18.2. Uniform electric fields
A subsection of Physics, 9702, through 18. Electric fields
Listing 10 of 62 questions
Two parallel vertical metal plates are connected to a power supply, as shown in . 16 mm + – metal plate metal plate The separation of the plates is 16 mm. On , draw at least six field lines to represent the electric field between the plates. An α-particle travels in a vacuum between the two plates. The electric field does work on the α-particle. The gain in kinetic energy of the α-particle is 15 keV. Calculate the electric field strength between the plates. electric field strength = V m–1
9702_s16_qp_22
THEORY
2016
Paper 2, Variant 2
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The diagram shows two parallel horizontal metal plates. The top plate is positively charged and the bottom plate is earthed. + liquid drop plates A small charged liquid drop, midway between the plates, is held in equilibrium by the combination of its weight and the electric force acting on it. The acceleration of free fall is g and the electric field strength is E. What is the polarity of the charge on the drop, and the ratio of charge to mass of the drop?
9702_s17_qp_11
MCQ
2017
Paper 1, Variant 1
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Two vertical metal plates in a vacuum are separated by a distance of 0.12 m. shows a side view of this arrangement. + 900 V 0 V 2.0 m 0.12 m 0.080 m metal plate X Y metal plate sand particle path of particle (not to scale) Each plate has a length of 2.0 m. The potential difference between the plates is 900 V. The electric field between the plates is uniform. A negatively charged sand particle is released from rest at point X, which is a horizontal distance of 0.080 m from the top of the positively charged plate. The particle then travels in a straight line and collides with the positively charged plate at its lowest point Y, as illustrated in . Describe the pattern of the field lines (lines of force) between the plates. State the names of the two forces acting on the particle as it moves from X to Y. By considering the vertical motion of the sand particle, show that the time taken for the particle to move from X to Y is 0.64 s. Calculate the horizontal component of the acceleration of the particle. horizontal component of acceleration = m s−2 Calculate the magnitude of the electric field strength. electric field strength = N C−1 The sand particle has mass m and charge q. Use your answers in and to determine the ratio q m. ratio = C kg−1 Another particle has a smaller magnitude of the ratio q m than the sand particle. This particle is also released from point X. For the movement of this particle, state the effect, if any, of the decreased magnitude of the ratio on: the vertical component of the acceleration the horizontal component of the acceleration.
9702_s19_qp_23
THEORY
2019
Paper 2, Variant 3
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A sphere of mass 1.6 × 10–10 kg has a charge of +0.27 nC. The sphere is in a uniform electric field that acts vertically upwards, as shown in the side view in . SIDE VIEW electric field lines sphere plane in which sphere moves The force exerted on the sphere by the electric field causes the sphere to remain at a fixed vertical height in a horizontal plane. There is a uniform magnetic field in the region of the electric field. The sphere moves at a speed of 0.78 m s–1 in the horizontal plane. The magnetic field causes the sphere to move in a circular path of radius 3.4 m, as shown in the view from above in . VIEW FROM ABOVE sphere 3.4 m electric field lines out of the page path of sphere Determine the direction of the uniform magnetic field. Explain why the motion of the sphere in the horizontal plane is circular. Calculate the strength of the uniform electric field. electric field strength = N C–1 By considering the magnetic force on the sphere, show that the flux density of the uniform magnetic field is 0.14 T.
9702_s22_qp_41
THEORY
2022
Paper 4, Variant 1
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A sphere of mass 1.6 × 10–10 kg has a charge of +0.27 nC. The sphere is in a uniform electric field that acts vertically upwards, as shown in the side view in . SIDE VIEW electric field lines sphere plane in which sphere moves The force exerted on the sphere by the electric field causes the sphere to remain at a fixed vertical height in a horizontal plane. There is a uniform magnetic field in the region of the electric field. The sphere moves at a speed of 0.78 m s–1 in the horizontal plane. The magnetic field causes the sphere to move in a circular path of radius 3.4 m, as shown in the view from above in . VIEW FROM ABOVE sphere 3.4 m electric field lines out of the page path of sphere Determine the direction of the uniform magnetic field. Explain why the motion of the sphere in the horizontal plane is circular. Calculate the strength of the uniform electric field. electric field strength = N C–1 By considering the magnetic force on the sphere, show that the flux density of the uniform magnetic field is 0.14 T.
9702_s22_qp_43
THEORY
2022
Paper 4, Variant 3
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Define electric field. shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of 6.7 cm and have a potential difference (p.d.) of 430 V between them. 6.7 cm +430 V 0 V electron, speed 2.6 × 107 m s–1 conducting plate conducting plate On , draw four field lines to represent the electric field between the plates. Determine the strength E of the electric field between the plates. E = N C–1 An electron travels at a speed of 2.6 × 107 m s–1 towards the region between the plates, as shown in . On , draw the path of the electron as it moves between and beyond the plates. A uniform magnetic field is now applied in the region of the electric field in , so that the electron in travels undeviated through the region. Determine the direction of the uniform magnetic field. Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated. Determine the flux density B of the uniform magnetic field. Give a unit with your answer. B = unit
9702_s24_qp_41
THEORY
2024
Paper 4, Variant 1
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Define electric field. shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of 6.7 cm and have a potential difference (p.d.) of 430 V between them. 6.7 cm +430 V 0 V electron, speed 2.6 × 107 m s–1 conducting plate conducting plate On , draw four field lines to represent the electric field between the plates. Determine the strength E of the electric field between the plates. E = N C–1 An electron travels at a speed of 2.6 × 107 m s–1 towards the region between the plates, as shown in . On , draw the path of the electron as it moves between and beyond the plates. A uniform magnetic field is now applied in the region of the electric field in , so that the electron in travels undeviated through the region. Determine the direction of the uniform magnetic field. Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated. Determine the flux density B of the uniform magnetic field. Give a unit with your answer. B = unit
9702_s24_qp_43
THEORY
2024
Paper 4, Variant 3
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Questions Discovered
62