Question: 7 (a) Explain the meaning of an electric field.
(b) There is a uniform electric field between two vertical metal plates AB and CD that are placed in parallel. The potential difference between the two metal plates is 450 V and the separation of the plates is 16 mm. An alpha particle is accelerated from plate AB to plate CD.
(i) Draw electric field lines between the plate AB and plate CD.
(ii) Determine the electric field between the plate AB and plate CD.
(iii) Determine the work done by the electric field on the alpha particle as it moves a distance of 0.016 m from plate AB to plate CD.
(iv) A beta particle has moved from plate AB to plate CD. Determine the ratio of the work done by the electric field on the alpha particle to the work done by the same electric field on the beta particle.
Mark Scheme:
7 (a) It is a region in space where a (stationary) charge experiences an (electric) force [1]
(b) (i) Students should draw at least four parallel straight lines that are perpendicular to the two plates. The lines representing electric field should also be equally spaced. [1]
There should be an arrow on the lines from left to right that indicates the direction of the electric field. [1]
(ii) Use the equation of electric field, E = V/d [1]
E = (450 / 16 × 10–3) = (28 125) V/m [1]
(iii) The work done, W = Eqd or Vq [1]
q = 3.2 × 10–19 C [1]
Thus, W = 450 × 3.2 × 10–19 = 1.4(4) × 10–16 J [1]
(iv) Ratio of work done = (450 × 3.2 ×10−19) / (450 × -1.6 ×10−19) = -2 [1]
Possible answers:
7 (a) An electric field is a region in space where an infinitely small test charge that is stationary experiences an electric (or electrostatic) force.
(b) (i) The electric field between the plate AB and plate CD is shown below:
(ii) The electric field between two plates is uniform.
Thus, E = V / d = (450 / 16 × 10–3) = 28,125 V/m
(iii) The work done by the electric field on the alpha particle can be calculated by using the formula, W = Eqd = qV.
Thus, W = qV = (3.2 × 10–19) × 450 = 1.44 × 10–16 J
(iv) The ratio of the “work done by the electric field on the alpha particle” to “work done by the same electric field on the beta particle” = qaV / qbV = qa / qb
Thus, the ratio = (3.2 ×10−19) / (-1.6 ×10−19) = -2
Feynman’s insights or goofs?:
In the words of Feynman, “the existence of the positive charge, in some sense, distorts, or creates a ‘condition’ in space, so that when we put the negative charge in, it feels a force. This potentiality for producing a force is called an electric field. When we put an electron in an electric field, we say it is “pulled.” We then have two rules: (a) charges make a field, and (b) charges in fields have forces on them and move (Feynman et al., 1963, section 2.2 Physics before 1920).” That is, Feynman visualizes a positively charged particle creates a “condition” in space such that when we put another charged particle there, it feels an electric force.
Furthermore, Feynman elaborates that the nature of electric field can be illustrated by an analogy as follows: “If we are in a pool of water and there is a floating cork very close by, we can move it “directly” by pushing the water with another cork. If you looked only at the two corks, all you would see would be that one moved immediately in response to the motion of the other—there is some kind of “interaction” between them. Of course, what we really do is to disturb the water; the water then disturbs the other cork (Feynman et al., 1963, section 2.2 Physics before 1920).” In other words, there is a disturbance in the water (an electric field) caused by a cork (a charged particle), and the water then disturbs the other cork directly.
References:
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
