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9702 · 18.4

Electric field of a point charge — practice questions

Practice and worked examples for 9702 Electric field of a point charge. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A point charge of +5.0 nC is placed in a vacuum. Calculate the electric field strength and the electric potential at a point 15 cm away from the charge. (Take 14πϵ0=9.0×109 N m2 C−2\frac{1}{4\pi\epsilon_0} = 9.0 \times 10^9 \text{ N m}^2 \text{ C}^{-2})

Show solution outline
  1. Convert units: Q=+5.0 nC=+5.0×10−9 CQ = +5.0 \text{ nC} = +5.0 \times 10^{-9} \text{ C} r=15 cm=0.15 mr = 15 \text{ cm} = 0.15 \text{ m}
  2. Calculate Electric Field Strength (E): E=14πϵ0Qr2E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} E=(9.0×109)5.0×10−9(0.15)2E = (9.0 \times 10^9) \frac{5.0 \times 10^{-9}}{(0.15)^2} E=2000 N C−1E = 2000 \text{ N C}^{-1} (or 2.0 kN C−12.0 \text{ kN C}^{-1}) The direction is radially outwards from the positive charge.
  3. Calculate Electric Potential (V): V=14πϵ0QrV = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} V=(9.0×109)5.0×10−90.15V = (9.0 \times 10^9) \frac{5.0 \times 10^{-9}}{0.15} V=300 VV = 300 \text{ V}

Worked example 2

Two point charges, QA=+4.0μCQ_A = +4.0 \mu C and QB=−2.0μCQ_B = -2.0 \mu C, are placed on a line 10.0 cm apart in a vacuum. QAQ_A is at x=0x=0 and QBQ_B is at x=0.10x=0.10 m. Point P is located on the line between them, at a distance of 6.0 cm from QAQ_A. Calculate: (a) The net electric potential at P. (b) The net electric field strength at P. (Take 14πϵ0=9.0×109 N m2 C−2\frac{1}{4\pi\epsilon_0} = 9.0 \times 10^9 \text{ N m}^2 \text{ C}^{-2})

Show solution outline
  1. Identify charges and distances: QA=+4.0×10−6 CQ_A = +4.0 \times 10^{-6} \text{ C} QB=−2.0×10−6 CQ_B = -2.0 \times 10^{-6} \text{ C} Distance from QAQ_A to P, rA=6.0 cm=0.060 mr_A = 6.0 \text{ cm} = 0.060 \text{ m} Distance from QBQ_B to P, rB=10.0 cm−6.0 cm=4.0 cm=0.040 mr_B = 10.0 \text{ cm} - 6.0 \text{ cm} = 4.0 \text{ cm} = 0.040 \text{ m}
  2. (a) Calculate Net Electric Potential (V): Potential is a scalar, so we sum the potentials from each charge algebraically. VA=14πϵ0QArA=(9.0×109)+4.0×10−60.060=+600,000 VV_A = \frac{1}{4\pi\epsilon_0} \frac{Q_A}{r_A} = (9.0 \times 10^9) \frac{+4.0 \times 10^{-6}}{0.060} = +600,000 \text{ V} VB=14πϵ0QBrB=(9.0×109)−2.0×10−60.040=−450,000 VV_B = \frac{1}{4\pi\epsilon_0} \frac{Q_B}{r_B} = (9.0 \times 10^9) \frac{-2.0 \times 10^{-6}}{0.040} = -450,000 \text{ V} Vnet=VA+VB=600,000+(−450,000)=+150,000 VV_{net} = V_A + V_B = 600,000 + (-450,000) = +150,000 \text{ V} Vnet=+1.5×105 VV_{net} = +1.5 \times 10^5 \text{ V}
  3. (b) Calculate Net Electric Field Strength (E): Field strength is a vector. At point P:
    • The field from QAQ_A (EAE_A) points to the right (repulsive).
    • The field from QBQ_B (EBE_B) also points to the right (attractive). So, the net field is the sum of their magnitudes. | EA=14πϵ0∣QA∣rA2=(9.0×109)4.0×10−6(0.060)2=1.0×107 N C−1E_A = \frac{1}{4\pi\epsilon_0} \frac{ | Q_A | }{r_A^2} = (9.0 \times 10^9) \frac{4.0 \times 10^{-6}}{(0.060)^2} = 1.0 \times 10^7 \text{ N C}^{-1} | | --- | --- | --- | | EB=14πϵ0∣QB∣rB2=(9.0×109)2.0×10−6(0.040)2=1.125×107 N C−1E_B = \frac{1}{4\pi\epsilon_0} \frac{ | Q_B | }{r_B^2} = (9.0 \times 10^9) \frac{2.0 \times 10^{-6}}{(0.040)^2} = 1.125 \times 10^7 \text{ N C}^{-1} | Enet=EA+EB=(1.0×107)+(1.125×107)=2.125×107 N C−1E_{net} = E_A + E_B = (1.0 \times 10^7) + (1.125 \times 10^7) = 2.125 \times 10^7 \text{ N C}^{-1} Enet≈2.1×107 N C−1E_{net} \approx 2.1 \times 10^7 \text{ N C}^{-1} to the right (towards QBQ_B).
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