Coulomb Law - Higher Physics - Solved Past Paper, Exams of Physics

Download Coulomb Law - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! Question 1 (Marks 15) (a) F+Q x -Q +Q d/2 d/2 s +q h θ θ Ftotal F–Q y The distance s is given by 22 )2/a(hs += Coulomb’s law for charges q1 and q2 is 2 21 0 r qqkF = which gives the magnitude of the force on +q 020QQ kh qQk === −+ FF and the magnitude of the total force is 1 due to +Q and –Q [ ]22 )2/a(s qQ + [ ] θ+=θ= sin)2/a(s qQk2sin h qQk2 22020totalF substituting 22 )2/a(s 2/a h 2/asin + ==θ [ ]220total )2/a(s qQk2 + =F 22 )2/a(s 2/a + [ ] 2/3220total )2/a(s aqQk + =F By symmetry see that is parallel to the y-axis, in direction so that totalF ĵ [ ] jF ˆ )2/a(s aqQk 2/3220total + = (b) The magnitude of the force between two charges is given by Coulomb’s law: 21 q,q 2 21 0 r qq 4 1F πε = (1) where is the permittivity of free space. The force can be seen to arise from the electric field around charge 0ε iq 2 i 0 r q 4 1 q FE πε == (2) 2 The total charge as a fn. of position is charge on nucleus Q plus the electronic contribution: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−+=+= 5 5 2 3 nuctotal R2 r3 R2 Qr5Q)r(qQ)r(q ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−=+= 5 5 2 3 nuctotal R2 r3 R2 r51Q)r(qQ)r(q The E-field around the nucleus is symmetric: rE ˆ)r(E= . Gauss’ law gives ∫ surface d. AE 0 2 surfacesurfacesurface q)r4)(r(Ed)r(Edˆd.ˆ ε Σ =π==== ∫∫∫ AAE(r)AE(r) rr (ii) In the region Rr0 << , 20 r )r(qk)r(E = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−= 5 5 2 3 20 R2 r3 R2 r51 r qk At radius Rr = the charge is 0 R2 r3 R2 r51Qq 5 5 2 3 R =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−= as it must be for a neutral atom. By Gauss’ law therefore, we must have 0)r(E = for Rr ≥ . 5 Question 3 (Marks 10) The diagram shows a circular ring of uniform electric charge of radius a. The total charge on the ring is Q coulombs. Derive an expression in terms of z and a for the electric potential at a point P vertically above the centre of the ring, 0, as shown. P • z r a 0 The potential at point P with position vector r is ∫= r dqkV 0 where 0 0 4 1k πε = . We note from the geometry that 22 zar += and then ∫∫ +πε == 22 0 0 za dq 4 1 r dqkV 22 0 xa Q 4 1 +πε = mV36.5V10x36.5V 2 10x58.7 33 rms ===ε − − 6