Two point charges are situated in a vacuum and separated by a distance R.

Question 2

(a) (i) State Coulomb’s law for the force between two point charges. [1]

(ii) Two point charges are situated in a vacuum and separated by a distance R. The force between the charges is FC.

On Fig. 5.1, sketch a graph to show the variation of the force F between the charges with separation x for values of x from x = R to x = 4R.

[3]

Fig. 5.1

(b) Two coils C and D are placed close to one another, as shown in Fig. 5.2.

Fig. 5.2

The variation with time t of the current I in coil C is shown in Fig. 5.3.

On Fig. 5.4, show the variation with time t of the e.m.f. E induced in coil D for time t = 0 to time t = t5.

Fig. 5.3

Fig. 5.4

[4]

 [Total: 8]

Reference: Past Exam Paper – November 2017 Paper 41 & 43 Q5

Solution:

(a) (i) Coulomb’s law states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the separation.

(ii)

{From Coulomb’s law, the force is inversely proportional to the square of the separation.

F ∝ 1 / x²

We are given, in the question, that the force is F_C and x = R. So, the graph starts at the point (R, F_C)}

curve starting at (R, FC)

{Also, from Coulomb’s law, the graph has the shape of the curve y = 1 / x².

From the proportion, when x becomes twice (i.e, x = 2R), the force becomes (= 1 / 2² = 1 / 4 = 0.25) a quarter.}

passing through (2R, 0.25FC)

{When x = 3R, force = 1 / 3² = 1 / 9 = 0.11}

{And when x = 4R, force = 1 / 4² = 1 / 16 = 0.06}

passing through (4R, 0.06FC)

  

{For a better graph, you could also include these:

When x = 1.5, force = 1 / 1.5² = 4.44

When x = 2.5, force = 1 / 2.5² = 1.6}

(b)

{The current in coil C produces a magnetic field around it. Since the current is changing (as in the graph), the magnetic field around it also changes.

As this magnetic field cuts the coil D, an e.m.f. is induced in the coil.

From Faraday’s law, the e.m.f. induced is proportional to the rate of change of magnetic flux.

If the magnetic flux is not changing, no e.m.f. is induced. The magnetic flux does not change (i.e. it remains constant) when the current is constant (i.e. it is NOT CHANGING).

Note that if the current is, say, 4A and it remains 4A, this means that it is NOT changing and so, the magnetic flux is also NOT changing and no e.m.f. is induced.

So, no e.m.f. is induced in coil D when the current in coil C is constant (in the graph, this is a horizontal line).}

graph: E = 0 when current constant (0 to t1, t2 to t3, t4 to t5)

{From t₁ to t₂, the current changes and so, an e.m.f. is induced. This appears as a step / peak.

Similarly, from t₃ to t₄, the current changes and hence, an e.m.f. is induced.}

stepped from t1 to t2 and t3 to t4

{However, from t₁ to t₂, the current changes from positive to negative and from t₃ to t₄, the current changes from negative to positive. That is, the changes are opposite. So, the peaks will also be opposite. One would be positive and the other negative.}

(steps) in opposite directions

{Also, the change in current from t₃ to t₄ is greater than that from t₁ to t₂. So, the change in magnetic flux will also be greater, resulting in a greater e.m.f. induced from t₃ to t₄.}

later one larger in magnitude