# Chapter 21 - Worked Solutions

### to selected extension questions from the OUP text *Senior Physics -
Concepts in Context* by Walding,
Rapkins and Rossiter

**Q6 :-**
If spheres A and B are touched, a nett charge of +5 mC is distributed in the ratio of the surface areas.

S.A sphere(A) = 4 x S.A sphere(B). as twice the radius, hence the charge on (A) is 4/5 x +5 mC = +4 mC. and the remainder of +1 mC will be on sphere(B).

**Q14 :-**
Voltage between parallel plates (V) = 300 V and the plate separation is 30.0 mm.

(a) Use **E = V/d** yields electric field strength of 1.0 x 10^{4} V.m^{-1}.

(b) The force on the charge **F = q.E** acts upwards and is calculated by substitution to be 6.0 x 10^{-2} N upwards

(c) Work done (W) = q.D V = energy gained, but as voltage difference = 200 V then then energy gained calculates to be 1.3 x 10^{-3} J.

**Q31 :-**

The data is plotted firstly as F v d, and then as F v 1/d^2. as shown below

(a)The data point in greatest error is the one furthest from the ideal inverse square curve.

(b) The slope of the 1/d^2 line will be(m)= k.Q.q, hence by calculating its value the Coulomb constant (k) can be found by substitution.

**Q32 :-**

Given electronic mass and charge m_{e}= 9.1x10^{-31}kg : and q_{e}= 1.6 x 10^{-19}C

(a) As V = 320V, use electron gun velocity v = (2qV/m)^{½} to obtain the beams horizontal velocity v = 1.06 x 10^{7} m.s^{-1}.

If the deflecting zone width is 0.022 m, then time of travel through the deflecting zone is (t)= d/v which yields a time of 2.1 x 10^{-9}s.

(b) Within the field **F = q.E** but as electric field E = V/d where V = 50V, hence F = qV/d which by substitution yields an upward deflecting force of 4.0 x 10^{-16}N.

To calculate the vertical displacement use kinematics :-

acceleration a = F/m yields the value of a = 4.4 x 10^{14} m.s^{-2}.

Hence vertical displacement d = ½.a.t^{2} yields d = 9.7 x 10^{-4} m in an upwards direction.

(c) To calculate the vertical component of the beam velocity at the exit of the deflection zone, use v = a.t which yields a value of 9.2 x 10 ^{5} m.s^{-1}.

Combining this vectorially with the horizontal component yields an actual velocity of

v = 1.06 x 10^{7} m.s^{-1} at an angle of 5^{0} to the horizontal as the beam hits the screen.

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