# CIE AS & A Level Physics : 18.1 Electric fields and field lines – Exam style question – Paper 4

### Question

(a)     State what is represented by an electric field line.

(b)    Two point charges P and Q are placed $$0.120$$ m apart as shown in Fig. 4.1. (i)      The charge of P is $$+4.0$$ nC and the charge of Q is $$–7.2$$ nC.

Determine the distance from P of the point on the line joining the two charges where the electric potential is zero.

(ii)     State and explain, without calculation, whether the electric field strength is zero at the same point at which the electric potential is zero.

(iii)    An electron is positioned at point X, equidistant from both P and Q, as shown in Fig. 4.2. On Fig. 4.2, draw an arrow to represent the direction of the resultant force acting on the electron.

Ans:

(a)     direction of force

force on a positive charge

(b)    (i)       $$V$$ = $$\frac{Q}{4\pi \varepsilon _{0}r}$$

$$\frac{4.0 \times 10^{-9}}{4\pi \varepsilon _{0}x} + \frac{-7.2\times 10^{-9}}{4\pi \varepsilon _{0}(0.120 – x)}$$

$$4(0.120 – x)$$ = $$7.2$$ x

x = $$0.043$$ m

(ii)    fields are in the same direction so no

(iii)   straight arrow drawn leftwards from X in direction between extended line joining Q and X and the horizontal ### Question

(a) State a similarity between the gravitational field lines around a point mass and the electric  field lines around a point charge.                                                                                                               
(b) The variation with radius r of the electric field strength E due to an isolated charged sphere in  a vacuum is shown in Fig. 6.1. Use data from Fig. 6.1 to:
(i) state the radius of the sphere

(ii) calculate the charge on the sphere.
charge = …………………………………………….. C 

(c) Using the formula for the electric potential due to an isolated point charge, determine the capacitance of the sphere in (b).
capacitance = ……………………………………………… F 
[Total: 7]

Ans

(a) (both have) radial field lines
(b) (i) 2.1 cm

(b) (ii) $$E=\frac{Q}{4\pi \varepsilon _{o}r^{2}}$$

e.g. r = 2.1 cm, E = 1.30 × 105 V m–1

$$Q=4\pi \varepsilon_{o}r^{2}E$$

$$=4\times \pi \times 8.85\times 10^{-12}\times 0.021^{2}\times 1.30\times 10^{5}$$

$$=6.4\times 10^{-9}C$$

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