12 exam-style questions on field strength, potential, orbits and capacitors. Tap Show answer for full working.
Calculate the gravitational field strength at a distance of \(2.0\times10^{7}\,\mathrm{m}\) from the centre of a planet of mass \(6.0\times10^{24}\,\mathrm{kg}\). (\(G = 6.67\times10^{-11}\,\mathrm{N\,m}^{2}\mathrm{kg}^{-2}\).)
\(g = GM/r^{2} = 6.67\times10^{-11}\times6.0\times10^{24}/(4.0\times10^{14}) = 1.00\,\mathrm{N\,kg}^{-1}\).
Two point charges of \(+4.0\,\mu\mathrm{C}\) and \(-2.0\,\mu\mathrm{C}\) are \(0.30\,\mathrm{m}\) apart in free space. Calculate the force between them. (\(\varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{F\,m}^{-1}\); \(k = 1/(4\pi\varepsilon_{0}) = 9.0\times10^{9}\).)
\(F = k|q_{1}q_{2}|/r^{2} = 9\times10^{9}\times4\times10^{-6}\times2\times10^{-6}/0.09 = 0.80\,\mathrm{N}\) (attractive).
A capacitor of capacitance \(47\,\mu\mathrm{F}\) is charged to \(12\,\mathrm{V}\). Calculate the charge stored and the energy stored.
\(Q = CV = 47\times10^{-6}\times12 = 5.64\times10^{-4}\,\mathrm{C}\).
\(E = \frac12 CV^{2} = 0.5\times47\times10^{-6}\times144 = 3.38\times10^{-3}\,\mathrm{J}\).
State the relationship between gravitational field strength \(g\) and gravitational potential \(V\). Sketch the variation of \(g\) and of \(V\) with distance \(r\) from the centre of a uniform sphere (outside the sphere).
\(g = -\mathrm{d}V/\mathrm{d}r\) (field is the negative gradient of potential).
\(g\) falls as \(1/r^{2}\); \(V\) falls as \(-1/r\) (approaches zero from below as \(r\to\infty\)).
A satellite orbits the Earth at a radius of \(7.5\times10^{6}\,\mathrm{m}\). Mass of Earth \(= 6.0\times10^{24}\,\mathrm{kg}\). Calculate the orbital speed and the period of the orbit. (\(G = 6.67\times10^{-11}\).)
\(v = \sqrt{GM/r} = \sqrt{6.67\times10^{-11}\times6.0\times10^{24}/7.5\times10^{6}} = 7300\,\mathrm{m\,s}^{-1}\).
\(T = 2\pi r/v = 2\pi\times7.5\times10^{6}/7300 = 6450\,\mathrm{s}\) (\(\approx 1.79\,\mathrm{h}\)).
The electric field strength between two parallel plates is \(2.5\times10^{4}\,\mathrm{V\,m}^{-1}\). The plates are \(8.0\,\mathrm{mm}\) apart. Calculate the potential difference between the plates and the force on an electron between the plates. (\(e = 1.6\times10^{-19}\,\mathrm{C}\).)
\(V = Ed = 2.5\times10^{4}\times8.0\times10^{-3} = 200\,\mathrm{V}\).
\(F = eE = 1.6\times10^{-19}\times2.5\times10^{4} = 4.0\times10^{-15}\,\mathrm{N}\).
A \(100\,\mu\mathrm{F}\) capacitor is charged to \(20\,\mathrm{V}\) and then connected across an uncharged \(50\,\mu\mathrm{F}\) capacitor. Calculate the final common potential difference and the energy lost as heat in the connecting wires.
Initial charge \(Q = 100\times10^{-6}\times20 = 2.0\times10^{-3}\,\mathrm{C}\).
Final \(C_{\mathrm{tot}} = 150\,\mu\mathrm{F}\); \(V = Q/C = 2.0\times10^{-3}/150\times10^{-6} = 13.3\,\mathrm{V}\).
Initial energy \(= \frac12\times100\times10^{-6}\times400 = 0.020\,\mathrm{J}\); final \(= \frac12\times150\times10^{-6}\times177 = 0.0133\,\mathrm{J}\); lost \(= 0.0067\,\mathrm{J}\).
Define gravitational potential at a point. Calculate the gravitational potential at the surface of a planet of mass \(5.0\times10^{24}\,\mathrm{kg}\) and radius \(4.0\times10^{6}\,\mathrm{m}\). (\(G = 6.67\times10^{-11}\).)
Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.
\(V = -GM/r = -6.67\times10^{-11}\times5.0\times10^{24}/4.0\times10^{6} = -8.34\times10^{7}\,\mathrm{J\,kg}^{-1}\).
Show that the escape speed from the surface of a planet is \(\sqrt{2GM/R}\). Calculate the escape speed from Earth (\(M = 6.0\times10^{24}\,\mathrm{kg}\), \(R = 6.4\times10^{6}\,\mathrm{m}\)).
Conservation of energy: \(\frac12 mv^{2} - GMm/R = 0\) (at infinity KE+PE=0) \(\Rightarrow v = \sqrt{2GM/R}\).
\(v = \sqrt{2\times6.67\times10^{-11}\times6.0\times10^{24}/6.4\times10^{6}} = 1.12\times10^{4}\,\mathrm{m\,s}^{-1}\).
Two capacitors of \(4.0\,\mu\mathrm{F}\) and \(6.0\,\mu\mathrm{F}\) are connected in series across a \(24\,\mathrm{V}\) supply. Calculate the charge on each capacitor, the PD across each, and the total energy stored.
\(1/C = 1/4 + 1/6 = 5/12 \Rightarrow C = 2.4\,\mu\mathrm{F}\).
\(Q = CV = 2.4\times10^{-6}\times24 = 5.76\times10^{-5}\,\mathrm{C}\) (same on both).
\(V_{4} = Q/C_{4} = 14.4\,\mathrm{V}\); \(V_{6} = 9.6\,\mathrm{V}\).
Energy \(= \frac12 CV^{2} = 0.5\times2.4\times10^{-6}\times576 = 6.91\times10^{-4}\,\mathrm{J}\).
A geostationary satellite orbits above the equator with a period of \(24\,\mathrm{h}\). Mass of Earth \(= 6.0\times10^{24}\,\mathrm{kg}\). Calculate the radius of the orbit and the height above the Earth’s surface (Earth radius \(= 6400\,\mathrm{km}\)).
\(T = 86400\,\mathrm{s}\); \(GM = \omega^{2}r^{3}\) with \(\omega = 2\pi/T\).
\(r^{3} = GMT^{2}/(4\pi^{2}) = 6.67\times10^{-11}\times6\times10^{24}\times(86400)^{2}/(4\pi^{2}) = 7.54\times10^{22}\).
\(r = 4.23\times10^{7}\,\mathrm{m}\); height \(= 4.23\times10^{7} - 6.4\times10^{6} = 3.59\times10^{7}\,\mathrm{m}\) (\(35900\,\mathrm{km}\)).
An electron is accelerated from rest through \(500\,\mathrm{V}\) and then enters a uniform electric field of strength \(2.0\times10^{4}\,\mathrm{V\,m}^{-1}\) directed vertically. Calculate the time taken for the electron to travel \(5.0\,\mathrm{cm}\) horizontally and the vertical deflection in that time. (\(e/m = 1.76\times10^{11}\,\mathrm{C\,kg}^{-1}\).)
\(\frac12 mv^{2} = eV \Rightarrow v = \sqrt{2(e/m)V} = \sqrt{2\times1.76\times10^{11}\times500} = 1.33\times10^{7}\,\mathrm{m\,s}^{-1}\).
\(t = x/v = 0.05/1.33\times10^{7} = 3.76\times10^{-9}\,\mathrm{s}\).
\(a = (e/m)E = 1.76\times10^{11}\times2\times10^{4} = 3.52\times10^{15}\,\mathrm{m\,s}^{-2}\).
\(y = \frac12 at^{2} = 0.5\times3.52\times10^{15}\times(3.76\times10^{-9})^{2} = 0.0249\,\mathrm{m}\) (\(2.5\,\mathrm{cm}\)).