NC-1 (but Vm-1 is preferred)

  8. 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. Uniform field Here you can see that the unit for E can also be Vm-1 Field lines are parallel – therefore the field is constant in intensity.

  10. Q2 Point charges are located at points A and B as shown in the diagram. Calculate the electric field strength at point C. eAIR = 8.9 pF m-1

  11. 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. A2 Let us find the magnitude of the field intensity at point C from each charge. lECAl = 3.6 x 10-9/(4p x 8.9 x 10-12 x (0.030)2) lECAl = 3.58 x 104 Vm-1 (P) lECBl = 3.2 x 10-9/(4p x 8.9 x 10-12 x (0.040)2) lECBl = 1.79 x 104 Vm-1 (P) (We are going to use these later so they need to be to one more sig fig than is given in the question)

  13. 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. A2 By Pythagoras E2 = ECA2 + ECB2 E2 = (3.58 x 104)2 + (1.79 x 104)2 lEl = 4.0 x 104Vm-1 (P) (Final answer – therefore to 2sf) Tan a =ECB/ECA = 1.79 x 104 / 3.58 x 104 = 0.5 • = 26.6o (P) E = 40 kVm-1 at 27o to CA (or 63o to CB) (P) (Final answer needs direction as E is a vector)

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