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9702 · 18.5

Electric potential — practice questions

Practice and worked examples for 9702 Electric potential. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A point charge of +5.0 nC is located at the origin. Calculate the absolute electric potential at a point 0.20 m away from the charge. (Constant k=14πϵ0=8.99×109 N m2 C−2k = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2})

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  1. Identify knowns: Q=+5.0×10−9 CQ = +5.0 \times 10^{-9} \text{ C} (remember nano = 10−910^{-9}), r=0.20 mr = 0.20 \text{ m}, k=8.99×109 N m2 C−2k = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}.
  2. Recall formula: The formula for electric potential due to a point charge is V=kQrV = k \frac{Q}{r}.
  3. Substitute values: V=(8.99×109)×+5.0×10−90.20V = (8.99 \times 10^9) \times \frac{+5.0 \times 10^{-9}}{0.20}.
  4. Calculate: V=8.99×109×2.5×10−8=+224.75 VV = 8.99 \times 10^9 \times 2.5 \times 10^{-8} = +224.75 \text{ V}.
  5. Final Answer: The absolute electric potential at 0.20 m is +225 V (to 3 significant figures).

Worked example 2

A fixed point charge Q=+2.0μCQ = +2.0 \mu C creates an electric field. Calculate the work done by the electric field when a proton (charge q=+1.60×10−19Cq = +1.60 \times 10^{-19} C) moves from a point A, 0.10 m from QQ, to a point B, 0.50 m from QQ. (Use k=8.99×109 N m2 C−2k = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2})

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  1. Strategy: First, calculate the electric potential at points A and B due to the source charge QQ. Then, find the potential difference ΔV\Delta V. Finally, use the relationship between work and potential energy to find the work done by the field.
  2. Calculate Potential at A (VAV_A): VA=kQrA=(8.99×109)×+2.0×10−60.10V_A = k \frac{Q}{r_A} = (8.99 \times 10^9) \times \frac{+2.0 \times 10^{-6}}{0.10} VA=+179,800 VV_A = +179,800 \text{ V}
  3. Calculate Potential at B (VBV_B): VB=kQrB=(8.99×109)×+2.0×10−60.50V_B = k \frac{Q}{r_B} = (8.99 \times 10^9) \times \frac{+2.0 \times 10^{-6}}{0.50} VB=+35,960 VV_B = +35,960 \text{ V}
  4. Calculate Potential Difference (ΔV\Delta V): The change in potential moving from A to B is: ΔV=VB−VA=35,960−179,800=−143,840 V\Delta V = V_B - V_A = 35,960 - 179,800 = -143,840 \text{ V}
  5. Calculate Change in Potential Energy (ΔU\Delta U): The change in electric potential energy is ΔU=qΔV\Delta U = q \Delta V. ΔU=(+1.60×10−19)×(−143,840)\Delta U = (+1.60 \times 10^{-19}) \times (-143,840) ΔU=−2.30144×10−14 J\Delta U = -2.30144 \times 10^{-14} \text{ J}
  6. Find Work Done by the Field (WfieldW_{field}): The work done by the electric field is the negative of the change in potential energy: Wfield=−ΔUW_{field} = -\Delta U. Wfield=−(−2.30144×10−14 J)=+2.30144×10−14 JW_{field} = -(-2.30144 \times 10^{-14} \text{ J}) = +2.30144 \times 10^{-14} \text{ J}
  7. Final Answer: The work done by the electric field is +2.3×10−14+2.3 \times 10^{-14} J (to 2 significant figures). The positive sign indicates the field did positive work, which is expected as the positive proton is repelled by the positive source charge and moves further away.
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