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