IBDP Physics -Unit 5. Electricity and magnetism -5.1 – Electric fields: IB Style Question Bank HL Paper 2


An electron is placed at a distance of 0.40 m from a fixed point charge of –6.0 mC.

(a) Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 108 N C–1. [2]

(b) (i) Calculate the magnitude of the initial acceleration of the electron. [2]

(ii) Describe the subsequent motion of the electron




E = \(\frac{k\times q}{r^2}\)

E = \(\frac{8.99\times 10^9\times 6.0\times 10^{-3}}{0.4^2}\) OR  E = 3.37 × 10 8 «NC-1 »

b i

F = q ×E  OR  F =  1.6 ×10-19 × 3.4 ×108 =  5.4 × 10-11 « N»

a = « \(\frac{5.4\times 10^{-11}}{9.1\times 10^{-31}}\) » 5.9 × 1019 « ms-1»

b ii

the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔


A vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mas     

(a) The charge per unit area on the surface of the wall is σ. It can be shown that the electric field strength E due to the charge on the wall is given by the equation

E = \tfrac{\sigma }{2_\varepsilon\ }

Demonstrate that the units of the quantities in this equation are consistent. [2]

(b)   (i) The thread makes an angle of 30° with the vertical wall. The ball has a mass of 0.025 kg. Determine the horizontal force that acts on the ball.     [3]

(ii) The charge on the ball is 1.2 × {10}^{-6} C. Determine σ.                          [2]

(c) The thread breaks. Explain the initial subsequent motion of the ball.       [3]

(d) The centre of the ball, still carrying a charge of 1.2 × {10}^{-6} C, is now placed 0.40 m from a point charge Q. The charge on the ball acts as a point charge at the centre of the ball. P is the point on the line joining the charges where the electric field strength is zero.


      1. Calculate the charge on Q. State your answer to an appropriate number of significant figures.  [3]

      2. Outline, without calculation, whether or not the electric potential at P is zero. [2]



a identifies units of σ as C m-2  \frac{C}{M^2}\times \frac{Nm^2}{c^2} seen and reduced to N C-1     Accept any analysis (eg dimensional) that yields answer correctly

b i horizontal force F on ball = T sin 30  T = \frac{mg}{cos30}  F «=mg tan 30 = 0.025 x 9.8 x tan 30» = 0.14 «N» Allow g= 10 N kg -1 Award  for a bald correct answer. Award for an answer of zero, interpreting that the horizontal force refers to the horizontal component of the net force.

b ii E= \frac{0.14}{1.2\times 10^-^6} « » σ = «\frac{2\times 8.85\times 10^1^2\times 0.14}{1.2\times 10^6}» = 2.1 x 10-6 «C m-2» Allow ECF from the calculated F in (b)(i) Award [2] for a bald correct answer.

c horizontal/repulsive force and vertical force/pull of gravity act on the ball  so ball has constant acceleration/constant net force  motion is in a straight line  at 30° to vertical away from wall/along original line of thread

d i \frac{Q}{0.22^2}=\frac{1.2\times 10^-^6}{0.18^2}  «+»1.8 x 10-6 «C» ✓ 2sf  Do not award MP2 if charge is negative Any answer given to 2 sig figs scores MP3

d ii work must be done to move a «positive» charge from infinity to P «as both  charges are positive» OR reference to both potentials positive and added OR identifies field as gradient of potential and with zero value therefore, point P is at a positive / non-zero potential  Award [0] for bald answer that P has non-zero potential