A small charged glass bead of weight 5.4 × 10⁻⁵N is initially at rest at point A in a vacuum. The bead then falls through a uniform horizontal electric field as it moves in a straight line to point B, as illustrated in Fig. 2.1. [Figure 2.1] The electric field strength is 1.3 × 10⁴ Vm⁻¹. The charge on the bead is –3.7 × 10⁻⁹C.

1. (a)Describe how two metal plates could be used to produce the electric field. Numerical values are not required. [2][2]
2. (b)Determine the magnitude of the electric force acting on the bead. electric force = .................................................... N [2][2]
3. (c)Use your answer in (b) and the weight of the bead to show that the resultant force acting on it is 7.2 × 10⁻⁵ N. [1][1]
4. (d)Explain why the resultant force on the bead of 7.2 × 10⁻⁵N is constant as the bead moves along path AB. [2][2]
5. (e)(i)Calculate the magnitude of the acceleration of the bead along the path AB. acceleration = ............................................... ms⁻² [2][2]
6. (e)(ii)The path AB has length 0.58 m. Use your answer in (i) to determine the speed of the bead at point B. speed = .................................................... ms⁻¹ [2][2]

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