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Physics Factsheet April 2002

www.curriculumpress.co.uk

Number 35

Comparison of Gravitational and Electric Fields This Factsheet will highlight the similarities and differences of gravitational and electric fields. Before working through this Factsheet, you should ensure you understand the two types of field individually.

These expressions can be used to determine an expression for the gravitational field strength due to a point mass m and the electric field strength due to a point charge Q :

Nature of field

Field Strength

Both electric and gravitational fields are vector quantities. • Gravitational fields act on all particles with mass.

Gravitational case

Electric case

r

They are always attractive.

r

F

M

•

q

m F

Q

Electric fields act on all particles with charge. They may be attractive or repulsive

F=G×

Field Strength or Intensity

Field strength F g = m = G× M2 Nkg-1 r

For both electric and graviational fields, there is a relationship between field strength and force on a body in the field.

•

For a gravitational field, F = mg (F = force; m = mass; g = gravitational field intensity)

•

For an electric field, F = qE (F = force; q = charge; E = electric field intensity)

Mm r2

Gravitational force F = mg

Qq F= 1 × 2 4πε0 r Field strength: Q F 1 F = q = 4πε × 2 NC-1 r 0 Electric force F = qE

The decline of either field with distance can be represented graphically: E or g

Exam Hint: - "g" is only equal to 9.81ms-2 at the earth's surface.

To measure the strength of a gravitational field we place a test mass m in the field and measure the force F on it . F, giving gravitation The intensity of the field is then defined to be the ratio m -1 field strength the units Nkg

r Note that these expressions for field strength assume that we are considering a point mass or point charge - which are useful approximations if the size of the body concerned is small compared to the distances being considered.

To measure the strength of an electric field, we place a positive test charge q in the field and then measure the force F on it. The intensity of the field is defined to be the ratio qF, giving electric field strength the units NC-1.

For a uniform sphere, the gravitational or electrical field outside it will be as if all of the body's mass or charge was concentrated at the centre of the body.

Variation of force and field with distance

What affects field strength?

The gravitational force Fg between two masses m1 and m2 separated by a distance r is given by Newton’s law: m1 m2 Fg = G r2 The electrostatic force Fe between two charges q1 and q2 separated by a distance r in a material of relative permittivity εr is given by Coulomb’s law: q q Fe = 1 × 12 2 4π εrε0 r Both forces follow an inverse square law. The constants of proportionality are G for gravitation 1 for electrostatic 4πεrε0

1

•

Field strength is always positive for a gravitational field, but for an electric field, may be positive (for the field produced by a positive charge) or negative (for a field produced by a negative charge).

•

For both electric and gravitational fields, field strength is affected by distance from the body producing the field - they follow inverse square laws.

•

Field strength is directly proportional to the mass of the body for gravitational fields, and the charge on the body for electric fields.

•

The strength of the gravitational field is not affected by the medium - it will be the same whether the body is in air, vacuum, water... The strength of the electric field is affected by the medium - it is inversely proportional to the relative permittivity of the medium.

Physics Factsheet

Comparison of Gravitational and Electric Fields

www.curriculumpress.co.uk Field lines, equipotentials and movement in fields Both types of field can be represented by field lines. • The closer together the lines are, the stronger the field.

Potential V For both gravitational and electrical fields, the potential at infinity is defined to be zero. • For a gravitational field, the potential is defined as the work done in bringing a test unit mass from infinity to that point. It is given by Vg = − GM r

•

•

The gravitational potential energy of a mass in a gravitational field is given by: (p.e)g = m × Vg ∆Vg The gravitational field strength is given by g = − ∆r For an electric field, the potential is defined as the work done in bringing a test unit positive charge from infinity to that point. Q 1 It is given by Ve = 4πε × r 0 The electrical potential energy of a charge in an electric field is

The direction of the lines indicates the direction a mass would move in a graviational field, and the direction a positive charge would move in an electric field (a negative charge would move in the opposite direction).

Equipotentials - surfaces joining points of equal potential - are at rightangles to the field lines in both cases.

An unsupported mass m is in a gravitational field will always move from a position of high gravitational potential to a point where the potential is lower - in the same direction as the field lines. Direction of V increasing (to zero at ∞)

given by: (p.e.)e = q × Ve ∆Ve The electric field strength is given by E = − ∆r M In both cases, potential is inversely proportional to r, and field strength is given by − gradient of potential.

Similarly, a positive charge that is free to move will always move from a position of high electrical potential to a point where the potential is lower - in the same direction as the field lines.

Comparing the sizes of gravitational and electrostatic forces Gravitational forces only really become important when very large objects are involved - you feel the gravitational force due to the Earth, but not due to other people, for example. Electrostatic forces, in contrast, are quite easily noticeable - you will have seen experiments with charged spheres demonstrating "like charges repel, unlike charges attract".

The electric field and associated potential due to a positively charged body will be positive, so the potential decreases to infinity.

To illustrate this, consider a hydrogen atom.

Direction of V increasing (from zero at ∞)

e p+

+Q

R

In this field, as we might expect, a positive test charge will tend to move outwards, and a negative charge inwards.

The proton and electron are oppositely charged; the magnitude of the 1 charge on each is 1.6 ×10-19C. 4πε is approximately 9 × 109 Fm-1 0 So the electrical force is given by: Fe = 9 × 109 ×

(1.6 × 10 ) 2.3 × 10 = R2 R2 -19 2

The electric field and associated potential due to a negatively charged body will be negative, so the potential increases to infinity.

-28

N Direction of V increasing (to zero at ∞)

The masses of the particles are mp = 1.7 ×10-27kg and me = 9.1 ×10-31kg. G is approximately 6.67×10-11 Nm2kg-2. So the gravitational force is given by: Fg = 6.67 × 10-11

positive charge

-Q

1.7 × 10-27 × 9.1 × 10-31 1.03 × 10-67 N = R2 R2

negative charge

To compare these forces we find their ratio: In this field, a positive test charge will tend to move inwards, and a negative charge outwards.

Fe

2.3 × 10-28 = ≈ 1039 Fg 1.03 × 10-67

This shows why it is acceptable, when calculating electrostatic forces between small objects, to ignore the effects of the gravitational force between them.

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Comparison of Gravitational and Electric Fields

Physics Factsheet www.curriculumpress.co.uk

Summary Gravitation

Newton's Law

G = 6.67 × 10-11 Nm2kg-2 m1 m2 F= G 2 r

Field Strength

g=

Constant

Potential Acts on Force

Constant Coulomb's Law

Gm r2 -Gm V= r

Field Strength Potential

General mass distribution F g= m F = mg

Force

General charge distribution F F = Eq E= q

Field

E=−

Acts on

∆Vg ∆r

Electrical 1 = 9 × 109 Fm-1 4πε0 q q Fe = 1 × 1 2 2 4πε0 r q 1 E = 4πε × r 2 0 q 1 V = 4πε × r 0

∆V ∆x

Field

g=−

Work done

W = mV

Work done

W = qV

Interaction

Attractive only

Interaction

Attractive or Repulsive

Medium

No permittivity effect

Medium

Permittivity important

most significant with massive objects such as planets, stars and galaxies

or

E=−

V d

holds atoms and molecules together

Exam Workshop

Questions

This is a typical student’s answer to an exam question. The comments explain what is wrong with the answers and how they can be improved. The examiner’s answer is given below.

1. (a) Write down expressions for field strength for (i) the gravitational field due to a point mass M (ii) the electric field due to a point charge Q

(a) Illustrate mathematically why we do not feel the gravitational force of attraction between ourselves and other people. [5] Take G = 6.67 × 10-11 Nm2kg-2 People have masses M and m, and are a distance r apart. m1 m2

m m So force on them is F = G 2 = 6.67 × 10-11 1 2 2 r r

(b) Sketch graphs to show how field strength declines with distance for the two cases in (a). 2. (a) Define potential for electrical and gravitational fields, and state its relationship to field strength.

2/5

(b) Explain what is meant by an equipotential.

The candidate has the correct idea, but without putting actual values in for m1, m2 and r, the argument is not convincing, so full marks cannot be awarded. Substitution of "sensible" values for the masses and distance would show that the force involved was too small to be felt.

3. State the factors that affect field strength for electric fields and gravitational fields. 4. Discuss the similarities and differences between electric and gravitational fields.

(b) (i) Estimate the mass of an object if you experienced a gravitational force of 100N at a distance of 1m from it. [3]
2 12 1/3 m1 = Fr /Gm2 = 1.5× 10 kg 

Answers Answers to questions 1, 2 and 3 can be found in the text.

The equation has been rearranged correctly, but this time the candidate has simply ignored m2, rather than substituting in a value. If the candidate had written down the step in which the substitution was carried out, s/he might have realised this mistake.

4. Differences: Gravitational forces can only be attractive; electric forces can be attractive and repulsive. Gravitational fields exist around masses; electrical around charges. Medium is important for electrical fields, not for gravitational fields

(ii) Calculate the charge that you and the object would each need, (assuming both charges are the same), to feel the same electrostatic force at this distance. Take 1 = 9 × 109 Fm-1 4πε0
q = √(4πεor2F) = 9.5 × 105 C  1/3

Similarities: Field strengths follow inverse square relationships with distance Work done in moving through the field is (mass or charge) × potential Potential is proportional to 1/distance

The problem here is that the candidate has taken 4πε0= 9 × 109 If s/he had shown more working, s/he might have gained a second mark, despite this mistake. Examiner's Answers. (a) F = Gm1m2/r2
For 2 people each of mass 60kg
, 1m
apart (or other sensible values) F= 6.67 × 10-11 × 60 × 60/1
= 0.24µN


Acknowledgements: This Physics Factsheet was researched and written by Keith Cooper The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

(b) (i) m1 = Fr2/Gm2
= 100 × 1/(G × 60)
= 2.5 × 1010kg (ii) q2 = F × 4πε0 × r2
q = √(100 × 4πε0 × 12)


q ≈ 10 –5C


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