State Coulomb’s law.

In a simple model of the hydrogen atom, the electron can be regarded as being in a circular orbit about the proton. The radius of the orbit is 2.0×10^–10 m.

(i) Determine the magnitude of the electric force between the proton and the electron.

(ii) Calculate the magnitude of the electric field strength E and state the direction of the electric field due to the proton at a distance of 2.0×10^–10 m from the proton.

(iii) The magnitude of the gravitational field due to the proton at a distance of 2.0×10^–10 m from the proton is H.
Show that the ratio \(\frac{H}{E}\) is of the order 10^–28C kg^–1.

(iv) The orbital electron is transferred from its orbit to a point where the potential is zero. The gain in potential energy of the electron is 5.4×10^–19J. Calculate the value of the potential difference through which the electron is moved.

An electric cell is a device that is used to transfer energy to electrons in a circuit. A particular circuit consists of a cell of emf ε and internal resistance r connected in series with a resistor of resistance 5.0 Ω.

(i) Define emf of a cell.

(ii) The energy supplied by the cell to one electron in transferring it around the circuit is 5.1×10^–19J. Show that the emf of the cell is 3.2V.

(iii) Each electron in the circuit transfers an energy of 4.0×10^–19 J to the 5.0 Ω resistor. Determine the value of the internal resistance r.

the force between two (point) charges;
is inversely proportional to the square of their separation and (directly) proportional to (the product of) their magnitudes;

Allow [2] for equation with F, Q and r defined.

(i) \(F = \left( {k\frac{{{q_1}{q_2}}}{{{r^2}}} = } \right)\frac{{9 \times {{10}^9} \times {{\left[ {1.6 \times {{10}^{ - 19}}} \right]}^2}}}{{4 \times {{10}^{ - 20}}}}\);
=5.8×10^-9(N);
Award [0] for use of masses in place of charges.

(ii) (\(\frac{{\left( b \right)\left( i \right)}}{{1.6 \times {{10}^{ - 19}}}}\) or 3.6 x 10¹⁰ (NC^-1) or (Vm^-1);
(directed) away from the proton;
Allow ECF from (b)(i).
Do not penalize use of masses in both (b)(i) and (b)(ii) – allow ECF.

(iii) \(H = \left( {G\frac{m}{{{r^2}}} = } \right)\frac{{6.67 \times {{10}^{ - 11}} \times 1.673 \times {{10}^{ - 27}}}}{{4 \times {{10}^{ - 20}}}} = 2.8 \times {10^{ - 18}}\left( {{\rm{Nk}}{{\rm{g}}^{ - 1}}} \right)\);

\(\frac{H}{E} = \frac{{2.8 \times {{10}^{ - 18}}}}{{3.6 \times {{10}^{10}}}}\) or 7.8×10^-29(Ckg^-1);
(≈10²⁸Ckg^-1)
Allow ECF from (b)(i).

(iv) 3.4(V); 

(i) power supplied per unit current / energy supplied per unit charge / work done per unit charge;

(ii) energy supplied per coulomb=\(\frac{{5.1 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}\) or 3.19(V);
(≈3.2V)

(iii) pd across 5.0Ω resistor=\(\left( {\frac{{4.0 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = } \right)2.5\left( {\rm{V}} \right)\);
pd across r=(3.2-2.5=)0.70(V);

and

either

current in circuit=\(\left( {\frac{{2.5}}{{5.0}} = } \right)0.5\left( {\rm{A}} \right)\);
resistance of r=\(\left( {\frac{{0.70}}{{0.50}} = } \right)1.4\left( \Omega  \right)\);

or

resistance of r=\(\frac{{0.70}}{{2.5}} \times 5.0\);
=1.4(Ω);

or

3.2=0.5(R+r);
resistance of r=1.4(Ω);
Award [4] for alternative working leading to correct answer.
Award [4] for a bald correct answer.

Many were able to state Coulomb’s law or to give the equation with explanations of the symbols. Some candidates however failed to define their symbols and lost marks.

(i) The electric force was calculated well by many.

(ii) The answer to (i) was well used to determine the magnitude of E. However, many candidates did not read the question and failed to state the direction of the field or gave it in an ambiguous way.

(iii) Calculations to show the order of magnitude of H/E were generally well done. The last step was often missing with the answer simply given as a fraction.

(iv) Many obtained this simple mark.

(i) Many candidates gave confused or incorrect definitions of the emf of a cell. Previous comments in this report on the memorizing of definitions apply. Too many had recourse to the next part and used this idea in their answer.

(ii) This was well done.

(iii) A large number of candidates completed this calculation stylishly, generally explaining steps (or at least writing down the algebra) in a logical way. There were many correct and original solutions that gained full marks.