*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.

**Answer/Explanation**

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**

**Question**

(a) Define electric potential at a point.

(b) Two point charges A and B are separated by a distance of 12.0cm in a vacuum, as illustrated

in Fig. 5.1.

The charge of A is \(+2.0 × 10^{–9}\)C.

A point P lies on the line joining charges A and B. Its distance from charge A is x.

The variation with distance x of the electric potential V at point P is shown in Fig. 5.2.

Use Fig. 5.2 to determine:

(i) the charge of B

charge = …………………………………………….. C

(ii) the change in electric potential when point P moves from the position where x = 9.0cm to the position where x = 3.0cm.

change = ……………………………………………… V

(c) An α-particle moves along the line joining point charges A and B in Fig. 5.1.

The α-particle moves from the position where x = 9.0cm and just reaches the position where

x = 3.0cm.

Use your answer in (b)(ii) to calculate the speed v of the α-particle at the position where

x = 9.0cm.

v = ………………………………….. \(ms^{-1}\)

**Answer/Explanation**

**Answer:**

(a) work done per unit charge

(work done on charge) moving positive charge from infinity

(b) (ii) \((2.0 × 10^{–9}) / 4 πε_0(4.0 × 10^{–2}) + Q / 4 πε_0(8.0 × 10^{–2}) = 0\)

\(Q = 4.0 × 10^{–9}\) C

Q given with negative sign

(ii) change = 1200 V

(c) \(1⁄2mv^2 = qV\)

\(1⁄2 × 4 × 1.66 × 10^{–27} × v^2 = 2 × 1.60 × 10^{–19} × 1200\)