AH 2.1.1 Electric Fields & Coulomb's law

Question 1

Three identical charges A, B and C are fixed at the positions shown in the right angled triangle below.

Each charge is +8 nC (i.e. +8.0 x 10-9 C) in magnitude.

(a) Calculate the forces exerted on charge A by charges B and C.
(b) Calculate the resultant force on charge A. (This means magnitude and direction)

Answer 1

Question 2

Fill in the missing words below.

For very small distances, for example between two ………………….. in a nucleus, there would be a strong
electrostatic force of repulsion. Why does the nucleus not fly apart? There is an even stronger attractive
…………………. nuclear force which only acts over extremely short distances. At distances above 1 x 10^-15 m the strong nuclear force becomes negligible and the ……………………………………. force becomes the dominant force. This is the force which basically binds atoms together and binds atoms to their neighbours.

Magnetic forces are due to ……………………. rather than stationary charges and are essentially the same force known as the ……………………………… force. At larger distances the ……………………………. force is dominant since most large scale objects are neutral or not significantly charged.

Answer 2

For very small distances, for example between two protons in a nucleus, there would be a strong
electrostatic force of repulsion. Why does the nucleus not fly apart? There is an even stronger attractive
strong nuclear force which only acts over extremely short distances. At distances above 1 x 10^-15 m the
strong nuclear force becomes negligible and the electrostatic force becomes the dominant force. This is the force which basically binds atoms together and binds atoms to their neighbours.

Magnetic forces are due to moving rather than stationary charges and are essentially the same force known as the electromagnetic force. At larger distances the gravitational force is dominant since most large scale
objects are neutral or not significantly charged.

Question 3

Explain the concept of an electric field.

Answer 3

A electric field is a region in space in which a test charge would feel and electrostatic force.

Question 4

Define electric field strength.

Answer 4

The electric field strength E at any point in an electric field is the force on a unit positive charge placed at that point.
If a test charge q_, placed at a point P in the electric field, experiences a force F then:

E = F/q

Electric Field strength therofre has units N C^-1

Question 5

What is the electric field strength at a point where a small object carrying a charge of 6.00 μC experiences a force of 0.0400 N?

Answer 5

E = F/q

= 0.0400 / 6.00 x 10^-6

E = 6670 N (to 3 s.f. as in data)

Question 6

Define electrostatic potential and state it’s unit.

Answer 6

The electrostatic potential V at a point in an electric field is defined as the work done by an external force in bringing a unit positive charge from infinity to that point.

So V = W/q

Electric potential thereofore has units J C^-1

Question 7

What equation gives the electrostatic potential V a distance r from a point charge Q?

Answer 7

V = (1/4πε₀ ) Q/r

Question 8

Calculate the electrostatic potential at a point, P, which is at a distance of 0.05 m from a point charge of + 3.0 x 10^-9 C.

Answer 8

V = (1/4πε₀ ) Q/r

= (1/4πε₀ ) x 3.0 x 10^-9 / 0.05

= (1/4π x 8.85 x 10^-12) x 3.0 x 10^-9 / 0.05

V = + 540 V

Remember to specify the sign of the potential (+ or -)

Question 9

By considering the uniform field between two oppositely charged parallel metal plates, derive the expression

V = Ed for a uniform electric field.

Answer 9

The work done by an external force in moving a charge q against the field is given by

W = qV

Fx d = qV as Work done = Force x distance

F/q = V/d

E = V/d

V=Ed

Question 10

a) What is the alternative expression for electric field strength when it describes a unform field E=V/d?
b) What is the unit of E in this case?
c) Why can this equation (E=V/d) not be used to find the field strength near a point charge?
d) Which equation does give the electric field srength near a point charge?

Answer 10

a) E = F/q
b) N C^-1
c) E=V/d only applies to unifrom fields (between paralel charged metal plates). The firld around apoint charge is non uniform - in fact, radial.

d)

E = F/q

And F = (1/4πε₀ ) Q q/r²

So E = (1/4πε₀ ) Q /r²

Question 11

a) Plate Y below is at a potential of +500 V. The metal plates are 80 mm apart.

Find the electric field strength between the plates?

b) Where between the plates does the field strength have this value?

Answer 11

a) E = V/d

= 500 / 0.080

E = 6250 Vm^-1

b) As this is a uniform field the field strength E = 6250 Vm^-1 at all points in the field bewteen the plates.

(Note that immediately outside the plates the field lines curve so E is not uniform outside the overlap of the plates).

Question 12

Define the electron-Volt.

Answer 12

The electron-volt is a non-SI unit of energy.

It is the amount of energy required to move an electron against the field through potential difference of 1 Volt and is given by

E_w=qV = eV = 1.6 x 10^-19 x 1 = 1.6 x 10^-19 J

Question 13

a) Sketch a positive main charge Q₁ and a positive test charge Q₂ a distance r from the centre of the main charge,
b) Write down an expression for the electrostatic potential due to Q₁ at this distance.
c) Write down an expression for the electrostatic potential energy of this system of two charges.

Answer 13

a) See diagram below.
b) V = (1/4πε0 ) Q₁/r

Note that the test charge is, by convention, positive.

Since the main charge is positive, the electrostatic potential V is positive. This inidcates that work must be done on the test charge to bring it to this point form infinity. That is, energy must be supplied to the system.

c) Electrostatic Potential Energy

E_p = (1/4πε₀ ) Q₁Q2/r

Question 14

The potential difference between the plates below is 200V. An electron is released from the lower plate X.

a) Describe what happens to the electron subsequently.
b) Calculate the velocity of the electron when it is midway bewteen the plates.

Answer 14

a) The electron accelerates uniformly towards the positive plate.
b) As this is a uniform field the potential halfway bewteen the plates is 1/2 x +200 = + 100V

So the electron has moved through a potetial difference of +100 V

The work done by the field on the electron is

Ew = qV = eV = 1.6 x 10^-19 x 100 = 1.6 x 10^-17 J

E_w = E_k = 1/2 m_e v²

1.6 x 10^-17 = 1/2 x 9.12 x 10^-31 x v²

v² = 1.6 x 10^-17 / 1/2 x 9.12 x 10^-31

v = 5.92 x 10⁶ ms^-1

Remember not to calculate intermediate answers to avoid rounding errors.

Question 15

An alpha particle and an electron are fired horizontally midway between two charged horizontal plates from the right, each travelling to the left at the same horizontal velocity. The diagram shows a plan view of the arrangement (i.e from above).

Sketch the path of both particles in the plane of the page on the same diagram when they are between the plates.

Explain the difference in the trajectories.

Answer 15

The alpha particle is positive so feels an unbalanced electrostatic force towards the negative plate.

The beta particle is negative so feels an unbalanced electrostatic force towards the 0V plate.

The electrostatic force on the alpha is F_a = qV = 2eV

The electrostatic force on the beta is F_{<span>b</span>} = qV = eV

So the alpha feels twice the electrostatic force of the beta.

However, the alpha has a mass of 6.64 x 10^-27 kg

The beta is an electron and has mass 9.1 x 10^-31 kg

So the alpha is of the order 10 000 times more massive.

Therefore, tha alpha curves much less than the beta, as shown in the diagram.

Question 16

Fast moving protons strike a glass screen with a speed of 2.0 x 10⁶ ms^-1. Glass is largely composed of silicon, which has an atomic number of 14. The diagram below shows the proton approaching the Silicon nucleus.

a) Describe the motion of the proton as it approaches the silicon nucleus.
b) Calculate the distance of closest approach that a proton could make in a head-on collision with a silicon nucleus.

Answer 16

The proton decelerates as it approaches the Silicon nucleus as both the proton and the nucleus carry a positive charge and so repulsion takes place.

The repulsive force is the Coulomb force given by

F = (1/4πε₀ ) Q_alphaQ_silicon/r²

Where r = separation ditance at any instant in time.

Clearly, as the proton gets closer, r decreases so the replusive force get’s stronger and the deceleration greater. So this is a non-uniform deceleration.

b) E_k lost by alpha particle = (electrostatic) E_p gained

1/2 mv² = (1/4πε₀ ) Q_alphaQ_silicon/r

1/2 x 9.11 x 10^-31 = [(1/4πε₀ ) x 2 x 1.6 x 10^-19 x 14 x 10^-19 ]/r

r = [(1/4πε₀ ) x 2 x 1.6 x 10^-19 x 14 x 10^-19 ]/1/2 x 9.11 x 10^-31

r = 9.7 x 10^-13 m

Question 17

The charge on the electron was measured by Millikan in an ingenious experiment. The method involved accurate measurements on charged oil drops moving between two parallel metal plates with a voltage across them.

In the diagram below the voltage V between the plates has been adjusted so that the oil drop is stationary, as observed through the microscope. The plates are distance d apart.

Using the potential gradient equation E=V/d and the forces shown in the diagram, find an expression for the charge on the oil drop.

Answer 17

Vertical forces are balanced, so

mg= qE = qV/d

q = mgd/V

Question 18

Explain how Millikan determined the charge on an electron from the results of his measuremnts of the charges on many different oil drops.

Answer 18

The charges on the oil drops were always an integer multiple of a basic charge. This basic charge was found to be = 1.6 x 10^-19 C = thereafter called the electronic charge, e.