Electric Fields: Exercise 2

An electron starts from rest from the bottom plate in Fig. 17.11. The potential difference across the plates is 1600 V and the separation of the plates is 15 mm. Calculate the time taken for the electron to reach the top plate. 

Solution:
F = Vq/d = (1600 x 1.6 x 10^-19) / 0.015 = 1.71 x 10^-14 N
a = force / mass = 1.71 x 10^-14 / 9.1 x 10^-31 = 1.88 x 10^16 ms^-2
s = ut + 0.5 at^2
0.015 = 0 + 0.5 (1.88 x 10^16) (t^2)
t^2 = 1.596 x 10^-18
t = 1.26 x 10^-9 s

For the electron in question 2  calculate:

a  the work done by the field on the electron,

b  the gain in kinetic energy,

c  the speed of the electron.

Solution:

a) work done = Fd = Eqd = Vq

W = 1.71 x 10^-14 x 0.015 = 2.565 x 10^-16 J

b) kinetic energy = Vq = 1600 x 1.6 x 10^-19 = 2.56 x 10^-16 J

c) v^2 = u^2 + 2as = 2 (1.88 x 10^16)(0.015) = 5.64 x 10^14

v = 2.37 x 10^7 ms^-1

Electric Fields: Exercise 1

Calculate the speed of an electron accelerated from rest through a distance of 40 mm by a uniform electric field of 3.0 × 103 N C−1.

Solution:

force = Eq = (3 × 103 × 1.6 x 10−19) = 4.8 × 10−16 N

acceleration = force/mass = 4.8 × 10−16 /9.1 × 10−31 = 5.27 × 1014 ms−2

v2 = 0 + 2as = 2 × 5.27 × 1014 ms−2 × 40 × 10−3 so speed v = 6.5 × 106 m s−1

9702/Oct Nov/42/2015/Q1

Solution:

Reference: Past Exam Paper – Oct Nov 2015 Paper 42 Q1

9702/May June/4/2003/Q1

(a) Define gravitational potential.

(b) Explain why values of gravitational potential near to an isolated mass are all negative.

(c) Earth may be assumed to be an isolated sphere of radius 6.4 × 10³ km with its mass of 6.0 × 10²⁴ kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 10⁴ km.

Calculate, for this object,

(i) change in gravitational potential,

(ii) speed of projection from the Earth’s surface, assuming air resistance is negligible.

(d) Suggest why the equation

v² = u² + 2as

is not appropriate for calculation in (c)(ii).

Solution:

(a) Gravitational potential (at a point) is defined as the work done in bringing/moving unit mass from infinity to the point.

(b) The potential at infinity is defined as being zero. The forces are always attractive, so work got out in moving to point (work is done on the mass when moving it to infinity).

(c)(i)

Gravitational potential, φ = - GM / R = - GM × (1/R)

{The distances should be converted in metre. Initial position is at the surface of the Earth, which is a distance of 6.4×10⁶m from the centre of the Earth. The final distance is 1.3×10⁷m (altitude = distance above surface) + 6.4×10⁶m (distance of surface of the centre) = 1.94×10⁷m}

Change in potential = (6.67×10^-11) (6.0×10²⁴) × ({6.4×10⁶}^-1 – {1.94×10⁷}^-1)

Change in potential = 4.19 × 10⁷ J kg^-1 (ignore sign)

(ii)

The kinetic energy is converted to gravitational potential energy as the height of the object from the surface of the Earth increases.

½ mv² = mΔφ

v² = 2 × 4.19×10⁷ = 8.38 × 10⁷

Speed v = 9150 m s^-1

(d) The acceleration is not constant.

Reference: Past Exam Paper – June 2003 Paper 4 Q1

9702/Oct Nov/11/2010/Q13

A rigid L-shaped lever arm is pivoted at point P.

Three forces act on the lever arm, as shown in the diagram.

What is the magnitude of the resultant moment of these forces about point P?

A 15 N m 

B 20 N m 

C 35 N m 

D 75 N m

Answer: A.

We need to consider which forces will cause a clockwise moment and which cause an anticlockwise moment, that is consider how the lever would rotate about P due to each force.

Clockwise moment is caused by the 5N upward and 10N horizontal forces.

Anticlockwise moment is caused by the 15N downward force.

As for the distances, we need to consider the perpendicular distances from the line of action of the forces from point P.

Perpendicular distance of 5N force: (3m + 1m) – 2m = 2m

Perpendicular distance of 10N force: 2m

Perpendicular distance of 15N force: 3m

Magnitude of resultant moment = Anticlockwise moment – Clockwise moment

Magnitude of resultant moment = (15 × 3) – [(10 × 2) + (5 × 2)] = 15 N m

Reference: PYQ - Oct/Nov 2010 Paper 11 Q13

Electric Fields tough question 1 (From Note)

In a simplified model, a uranium nucleus is a sphere of radius 8.0 × 10⁻¹⁵ m. The nucleus contains 92 protons (and rather more neutrons). The charge on a proton is 1.6 × 10⁻¹⁹ C. It can be assumed that the charge of these protons acts as if it were all concentrated at the centre of the nucleus. The nucleus releases an α particle containing two protons (and two neutrons) at the surface of the nucleus. Calculate

a)  the electric field strength at the surface of the nucleus before emission of the α-particle,

b)  the electric force on the α-particle at the surface of the nucleus,

c)  the electric potential at the surface of the nucleus before emission of the α-particle,

d)  the electric potential energy of the α-particle when it is at the surface of the nucleus.

Solution:

9702/May Jun/42/2010/Q10

(a) State the name of an electrical sensing device that will respond to changes in
(i) length,
(ii) pressure.

(b) A relay is sometimes used as the output of a sensing circuit.
The output of a particular sensing circuit is either + 2 V or – 2 V.
On Fig. 10.1, draw symbols for a relay and any other necessary component so that the
external circuit is switched on only when the output from the sensing circuit is + 2 V.

Solution:

(a)

(i) Strain gauge

(ii) Piezo-electric / quartz crystal / transducer

(b) For the circuit, a coil of relay is connected between the sensing circuit output and earth and the switch is across the terminals of the external circuit.

A diode is connected in series with the coil with a correct polarity for the diode (downwards). A second diode with correct polarity is connected (upwards and parallel to coil of relay).

{Note that the following explanations have been taken directly from the application booklet.

The diode D₁ conducts only when the output is positive with respect to earth and thus the relay coil is energized only when the output is positive. When the current in the relay coil is switched off, a back e.m.f. is generated in the coil that could damage the sensing circuit. A diode D₂ is connected across the coil to protect the sensing circuit from this back e.m.f.}

Reference: PYQ - May/June 2010 Paper 42 Q10

Reference links:

https://electronics.stackexchange.com/questions/100134/why-is-there-a-diode-connected-in-parallel-to-a-relay-coil
https://electronics.stackexchange.com/questions/56322/do-i-need-a-flyback-diode-with-an-automotive-relay/56323#56323