| Date | May 2012 | Marks available | 4 | Reference code | 12M.2.HL.TZ2.7 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | State | Question number | 7 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about fields, electric potential difference and electric
circuits. Part 2 is about thermodynamic cycles.
Part 1 Fields, electric potential difference and electric circuits
The magnitude of gravitational field strength g is defined from the equation shown below.
\[g = \frac{{{F_g}}}{m}\]
The magnitude of electric field strength E is defined from the equation shown below.
\[E = \frac{{{F_E}}}{q}\]
For each of these defining equations, state the meaning of the symbols
(i) Fg.
(ii) FE.
(iii) m.
(iv) q.
In a simple model of the hydrogen atom, the electron is regarded as being in a circular orbit about the proton. The magnitude of the electric field strength at the electron due to the proton is Ep . The magnitude of the gravitational field strength at the electron due to the proton is gp.
Determine the order of magnitude of the ratio shown below.
\[\frac{{{E_p}}}{{{g_p}}}\]
Markscheme
(i) the force exerted on a small/test/point mass;
Do not allow bald “gravitational force”.
(ii) the force exerted on a small/point/test positive charge;
To award [1] “positive” is required.
Do not allow bald “electric force”.
(iii) the size/magnitude/value of the small/point mass;
Do not accept bald “mass”.
(iv) the magnitude/size/value of the small/point/test (positive) charge;
Do not accept bald “charge”.
In part (a) only penalize lack of “small/test/point” once, annotate as ECF.
It must be clear that the mass/charge in (iii) & (iv) refer to the object in (i) and (ii).
\({E_p} = \frac{e}{{4\pi {\varepsilon _0}{r^2}}}\) and \({g_p} = \frac{{G{m_p}}}{{{r^2}}}\); (both needed)
\(\frac{e}{{4\pi {\varepsilon _0}G{m_p}}}\left( { = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}}}}{{6.7 \times {{10}^{ - 11}} \times 1.7 \times {{10}^{ - 27}}}}} \right)\);
≈1028;
Award [2 max] if response calculates ratio of force as this is an ECF from the first marking point (1039) .
Award [3] for solution that correctly evaluates field strengths separately and then divides.
Examiners report
In this part candidates were completely at a loss and could not state the meanings of the symbols in the definitions of gravitational or electric field strengths. This was a disappointing failure in what was meant to be an easy opener to the whole question.
Following (a) candidates failed widely on this part too. They often had little idea which data to use (mass and charge were frequently confused) and sometimes the meaning of the constants in the equations failed them too. This was compounded by arithmetic errors to make a straightforward calculation very hard for many.
