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Questions on Coulomb's law and electric field strength
Coulomb's law gives the force between two point charges as Q₁Q₂/4πε₀r², repulsive for two like charges and attractive for two unlike ones. Electric field strength is the force per unit positive charge, E = F/Q, uniform at E = V/d between plates and radial around a point charge. A charge entering that uniform field at right angles follows a parabola.
19 original questions · 52 marks · the Coulomb's law and electric field strength notes · Electric fields
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State Coulomb's law for the force between two point charges.
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The force is proportional to the product of the charges (1); and inversely proportional to the square of their separation, F = kQ1Q2/r2 where k = 1/4πε0 (1).Two point charges, each of +2.0 μC, are 0.10 m apart. Calculate the force between them.
k = 8.99 × 109 N m2 C−2Mark scheme
F = kQ1Q2/r2 = (8.99 × 109 × (2.0 × 10−6)2)/0.102 (1)
F = 3.60 N (repulsive) (1)A charge of 2.0 × 10−6 C experiences a force of 6.0 × 10−3 N at a point in an electric field. Calculate the electric field strength at that point.
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E = F/Q = (6.0 × 10−3)/(2.0 × 10−6) (1)
E = 3000 N C−1 (1)The electrostatic force between two point charges is F. The separation of the charges is doubled. State the new force in terms of F.
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F/4, because the force follows an inverse-square law (1).Electric field strength can be given in N C−1 or in V m−1. Show that these two units are equivalent.
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1 V = 1 J C−1, so 1 V m−1 = 1 J C−1 m−1 (1). 1 J = 1 N m, so 1 J C−1 m−1 = 1 N m C−1 m−1 = 1 N C−1 (1).Two charged conducting spheres, each of radius 2.0 cm, are placed with their surfaces 4.0 cm apart. State the separation r that should be used in Coulomb's law for the force between them, and the property of charged spheres that justifies it.
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r = 8.0 cm (0.080 m), the distance between the centres (1). A charged sphere behaves as if all its charge were concentrated at its centre (1).In a hydrogen atom the electron and the proton are separated by 5.3 × 10−11 m. Calculate the electrostatic force between them.
e = 1.60 × 10−19 CMark scheme
F = ke2/r2 = (8.99 × 109 × (1.60 × 10−19)2)/(5.3 × 10−11)2 (1)
F = 8.19 × 10−8 N (1)Two parallel plates separated by 0.050 m have a potential difference of 2000 V across them. Calculate the electric field strength between the plates and the force on a charge of 3.0 × 10−6 C placed between them.
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E = V/d = 2000/0.050 = 40000 V m−1 (2)
F = EQ = 40000 × 3.0 × 10−6 = 0.12 N (1)Calculate the electric field strength at a distance of 0.20 m from a point charge of 5.0 × 10−9 C.
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E = kQ/r2 = (8.99 × 109 × 5.0 × 10−9)/0.202 (1)
E = 1.12 × 103 N C−1 (1)A +3.0 μC charge and a −2.0 μC charge are 0.15 m apart. Calculate the magnitude of the force between them and state whether it is attractive or repulsive.
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F = kQ1Q2/r2 = (8.99 × 109 × 3.0 × 10−6 × 2.0 × 10−6)/0.152 (1)
F = 2.40 N (1)
Attractive, because the charges have opposite signs (1).Determine the separation at which the electrostatic repulsion between two protons is 2.3 × 10−4 N.
k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 CMark scheme
F = ke2/r2, so r = √(ke2/F) (1)
r = √((8.99 × 109 × (1.60 × 10−19)2)/(2.3 × 10−4)) (1)
r = 1.0 × 10−12 m (1)At a distance of 0.15 m from the centre of a small charged sphere, the electric field strength is 8.0 × 103 N C−1. Determine the charge on the sphere (k = 8.99 × 109 N m2 C−2).
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E = kQ/r2, so Q = Er2/k (1)
Q = (8.0 × 103 × 0.152)/(8.99 × 109) (1)
Q = 2.0 × 10−8 C (20 nC) (1)The charge on each of two point charges is doubled and their separation is halved. Determine the factor by which the electrostatic force between them changes.
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F ∝ Q1Q2/r2: doubling both charges multiplies F by 4, and halving r multiplies it by another 4 (1). The force increases by a factor of 16 (1).An electron enters a uniform electric field of strength 1.0 × 104 N C−1. Calculate the acceleration of the electron.
e = 1.60 × 10−19 C, me = 9.11 × 10−31 kgMark scheme
F = eE (1)
a = eE/m = (1.60 × 10−19 × 1.0 × 104)/(9.11 × 10−31) (1)
a = 1.76 × 1015 m s−2 (1)A point charge of 8.0 × 10−9 C is fixed in position. Calculate the electric field strength at a point 0.30 m from the charge, and hence the force on a +2.0 × 10−9 C charge placed at that point.
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E = kQ/r2 = (8.99 × 109 × 8.0 × 10−9)/0.302 (1)
E = 799 N C−1 (1)
F = EQ = 799 × 2.0 × 10−9 (1)
F = 1.60 × 10−6 N (repulsive) (1)A charged particle enters a uniform electric field at right angles to the field lines. Describe the path of the particle in the field and explain why it has this shape.
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The path is parabolic (1). The field exerts a constant force, and so a constant acceleration, parallel to the field lines (1); the velocity component perpendicular to the field is unchanged, so the motion is like that of a projectile (1).An electron enters the uniform field between two horizontal parallel plates, midway between them and moving parallel to them at 2.0 × 107 m s−1. The plates are 40 mm long and 20 mm apart, with a pd of 80 V across them. Assume that gravitational forces are negligible. Determine the deflection of the electron, perpendicular to the plates, by the time it leaves the field.
e = 1.60 × 10−19 C, me = 9.11 × 10−31 kgMark scheme
E = V/d = 80/0.020 = 4000 V m−1 (1)
a = eE/m = (1.60 × 10−19 × 4000)/(9.11 × 10−31) = 7.03 × 1014 m s−2 (1)
time in the field t = L/v = 0.040/(2.0 × 107) = 2.0 × 10−9 s (1)
deflection s = ½at2 (1)
s = 1.4 mm (1)Air conducts, producing a spark, wherever the electric field strength exceeds 3.0 × 106 V m−1. The dome of a school Van de Graaff generator is a sphere of radius 15 cm carrying a charge of 8.0 μC. Deduce whether the air at the surface of the dome will break down.
k = 8.99 × 109 N m2 C−2Mark scheme
The dome acts as a point charge at its centre, so at the surface E = kQ/r2 (1)
E = (8.99 × 109 × 8.0 × 10−6)/0.152 (1)
E = 3.2 × 106 V m−1 (1)
This exceeds 3.0 × 106 V m−1, so the air breaks down and a spark is likely (1)Describe the electric field (a) between two oppositely charged parallel plates and (b) around an isolated positive point charge. In each case state how the field strength varies from place to place.
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(a) A uniform field: the field lines are parallel, equally spaced, and run from the positive plate to the negative plate (1); the field strength E = V/d is the same everywhere between the plates (1). (b) A radial field: the lines point outward from the charge in every direction (1); the field strength falls with the inverse square of the distance, E = kQ/r2 (1).
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