AS & A Level Physics (9702)
Electric fields Banker
29 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
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Q1Electric potential & potential energy · Core · Paper 4 style · 4 marksTake ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Define electric potential at a point. [2]
- (b) Calculate the electric potential 0.30 m from a point charge of +5.0 nC. [2]
Show answer
- (a) The work done per unit positive charge in bringing a small test charge from infinity to the point.
B1 work done per unit positive charge; B1 from infinity
- (b) 150 V
V = Q/(4πε₀r) = 8.99 × 10⁹ × 5.0 × 10⁻⁹/0.30 (C1) = 150 V (A1)
Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Explain what is meant by an equipotential surface. [1]
- (b) State the angle between field lines and equipotential surfaces, and explain why no work is done moving a charge along an equipotential. [2]
- (c) Calculate the potential difference between two points at 0.10 m and 0.20 m from a point charge of +4.0 nC. [2]
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- (a) A surface on which every point has the same electric potential.
B1
- (b) 90°; there is no potential difference between points on it, so W = qΔV = 0.
B1 90°; B1 ΔV = 0
- (c) 180 V
V = 8.99 × 10⁹ × 4.0 × 10⁻⁹ × (1/0.10 − 1/0.20) (C1) = 180 V (A1)
Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Explain why the electric potential near an isolated negative charge is negative. [2]
- (b) Calculate the electric potential energy of an electron and a proton separated by 1.0 × 10⁻¹⁰ m. [2]
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- (a) Potential is zero at infinity; a positive test charge is attracted, so work is done by the field (negative work done on it) as it is brought in from infinity.
B1 zero at infinity; B1 attraction → work done on charge is negative
- (b) −2.3 × 10⁻¹⁸ J
Eₚ = −e²/(4πε₀r) (C1) = −2.3 × 10⁻¹⁸ J (A1)
An isolated metal sphere of radius 0.10 m carries a charge of +8.0 nC. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the potential at the surface of the sphere. [1]
- (b) Calculate the potential at a point 0.40 m from the centre of the sphere. [1]
- (c) Calculate the work done in moving a charge of +2.0 nC from the point 0.40 m from the centre to the surface of the sphere. [2]
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- (a) 719 V
V = 8.99 × 10⁹ × 8.0 × 10⁻⁹/0.10 = 719 V (B1)
- (b) 180 V
8.99 × 10⁹ × 8.0 × 10⁻⁹/0.40 = 180 V (B1)
- (c) 1.1 × 10⁻⁶ J
W = qΔV = 2.0 × 10⁻⁹ × (719 − 180) (C1) = 1.08 × 10⁻⁶ J (A1)
An alpha particle with kinetic energy 5.0 MeV travels directly towards a gold nucleus. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Explain why the alpha particle slows down as it approaches the nucleus. [1]
- (b) Calculate the distance of closest approach. The charge of the gold nucleus is +79e and of the alpha particle +2e. Treat the nucleus as fixed. [3]
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- (a) It is repelled by the positive nucleus; its kinetic energy is converted to electric potential energy.
B1
- (b) 4.5 × 10⁻¹⁴ m
KE = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J (C1); KE = (2e)(79e)/(4πε₀d) (C1); d = 4.5 × 10⁻¹⁴ m (A1)
The electric potential is 900 V at a distance of 0.20 m from a point charge and 600 V at 0.30 m from it. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Show that the data are consistent with a point charge, and calculate the charge. [3]
- (b) Use the relationship between field strength and potential gradient to calculate the field strength at 0.20 m from the charge. [2]
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- (a) 2.0 × 10⁻⁸ C
Vr = 900 × 0.20 = 600 × 0.30 = 180 V m, constant (M1); Q = 4πε₀ × 180 (C1) = 2.0 × 10⁻⁸ C (A1)
- (b) 4.5 × 10³ V m⁻¹
E = −dV/dr = Q/(4πε₀r²) = V/r (C1) = 900/0.20 = 4.5 × 10³ V m⁻¹ (A1)
An electron starts from rest at a point where the electric potential is −200 V and moves, under electric forces only, to a point where the potential is +300 V. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the gain in kinetic energy of the electron. [2]
- (b) Calculate its final speed. [2]
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- (a) 8.0 × 10⁻¹⁷ J
ΔE = eΔV = 1.60 × 10⁻¹⁹ × 500 (C1) = 8.0 × 10⁻¹⁷ J (A1)
- (b) 1.3 × 10⁷ m s⁻¹
½mv² = 8.0 × 10⁻¹⁷ (C1); v = 1.3 × 10⁷ m s⁻¹ (A1)
Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) State one similarity and one difference between gravitational potential near a mass and electric potential near a positive point charge. [2]
- (b) Calculate the work done in bringing a charge of +1.5 µC from infinity to a point 0.25 m from a fixed point charge of +40 nC. [2]
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- (a) Similarity: both ∝ 1/r and both zero at infinity. Difference: gravitational potential is always negative; potential near a positive charge is positive.
B1 each
- (b) 2.2 × 10⁻³ J
W = qV = 1.5 × 10⁻⁶ × 8.99 × 10⁹ × 40 × 10⁻⁹/0.25 (C1) = 2.2 × 10⁻³ J (A1)
Two small spheres, each of mass 2.0 g and charge +3.0 µC, are held 0.10 m apart. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the electric potential energy of the system at the start. [2]
- (b) The spheres are released from rest and move apart along a frictionless horizontal track. Calculate the speed of each sphere when they are very far apart. [3]
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- (a) 0.81 J
Eₚ = Q₁Q₂/(4πε₀r) = 8.99 × 10⁹ × (3.0 × 10⁻⁶)²/0.10 (C1) = 0.81 J (A1)
- (b) 20 m s⁻¹
by symmetry each gets half the energy: 0.405 J (C1); ½ × 0.0020 × v² = 0.405 (C1); v = 20 m s⁻¹ (A1)
A point charge of +4.0 nC is fixed at A and a point charge of −1.0 nC is fixed at B, 0.30 m from A.
- (a) Calculate the distance from A of the point between A and B where the electric potential is zero. [3]
- (b) Explain why the electric field strength at this point is not zero. [2]
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- (a) 0.24 m
4.0/x = 1.0/(0.30 − x) (C1); 1.20 − 4.0x = x (C1); x = 0.24 m (A1)
- (b) E is the (negative) potential gradient, not the potential; between opposite charges the fields of both point in the same direction (from A towards B), so they add.
B1 E = −dV/dx / V = 0 does not mean gradient zero; B1 fields in same direction
An isolated conducting sphere of radius 0.050 m carries a charge of +6.0 nC in a vacuum. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the electric potential at the surface of the sphere. [1]
- (b) A proton is released from rest at the surface. Calculate its speed when it is very far from the sphere. [3]
- (c) Explain why the proton's acceleration decreases as it moves away. [1]
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- (a) 1.1 × 10³ V
V = 8.99 × 10⁹ × 6.0 × 10⁻⁹/0.050 = 1.08 × 10³ V (B1)
- (b) 4.5 × 10⁵ m s⁻¹
loss of PE = eV = 1.60 × 10⁻¹⁹ × 1079 = 1.73 × 10⁻¹⁶ J (C1); ½mₚv² = 1.73 × 10⁻¹⁶ (C1); v = 4.5 × 10⁵ m s⁻¹ (A1)
- (c) The field strength (force) decreases with distance as 1/r².
B1
A hollow metal sphere of radius 0.12 m is charged so that the potential at its surface is 6000 V. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the charge on the sphere. [2]
- (b) Calculate the electric field strength just outside the surface. [1]
- (c) State the electric potential and the field strength at the centre of the sphere. [2]
- (d) Calculate the potential at 0.30 m from the centre. [1]
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- (a) 8.0 × 10⁻⁸ C
Q = 4πε₀rV = 4π × 8.85 × 10⁻¹² × 0.12 × 6000 (C1) = 8.0 × 10⁻⁸ C (A1)
- (b) 5.0 × 10⁴ V m⁻¹
E = V/r = 6000/0.12 (B1)
- (c) 6000 V; zero
B1 V same as surface (6000 V); B1 E = 0 inside a conductor
- (d) 2400 V
V ∝ 1/r: 6000 × 0.12/0.30 (B1)
Two small charged spheres, of charges +3.0 nC and −5.0 nC, are held with their centres 4.0 cm apart in air. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) State Coulomb's law. [2]
- (b) Calculate the magnitude of the force between the charges and state whether it is attractive or repulsive. [2]
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- (a) The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation.
B1 ∝ product of charges; B1 ∝ 1/separation² (point charges)
- (b) 8.4 × 10⁻⁵ N, attractive
F = Q₁Q₂/(4πε₀r²) = (3.0 × 10⁻⁹ × 5.0 × 10⁻⁹)/(4π × 8.85 × 10⁻¹² × 0.040²) (C1) = 8.4 × 10⁻⁵ N, attractive (A1)
An isolated point charge of +12 nC is in a vacuum. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Define electric field strength. [1]
- (b) Calculate the electric field strength at a point 0.25 m from the charge. [2]
- (c) Calculate the magnitude of the force on an electron at this point and state its direction. [2]
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- (a) Force per unit positive charge (acting on a small stationary charge).
B1
- (b) 1.7 × 10³ N C⁻¹
E = Q/(4πε₀r²) = 12 × 10⁻⁹ × 8.99 × 10⁹/0.25² (C1) = 1.73 × 10³ N C⁻¹ (A1)
- (c) 2.8 × 10⁻¹⁶ N, towards the charge
F = eE = 1.60 × 10⁻¹⁹ × 1.73 × 10³ = 2.8 × 10⁻¹⁶ N (B1); towards the positive charge (B1)
- (a) Describe the pattern of electric field lines around an isolated positive point charge. [2]
- (b) State how the field lines show where the field is strongest. [1]
- (c) The field strength at a distance r from the charge is E. Calculate the field strength at distance 3r, in terms of E. [1]
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- (a) Straight radial lines, pointing outwards from the charge, getting further apart with distance.
B1 radial/straight lines outwards; B1 spacing increases with distance
- (b) The field is strongest where the lines are closest together (near the charge).
B1
- (c) E/9
E ∝ 1/r² (B1)
A thundercloud model treats a region of charge as a point charge of −6.0 µC. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the electric field strength 2.0 m from the charge. [2]
- (b) State the direction of the field at this point. [1]
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- (a) 1.35 × 10⁴ N C⁻¹
E = 8.99 × 10⁹ × 6.0 × 10⁻⁶/2.0² (C1) = 1.35 × 10⁴ N C⁻¹ (A1)
- (b) Towards the charge (radially inwards).
B1
The spherical metal dome of a Van de Graaff generator in a school science club has radius 0.15 m. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Explain why the charge on the sphere may be treated as a point charge at its centre for points outside the sphere. [1]
- (b) Air breaks down when the field strength exceeds 3.0 × 10⁶ V m⁻¹. Calculate the maximum charge the dome can hold. [2]
- (c) With this maximum charge, calculate the field strength at a distance of 0.45 m from the centre of the dome. [2]
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- (a) The charge is distributed uniformly over the surface (by symmetry), so outside the sphere the field is the same as for a point charge at the centre.
B1
- (b) 7.5 × 10⁻⁶ C
Q = 4πε₀r²E = 4π × 8.85 × 10⁻¹² × 0.15² × 3.0 × 10⁶ (C1) = 7.5 × 10⁻⁶ C (A1)
- (c) 3.3 × 10⁵ V m⁻¹
r is 3 times larger, E ∝ 1/r² (C1); E = 3.0 × 10⁶/9 = 3.3 × 10⁵ V m⁻¹ (A1)
A point charge of +4.0 µC is fixed at point A and a point charge of +1.0 µC is fixed at point B, 0.30 m from A.
- (a) Explain why there is a point between the charges where the electric field strength is zero. [1]
- (b) Calculate the distance of this point from A. [3]
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- (a) Both charges are positive, so their fields point in opposite directions between them and can cancel.
B1
- (b) 0.20 m
4.0/x² = 1.0/(0.30 − x)² (C1); 2/x = 1/(0.30 − x) (C1); x = 0.20 m (A1)
Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Show that the ratio (electric force)/(gravitational force) between two protons is independent of their separation. [1]
- (b) Calculate this ratio. The gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻². [2]
- (c) Suggest why gravitational forces, not electric forces, dominate the motion of planets. [1]
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- (a) Both forces are proportional to 1/r², so r cancels in the ratio.
B1
- (b) 1.2 × 10³⁶
ratio = e²/(4πε₀Gmₚ²) (C1) = 1.2 × 10³⁶ (A1)
- (c) Planets are electrically neutral overall, so the net electric force is negligible, but mass is always positive so gravity always attracts.
B1
Two identical small conducting spheres, each of mass 0.20 g, hang from the same point on insulating threads. They are given equal charges and come to rest with each thread at 10° to the vertical. The distance between the centres of the spheres is then 5.0 cm. g = 9.81 m s⁻². Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Show that the electric force between the spheres is about 3.5 × 10⁻⁴ N. [2]
- (b) Calculate the charge on each sphere. [2]
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- (a) 3.46 × 10⁻⁴ N
tan 10° = F/mg (C1); F = 0.20 × 10⁻³ × 9.81 × tan 10° = 3.46 × 10⁻⁴ N (A1)
- (b) 9.8 × 10⁻⁹ C
F = q²/(4πε₀r²) (C1); q = 0.050 × √(3.46 × 10⁻⁴ × 4πε₀) = 9.8 × 10⁻⁹ C (A1)
Point charges of +2.0 nC and −2.0 nC are fixed at points A and B respectively, 0.10 m apart in a vacuum. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the electric field strength at the midpoint M of AB and state its direction. [3]
- (b) Point P lies on the perpendicular bisector of AB, 0.12 m from M. Determine the electric field strength at P. [3]
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- (a) 1.44 × 10⁴ N C⁻¹ towards B
each field = 8.99 × 10⁹ × 2.0 × 10⁻⁹/0.050² = 7.19 × 10³ N C⁻¹ (C1); both point from A towards B, so they add (C1); 1.44 × 10⁴ N C⁻¹ towards B (A1)
- (b) 8.2 × 10² N C⁻¹, parallel to AB, towards B
AP = BP = 0.13 m; each field = 8.99 × 10⁹ × 2.0 × 10⁻⁹/0.13² = 1.06 × 10³ N C⁻¹ (C1); perpendicular components cancel, parallel components add: 2 × 1.06 × 10³ × (0.050/0.13) (C1) = 8.2 × 10² N C⁻¹ parallel to AB towards B (A1)
In a simple model of the hydrogen atom, an electron moves in a circular orbit of radius 5.3 × 10⁻¹¹ m around a proton. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the electric force on the electron. [2]
- (b) This force provides the centripetal force. Calculate the speed of the electron. [2]
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- (a) 8.2 × 10⁻⁸ N
F = e²/(4πε₀r²) = 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)²/(5.3 × 10⁻¹¹)² (C1) = 8.2 × 10⁻⁸ N (A1)
- (b) 2.2 × 10⁶ m s⁻¹
mv²/r = F → v = √(Fr/m) = √(8.2 × 10⁻⁸ × 5.3 × 10⁻¹¹/9.11 × 10⁻³¹) (C1) = 2.2 × 10⁶ m s⁻¹ (A1)
A probe measures the electric field strength E at distances r from the centre of a charged metal sphere.r / m: 0.20, 0.30, 0.40E / 10⁴ V m⁻¹: 4.50, 2.00, 1.125Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Show that the data are consistent with the field of a point charge at the centre of the sphere. [2]
- (b) Determine the charge on the sphere. [2]
- (c) Calculate the distance from the centre at which the field strength would be 5.0 × 10³ V m⁻¹. [1]
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- (a) Er² = 1800 V m in each case, so E ∝ 1/r².
M1 calculates Er² (or ratio) for all three; A1 constant 1.80 × 10³ V m so inverse-square
- (b) 2.0 × 10⁻⁷ C
Q = 4πε₀ × Er² (C1) = 4π × 8.85 × 10⁻¹² × 1800 = 2.0 × 10⁻⁷ C (A1)
- (c) 0.60 m
r = √(1800/5000) = 0.60 m (B1)
Two parallel horizontal metal plates are 4.0 cm apart. The potential difference between them is 600 V. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the electric field strength between the plates. [2]
- (b) Calculate the force on an oil drop carrying a charge of +3e between the plates. [2]
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- (a) 1.5 × 10⁴ V m⁻¹
E = V/d = 600/0.040 (C1) = 1.5 × 10⁴ V m⁻¹ (A1)
- (b) 7.2 × 10⁻¹⁵ N
F = qE = 3 × 1.60 × 10⁻¹⁹ × 1.5 × 10⁴ (C1) = 7.2 × 10⁻¹⁵ N (A1)
- (a) Describe the electric field between two parallel charged plates, away from the edges. [2]
- (b) The plates are 2.5 mm apart in an air-filled capacitor with a p.d. of 9.0 V. Calculate the field strength. [1]
- (c) State how the force on a charged particle varies as it moves from one plate to the other. [1]
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- (a) Uniform: field lines parallel, equally spaced, perpendicular to the plates, from positive to negative plate.
B1 parallel/equally spaced (uniform); B1 perpendicular to plates, + to −
- (b) 3.6 × 10³ V m⁻¹
E = 9.0/0.0025 (B1)
- (c) It stays constant (uniform field).
B1
An oil drop of mass 2.4 × 10⁻¹⁵ kg is held stationary between two horizontal plates 6.0 mm apart when the potential difference between them is 294 V. g = 9.81 m s⁻². Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) State the direction of the electric force on the drop and explain your answer. [1]
- (b) Calculate the charge on the drop. [3]
- (c) Deduce the number of excess electrons or missing electrons on the drop, and state which it is if the upper plate is positive. [2]
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- (a) Upwards, to balance its weight (resultant force zero).
B1
- (b) 4.8 × 10⁻¹⁹ C
qV/d = mg (C1); q = mgd/V = 2.4 × 10⁻¹⁵ × 9.81 × 6.0 × 10⁻³/294 (C1) = 4.8 × 10⁻¹⁹ C (A1)
- (c) 3 excess electrons
4.8 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 3 (B1); force upward towards positive plate, so negative: 3 excess electrons (B1)
An electron is accelerated from rest through a potential difference of 250 V in an electron gun. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the kinetic energy gained by the electron. [1]
- (b) Calculate the final speed of the electron. [2]
- (c) State the effect on the final speed of doubling the accelerating potential difference. [1]
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- (a) 4.0 × 10⁻¹⁷ J
eV = 1.60 × 10⁻¹⁹ × 250 (B1)
- (b) 9.4 × 10⁶ m s⁻¹
½mv² = eV (C1); v = √(2 × 4.0 × 10⁻¹⁷/9.11 × 10⁻³¹) = 9.4 × 10⁶ m s⁻¹ (A1)
- (c) Speed increases by a factor of √2.
B1 v ∝ √V
In an electrostatic dust collector in a cement factory chimney, two vertical parallel plates 0.20 m apart have a potential difference of 40 kV between them. A dust particle of mass 1.0 × 10⁻¹² kg carries a charge of 8.0 × 10⁻¹⁷ C.
- (a) Calculate the electric field strength between the plates. [1]
- (b) Calculate the acceleration of the dust particle, ignoring gravity and air resistance. [2]
- (c) The particle starts from rest midway between the plates. Calculate the time it takes to reach a plate. [2]
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- (a) 2.0 × 10⁵ V m⁻¹
40 × 10³/0.20 (B1)
- (b) 16 m s⁻²
F = qE = 8.0 × 10⁻¹⁷ × 2.0 × 10⁵ = 1.6 × 10⁻¹¹ N (C1); a = F/m = 16 m s⁻² (A1)
- (c) 0.11 s
0.10 = ½ × 16 × t² (C1); t = 0.11 s (A1)
An electron moving horizontally at 2.0 × 10⁷ m s⁻¹ enters the uniform field between two horizontal parallel plates midway between them. The plates are 5.0 cm long and 2.0 cm apart, with a potential difference of 150 V between them. Ignore gravity. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg where needed.
- (a) Calculate the acceleration of the electron between the plates. [2]
- (b) Calculate the time the electron spends between the plates. [1]
- (c) Calculate the vertical deflection of the electron as it leaves the plates, and state whether it hits a plate. [2]
- (d) Describe the shape of the path of the electron between the plates. [1]
Show answer
- (a) 1.3 × 10¹⁵ m s⁻²
E = 150/0.020 = 7500 V m⁻¹ (C1); a = eE/m = 1.3 × 10¹⁵ m s⁻² (A1)
- (b) 2.5 × 10⁻⁹ s
t = 0.050/2.0 × 10⁷ (B1)
- (c) 4.1 mm; it does not hit a plate (it is 10 mm from each plate)
y = ½at² = ½ × 1.32 × 10¹⁵ × (2.5 × 10⁻⁹)² = 4.1 × 10⁻³ m (C1); 4.1 mm < 10 mm so it emerges (A1)
- (d) A parabola (constant horizontal velocity, constant vertical acceleration).
B1
A proton and an alpha particle, both initially at rest, are accelerated through the same potential difference V.
- (a) Show that the speed of a charged particle accelerated from rest through a p.d. V is v = √(2qV/m). [1]
- (b) Calculate the ratio (speed of proton)/(speed of alpha particle). Take the alpha particle mass as 4mₚ and charge 2e. [2]
- (c) Calculate the ratio (kinetic energy of proton)/(kinetic energy of alpha particle). [1]
Show answer
- (a) qV = ½mv² → v = √(2qV/m)
B1
- (b) 1.41 (√2)
v ∝ √(q/m) (C1); ratio = √((e/mₚ)/(2e/4mₚ)) = √2 = 1.41 (A1)
- (c) 0.5
KE = qV, so ratio = e/2e = 0.5 (B1)
