Lesson 7.2.1
7.2.1 Electric potential, field lines and equipotentials Quiz: Pearson Edexcel Physics, Unit 7
20 questions
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Lesson 7.2.1, Electric potential, field lines and equipotentials: 20 multiple choice questions for the Pearson Edexcel Physics (9PH0), Unit 7: Electric and Magnetic Fields, written with Revision Ninja.
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The 20 questions
-
How is the electric potential at a point defined?
- The work done per unit positive charge in bringing a small positive charge from infinity to that point
- The energy stored per unit volume of the electric field at that point
- The force per unit positive charge acting at that point in the field
- The work done per unit time in moving a charge through a conductor
-
What is the SI unit of electric potential?
- The farad, equivalent to C V^-1
- The newton per coulomb, equivalent to N C
- The coulomb per second, equivalent to A
- The volt, equivalent to J C^-1
-
What is the electric potential at a distance r from a point charge Q in a vacuum?
- V = 4 pi epsilon0 Q / r
- V = Q / (4 pi epsilon0 r)
- V = Q / (4 pi epsilon0 r^2)
- V = Q r / (4 pi epsilon0)
-
What is the relationship between the direction of electric field lines and equipotential lines or surfaces?
- Field lines are perpendicular to equipotential surfaces at every point
- Field lines are at 45 degrees to equipotential surfaces at every point
- Field lines are parallel to equipotential surfaces at every point
- Field lines are unrelated to the shape of equipotential surfaces
-
The electric field is related to the potential gradient. Which equation expresses this relationship for a uniform field between parallel plates?
- E = V / d
- E = V^2 / d
- E = V d
- E = d / V
-
A point charge is surrounded by equipotential surfaces. What shape are these surfaces for an isolated point charge?
- Concentric cylinders centred on the charge
- Straight lines radiating outwards from the charge
- Parallel flat planes perpendicular to the charge
- Concentric spheres centred on the charge
-
What is the electric potential at a point 0.20 m from a point charge of +4.0 nanocoulomb?
- 720 V
- -180 V
- 45 V
- 180 V
-
A charge of 2.0 microcoulomb is moved through a potential difference of 50 V. What work is done?
- 4.0 x 10^-5 J
- 2.5 x 10^-8 J
- 1.0 x 10^-4 J
- 1.0 x 10^-2 J
-
The potential difference between two parallel plates is 40 V and their separation is 0.050 m. What is the electric field strength between them?
- 2000 V m^-1
- 800 V m^-1
- 0.00125 V m^-1
- 8.0 V m^-1
-
An electron is accelerated from rest through a potential difference of 100 V. What kinetic energy does it gain?
- 1.6 x 10^-19 J
- 1.6 x 10^-17 J
- 1.0 x 10^2 J
- 6.3 x 10^-19 J
-
A proton is accelerated from rest through a potential difference of 2.0 kV. What is its kinetic energy?
- 3.2 x 10^-16 J
- 2.0 x 10^3 J
- 1.6 x 10^-19 J
- 1.3 x 10^16 J
-
A potential falls by 6.0 V over a distance of 0.030 m in a uniform field. What is the field strength?
- 600 V m^-1
- 0.0050 V m^-1
- 200 V m^-1
- 18 V m^-1
-
A charge is moved along an equipotential surface. How much work is done on the charge?
- Zero, because there is no potential difference along the surface
- Equal to the charge multiplied by the sum of all potentials on the surface
- Equal to the charge multiplied by the potential at the starting point
- Equal to the charge multiplied by the field strength and the distance moved
-
The potential due to a point charge is measured at distance r and then at distance 2r. How does the second potential compare with the first?
- It is unchanged
- It is half as large
- It is twice as large
- It is a quarter as large
-
A +3.0 nC charge and a -3.0 nC charge are 0.60 m apart. What is the potential at the midpoint between them?
- 180 V
- -90 V
- 90 V
- Zero
-
A point charge of +2.0 microcoulomb is fixed in place. What is the work done by an external agent in moving a charge of +1.0 microcoulomb from 0.40 m to 0.10 m from it?
- 0.0045 J
- 0.27 J
- 1.35 J
- 0.135 J
-
A charge of +2.0 C moves from a point at 10 V to a point at 30 V. What is the work done by the electric field on the charge?
- +40 J
- +20 J
- -20 J
- -40 J
-
Why are equipotential surfaces always perpendicular to electric field lines?
- The field is zero along equipotentials, so the lines must meet them at a right angle
- Equipotential surfaces carry no charge and therefore must lie at right angles to the field
- If they were not, the field would have a component along the surface, so moving a charge along it would do work and change its potential
- Electric field lines are always curved and so must cross equipotentials at right angles by definition
-
A student says that wherever the electric field is zero, the electric potential must also be zero. Which response is correct?
- The student is wrong because potential is a vector quantity that cannot be zero anywhere
- The student is wrong: at the midpoint between two equal like charges the field is zero but the potential is not zero
- The student is right: zero field always means zero potential everywhere in space
- The student is right, but only when the charges are equal and opposite
-
What is the electric potential at a point 0.50 m from a point charge of +4.0 nanocoulomb?
- -72 V
- 72 V
- 144 V
- 18 V
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