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Questions on comparing electric and gravitational fields
Every gravitational equation has an electric twin of the same form, with charge for mass and 1/4πε₀ for G, and both are inverse square laws. Charge has two signs and mass has one, so electric forces can repel while gravity cannot. Between two protons the electric repulsion is about 10³⁶ times the gravitational attraction, whatever their separation.
18 original questions · 51 marks · the comparing electric and gravitational fields notes · Electric fields
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Write down the equation for the gravitational force between two point masses and the equation for the electrostatic force between two point charges.
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Gravitational: F = GMm/r2, depending on the product of the masses (1). Electrostatic: F = kQ1Q2/r2, depending on the product of the charges; both are inverse-square laws (1).Write down the equation for the field strength in a radial gravitational field and in the radial electric field of a point charge.
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Gravitational: g = GM/r2 (1). Electric: E = kQ/r2 (1).State one fundamental difference between the gravitational force and the electrostatic force between two point objects.
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The gravitational force is only ever attractive, as there is one type of mass (1); the electrostatic force can be attractive or repulsive, as charge exists in two types (1).State the SI unit of gravitational field strength and the SI unit of electric field strength.
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Gravitational field strength: N kg−1 (1). Electric field strength: N C−1, equivalently V m−1 (1).The ratio of the electrostatic force to the gravitational force between two protons is the same whatever their separation. State why.
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Both forces follow an inverse-square law, so the r2 factors cancel in the ratio (1).Write down the equation for the potential in a radial gravitational field and in a radial electric field, and state one difference between the two potentials.
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Gravitational: V = −GM/r (1). Electric: V = kQ/r (1). Gravitational potential is always negative, whereas electric potential is positive near a positive charge and negative near a negative charge (1).Calculate the ratio of the electrostatic force to the gravitational force between a proton and an electron.
e = 1.60 × 10−19 C, mp = 1.67 × 10−27 kg, me = 9.11 × 10−31 kgMark scheme
The separation cancels: ratio = ke2/(Gmpme) (1)
= (8.99 × 109 × (1.60 × 10−19)2)/(6.67 × 10−11 × 1.67 × 10−27 × 9.11 × 10−31) (1)
ratio = 2.27 × 1039 (1)The electrostatic force between charged particles is very much stronger than the gravitational force between them. Explain why gravity is nevertheless the dominant force between planets and stars.
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Astronomical bodies are almost exactly electrically neutral (1); their positive and negative charges cancel, leaving almost no net electrostatic force (1). Mass cannot cancel, so with very large masses the gravitational attraction dominates (1).State two similarities between gravitational and electric fields.
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Any two of the following, one mark each (2): both obey inverse-square force laws; both are radial around a point source; field strength is defined as force per unit mass or per unit charge; potential in both has its zero at infinity.Near the Earth's surface, g = 9.81 N kg−1. Calculate the electric field strength that would support an electron against its weight, and comment on your answer.
e = 1.60 × 10−19 C, me = 9.11 × 10−31 kgMark scheme
Weight = mg = 9.11 × 10−31 × 9.81 = 8.9 × 10−30 N (1)
For balance EQ = mg, so E = mg/e = 5.6 × 10−11 V m−1 (1)
An extremely small field balances gravity, which shows how weak gravity is on a subatomic particle: in laboratory fields the electron's weight is negligible (1)Calculate the gravitational field strength and the electric field strength at a distance of 5.3 × 10−11 m from a proton.
G = 6.67 × 10−11 N m2 kg−2, k = 8.99 × 109 N m2 C−2, mp = 1.67 × 10−27 kg, e = 1.60 × 10−19 CMark scheme
g = Gmp/r2 = (6.67 × 10−11 × 1.67 × 10−27)/(5.3 × 10−11)2 (1)
g = 4.0 × 10−17 N kg−1 (1)
E = ke/r2 = (8.99 × 109 × 1.60 × 10−19)/(5.3 × 10−11)2 (1)
E = 5.1 × 1011 N C−1 (1)Explain why gravitational potential is always negative, while the electric potential near a proton is positive.
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Gravity is attractive, so the field itself does the work of bringing a mass in from infinity: the work done per unit mass by an external agent is negative (1). A positive test charge is repelled by a proton, so external work must be done on it to bring it in from infinity, making the potential positive (1).Two protons in a nucleus are separated by 1.0 × 10−15 m. Calculate the electrostatic force and the gravitational force between them.
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Felec = ke2/r2 = (8.99 × 109 × (1.60 × 10−19)2)/(1.0 × 10−15)2 (1)
Felec = 230 N (repulsive) (1)
Fgrav = Gmp2/r2 (1)
Fgrav = 1.86 × 10−34 N (attractive) (1)Compare gravitational and electric fields under the following headings: the source of the field, the force law, whether attraction and repulsion occur, and the sign of the potential.
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Source: mass, compared with charge (1). Force law: F = GMm/r2 compared with F = kQ1Q2/r2, both inverse-square (1). Gravity gives attraction only; the electric force gives attraction or repulsion (1). Potential: V = −GM/r is always negative; V = kQ/r takes either sign (1).Explain why gravity is noticeable in everyday life while the much stronger electric force between everyday objects is not.
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Everyday objects are electrically neutral overall (1); their positive and negative charges cancel, so they exert almost no net electric force on each other (1). Mass does not cancel, so the gravitational pull of the very large mass of the Earth is easily noticed (1).Two identical conducting spheres, each of mass 1.0 kg, have their centres 1.0 m apart. (a) Calculate the gravitational force between them. (b) Determine the charge each sphere would need, equal on both, for the electrostatic repulsion to balance the gravitational attraction, and comment on the size of this charge.
G = 6.67 × 10−11 N m2 kg−2, k = 8.99 × 109 N m2 C−2Mark scheme
(a) F = Gm2/r2 = 6.67 × 10−11 N (1)
(b) For balance, kQ2/r2 = Gm2/r2 (1)
Q = m√(G/k) (1)
Q = 8.6 × 10−11 C (1)
A charge below a tenth of a nanocoulomb, about 5.4 × 108 electrons' worth, cancels the entire gravitational attraction: the electrostatic force is enormously the stronger (1)A satellite's instruments can be shielded from external electric fields by enclosing them in a metal case. Suggest why nothing can shield them from a gravitational field.
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Shielding requires a second field that cancels the first (1). Charge comes in two signs, so the induced charges on the case set up a field that cancels the external electric field inside (1). Mass comes in one sign only, and there is no negative mass, so no arrangement of matter can produce a cancelling gravitational field (1).Two protons in a nucleus, separated by 1.0 × 10−15 m, repel each other with an electrostatic force of about 230 N, and the gravitational attraction between them is smaller by a factor of about 1036. Explain what can be concluded about the forces acting in a stable nucleus.
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The gravitational attraction between the protons is far too small to balance the 230 N of electrostatic repulsion (1); yet stable nuclei exist, so a further attractive force must act between nucleons, the strong nuclear force (1); at this separation it must be stronger than the electrostatic repulsion it has to beat (1).
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