Define electric potential at a point in an electric field.

A positive point charge is moving towards a small, charged metal sphere along a radial path.

At the position shown in the diagram, the point charge has a speed of 3.50 m s^–1 and is at a distance of 2.78 m from the centre of the metal sphere. The charge on the sphere is +4.00 μC.

(i) State the direction of the velocity of the point charge with respect to an equipotential surface due to the metal sphere.

(ii) Show that the electric potential V due to the charged sphere at a distance of 2.78 m from its centre is 1.29 ×10⁴ V.

(iii) The electric potential at the surface of the sphere is 7.20 ×10⁴ V. The point charge has a charge of +0.0320 μC and its mass is 1.20 ×10^{– 4} kg. Determine if the point charge will collide with the metal sphere.

the work done per unit charge;
when a small/test/point positive charge; (charge sign is essential)
is moved from infinity to the point;

(i) perpendicular / at right angles / at 90° / normal;

(ii) \(V = \frac{{8.99 \times {{10}^9} \times 4.00 \times {{10}^{ - 6}}}}{{2.78}}\) or \(1.2935 \times {10^4}{\rm{V}}\); (use of \(\frac{1}{{4\pi {\varepsilon _0}}}\) gives 1.29378×10⁴)
(≈1.29×10⁴V)

There were three marks for this question and most scored 1 or 2. The full definition of electric potential at a point was simply not well enough remembered. Many forgot that it is (i) the work done per unit charge on a (ii) positive (test) charge (iii) as the charge is moved from infinity to the point in question.

(i) Most failed to read the question and did not realize that the question required the direction with respect to an equipotential surface. Most gave the direction relative to the diagram.
(ii) This was well done by a great many, however examiners did expect to see a full substitution, it was not acceptable to leave the permittivity of free space as a symbol in this “show that” question.
(iii) Even where candidates could negotiate a path through this testing question, solutions were seldom clear.