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Comparing electric and gravitational fields questions
Set the two field theories side by side and the equations match line for line, the same mathematics written twice with different constants. The differences are that charge comes in two signs while mass comes in one, and that between two protons the electrostatic repulsion is about 10³⁶ times the gravitational attraction.
17 original questions · 48 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.
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 kgCalculate 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 CExplain why gravitational potential is always negative, while the electric potential near a proton is positive.
Two protons in a nucleus are separated by 1.0 × 10−15 m. Calculate the electrostatic force and the gravitational force between them.
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.
Explain why gravity is noticeable in everyday life while the much stronger electric force between everyday objects is not.
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−2A 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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