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).
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 104 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 1.2 x 10-3 J.
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.
Given electronic mass and charge me= 9.1x10-31kg : and qe= 1.6 x 10-19C
(a) As V = 320V, use electron gun velocity v = (2qV/m)½ to obtain the beams horizontal velocity v = 1.06 x 107 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 10-9s.
(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 10-16N.
To calculate the vertical displacement use kinematics :-
acceleration a = F/m yields the value of a = 4.4 x 1014 m.s-2.
Hence vertical displacement d = ½.a.t2 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 107 m.s-1 at an angle of 50 to the horizontal as the beam hits the screen.