Physics · OCR H556 · Topic 6.2

Electric fields

Point and spherical charges, Coulomb's law, uniform electric field, electric potential and energy.

  • 4 sub-topics
  • Taught in Year 13
  • 50 practice questions
  • 4 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 6.2.1Point and spherical chargesY13
  2. 6.2.2Coulomb's lawY13
  3. 6.2.3Uniform electric fieldY13
  4. 6.2.4Electric potential and energyY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 50 questions in this topic, free, no account needed.

Electric and magnetic fields

Not needed for Year 12Easy

Electric field strength · Year 13 · 2 marks

A charge of 1.6×10−191.6 \times 10^{-19} C experiences a force of 3.2×10−153.2 \times 10^{-15} N in an electric field.
  1. (a)
    Calculate the electric field strength.
    (1)
  2. (b)
    State the force on a charge twice as large, of opposite sign.
    (1)

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Part (a)1 mark

  1. E=20000 N C−1E = 20000\ \mathrm{N\,C^{-1}}

Part (b)1 mark

  1. 6.40×10−156.40 \times 10^{-15} N, in the opposite direction.

Total2

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Electric and magnetic fields

Not needed for Year 12Medium

Coulomb's law · Year 13 · 4 marks

Two point charges of 2.0×10−62.0 \times 10^{-6} C and 4.0×10−64.0 \times 10^{-6} C are 0.3 m apart.
  1. (a)
    Calculate the force between them, and state whether it is attractive or repulsive.
    (3)
  2. (b)
    State the force if the separation is halved.
    (1)

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Part (a)3 marks

  1. FF=Q1Q24πε0r2{}= \frac{Q_1Q_2}{4\pi\varepsilon_0r^2}
  2. F=0.799 NF = 0.799\ \mathrm{N}
  3. Repulsive.

Part (b)1 mark

  1. F=3.20 NF = 3.20\ \mathrm{N}

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Fields and their consequences

Not needed for Year 12Hard

Uniform electric field · Year 13 · 11 marks

Figure 1 shows two parallel plates a distance d = 20 mm apart with a p.d. of 400 V between them. An electron is released from rest at the negative plate. (e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg)
+400 V0 V20 mm
Figure 1
  1. (a)
    Calculate the electric field strength between the plates.
    (2)
  2. (b)
    Calculate the force on the electron.
    (2)
  3. (c)
    Calculate the acceleration of the electron.
    (2)
  4. (d)
    Calculate the speed of the electron as it reaches the positive plate.
    (3)
  5. (e)
    The plate separation is doubled with the same p.d. State and explain the effect on your answer to the last part.
    (2)

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Part (a)2 marks

  1. The field between parallel plates is uniform. Convert mm to m.EE=Vd{}= \frac{V}{d}=40020×10−3{}= \frac{400}{20 \times 10^{-3}}
  2. E=20000 V m−1E = 20000\ \mathrm{V\,m^{-1}}

Part (b)2 marks

  1. Force on a charge in a field.F=EQF = EQ
  2. FF=3.20×10−15 N{}= 3.20 \times 10^{-15}\ \mathrm{N}

Part (c)2 marks

  1. Newton's second law.a=Fma = \frac{F}{m}
  2. aa=3.51×1015 m s−2{}= 3.51 \times 10^{15}\ \mathrm{m\,s^{-2}}

Part (d)3 marks

  1. The work done by the field becomes kinetic energy.eV=12mv2eV = \tfrac{1}{2}mv^2
  2. Rearrange for v.vv=2×1.60×10−19×4009.11×10−31{}= \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 400}{9.11 \times 10^{-31}}}
  3. vv=1.19×107 m s−1{}= 1.19 \times 10^{7}\ \mathrm{m\,s^{-1}}

Part (e)2 marks

  1. State the effect.The speed is unchanged.
  2. Half the force acts over twice the distance.The energy gained is eV, which depends only on the p.d., not on the separation.

Total11

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