(i) On the diagram above, draw an arrow to show the direction of the electric field at point X.

(ii) Assuming the nucleus acts as a point charge, determine the magnitude of the electric field strength at point X.

(d) The diagram shows part of a potential divider circuit used to measure the current-potential difference (I–V) characteristic of the bulb.

(i) Draw the complete circuit showing the correct position of the bulb, ammeter and voltmeter.

(ii) On the axes provided, sketch a graph to show how the current in the bulb varies with the potential difference.

A student sets up a different circuit to measure the I–V graph. The cell has an emf of 6.0 V and negligible internal resistance. The variable resistor has a minimum resistance of zero and a maximum resistance of 5.0 Ω.

Explain, with a calculation, why this circuit will not allow for a full range of potential difference from 0 V to 6 V across the bulb. Assume that the resistance of the lamp remains constant at a value of 2.4 Ω.

(i) arrow pointing away from nucleus;

(ii) \(E = \frac{{74 \times 1.6 \times {{10}^{ - 19}}}}{{4\pi  \times 8.85 \times {{10}^{ - 12}} \times {{\left[ {1.4 \times {{10}^{ - 10}}} \right]}^2}}}\);
5.4×10¹²Vm^-1 or NC^-1;
Award [2] for a bald correct answer.

(i) lamp connected so that pd can be varied;
ammeter in series with lamp and voltmeter in parallel with lamp; (both needed)
Award [0] if lamp cannot light.

(ii) through origin;
correct shape;

minimum pd across bulb is when R=5.0Ω
;
pd across bulb\( = 6.0 \times \frac{{2.4}}{{7.4}}\)
=1.9 (V) or 2.0 (V);
so range −1.9−6.0V or 0V across lamp cannot be obtained;