Coulomb´s law is the electrical equivalent of Newton's law of gravitation.
The universe is a simple place really!
Key Concepts
The force between two point charges is proportional to the product of their charge and inversely proportional to their separation squared:
\(F \propto {Qq \over r^2}\)
NB: Force is measured in Newtons.
You don't need to go inside spheres of charge for the IB (we can assume that all of the charge acts at the centre)... but in your spare time you can go where you like.
A sphere of charge acts like all the charge was at its centre. Inside the sphere the charge is zero.
Use quizzes to practise application of theory.
START QUIZ!
F = kQ1Q2/r2
k2Q1Q2/r2 = 2F
The force between 2 charged spheres is F. What is the force if their separation is reduced by a factor 4
F ∝ 1/r2
1/(r/4)2 = 16/r2
The diagram shows two sphere of charge. The force experienced by sphere A is F.
The next diagram shows the same bodies but arranged differently
The force on A is now
F = 4Q x 4Q/r2
F´=4Q x 2Q /r2
The diagram shows two sphere of charge. The force experienced by sphere A is F.
The next diagram shows the same bodies but arranged differently
The force is now
F = 4Q x 4Q/r2
F´=2Q x -2Q /r2
The diagram shows two sphere of charge. The force experienced by sphere A is F.
The next diagram shows the same bodies but arranged differently
F = 4Q x 4Q/r2
F´=2Q x 4Q /(2r)2
The unit of field strength is
E is force per unit charge
The field strength at a point is 1000 NC-1
The force experienced by a 1 μC charge placed at that point is:
1000 x 1 x 10-6 = 1 x 10-3
The diagram represents two points in an electric field
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If the field at A has magnitude E, the field at B ill have magnitude:
The field is radial so
E is inversely proportional to distance from centre
Twice the distance from the centre so 1/4 the field
The diagram represents two points in an electric field
If the field at A is E the field at B is:
Uniform field is the same everywhere
The diagram represents two points in an electric field
If the field at A is E the field at B is:
Field strength is proportional to line density.
The diagram represents two charges.
Which arrow gives the resultant field at the point shown?
The diagram represents 2 charges
Which arrow gives the resultant field at the point shown?
The diagram represents 2 charges
The red arrow shows the fiekd due to the positive charge.
Which arrow represents the resultant?
E is proportional to Q/r2 so the field due to the negative charge is 2x the field due to the positive
The diagram represents a positive test charge in a hollow sphere.
The direction of the force experienced by the test charge is:
The field inside a hollow sphere of charge is zero
The diagram represents 4 charges placed on the corners of a square and a test charge.
The direction of force experienced by the test charge is:
Vertical components cancel but the test charge is closer to the charges on the left so the force right will be greater.
The field inside a hollow metal sphere of charge is zero because:
How much of Coulombs law and field strength have you understood?
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