Let's solve each part **step by step**: --- ## (a) At which of the labeled points is the electric potential lowest? - The **electric potential** is marked on the equipotential lines. The outermost line (next to the grounded plate) is ** V**. - **Point D** is located on the V line. **Final Answer (a):** > The electric potential is lowest at **D**. --- ## (b) At which of the labeled points is the magnitude of the electric field the least? - The **electric field** is strongest where equipotential lines are **closest together** (largest change in potential per distance), and weakest where they are **furthest apart**. - **Point A** is where the lines are farthest apart (around the 60 V region). **Final Answer (b):** > The magnitude of the electric field is least at **A**. **Explanation:** The electric field is proportional to the rate of change of potential (E = -dV/dr). Where the equipotential lines are spaced far apart (as at A), the potential changes slowly with distance, so the electric field is small. --- ## (c) Small object with charge \(+7\) nC at point B ### (i) If moved from B to E, the electrical potential energy would... - Potential at B = 40 V, E = 60 V. - \( U = qV \) - Moving from lower to higher potential with a positive charge **increases** potential energy. **Answer:** > **increase** --- ### (ii) If moved from B to the grounded plate ( V), the electrical potential energy would... - Potential at B = 40 V, grounded plate = V. - \( U = qV \) - Moving from higher to lower potential with a positive charge **decreases** potential energy. **Answer:** > **decrease** --- ### (iii) Quantitative calculation of the change in electrical potential energy - \( q = +7\,\text{nC} = 7 \times 10^{-9}\,\text{C} \) - \( V_B = 40\,\text{V} \) - \( V_{\text{plate}} = \,\text{V} \) - Change in potential energy: \( \Delta U = q (V_{\text{final}} - V_{\text{initial}}) \) - Here, final = V, initial = 40 V \[ \Delta U = 7 \times 10^{-9} \times ( - 40) = -2.8 \times 10^{-7}\,\text{J} \] **Answer:** > The electrical potential energy decreases by \(2.8 \times 10^{-7}\) J. --- ## **Summary Table** | Part | Answer | Explanation | |------|--------|-------------| | (a) | D | D is on the V equipotential, the lowest. | | (b) | A | Equipotentials are farthest apart at A; weakest E field. | | (c-i) | Increase | Moving to higher V with positive charge increases U. | | (c-ii) | Decrease | Moving to lower V with positive charge decreases U. | | (c-iii)| \(-2.8 \times 10^{-7}\) J | \( \Delta U = 7 \times 10^{-9} \times ( - 40) \) | --- Let me know if you need a diagram or further explanation!