Electric Charges

Reading cue. The rings crowd towards the surface, so the density rises with radius: r runs from the centre out to P, and the charge inside r grows faster than r³.

1. aDetermine Determine the volume charge density at the centre of the sphere, at the point P marked at radius r, and at the surface, in terms of ρ₀.

   recall3 marks

   Check answer · 3 marks

   1. zero at the centre
   2. ρ₀r/R at P
   3. ρ₀ at the surface, which is why the rings are drawn closest together there

2. bDetermine Determine the fraction of the sphere's total charge that lies within a radius R/2 of the centre.

   routine4 marks

   Check answer · 4 marks

   1. q(r) = ∫₀^r (ρ₀s/R)4πs² ds = πρ₀r⁴/R
   2. so q ∝ r⁴, and the total charge is Q = πρ₀R³
   3. q(R/2)/Q = (1/2)⁴ = 1/16, that is 6.25%
   4. much less than the 1/8 a uniform sphere would give, because the charge is pushed outwards

3. cSketch Sketch a graph of the magnitude of the electric field against distance from the centre, from the centre out to 3R, marking the position and the value of the maximum.

   demanding5 marks

   Check answer · 5 marks

   1. for r < R, E = q(r)/4πε₀r² = ρ₀r²/4ε₀R, drawn as a curve rising from the origin with increasing gradient, not a straight line
   2. for r > R, E = Q/4πε₀r², falling as 1/r²
   3. the two expressions agree at r = R, so the curve has no step in it
   4. maximum at the surface, r = R, of value ρ₀R/4ε₀
   5. at r = 3R the field has fallen to one ninth of the maximum

4. dDerive Using an energy density of ½ε₀E², derive an expression for the total electric energy stored in the field outside the sphere, and explain why the same calculation cannot be carried through for a point charge.

   top of the paper5 marks

   Check answer · 5 marks

   1. dU = ½ε₀E²(4πr² dr) with E = Q/4πε₀r² for r > R
   2. U = (Q²/8πε₀)∫_R^∞ dr/r² = Q²/8πε₀R
   3. substituting Q = πρ₀R³ gives U = πρ₀²R⁵/8ε₀
   4. for a point charge the lower limit becomes r = 0 and the integral diverges
   5. so a point charge implies infinite field energy, which is why a finite size has to be assumed in such calculations