Physics 9702 · AS & A Level · Potential difference and power

Potential difference and power — practice question

A charged oil drop is suspended in a vacuum between two horizontal metal plates. A potential difference of $1340\,\text{V}$ is applied across the plates, producing a uniform electric field, as illustrated in Fig. 2.1. The plate separation is $1.4 \times 10^{-2}\,\text{m}$. The oil drop, with weight $4.6 \times 10^{-14}\,\text{N}$, stays at rest halfway between the plates.
(a(i))[2]

Calculate the field strength magnitude.

(a(ii))[3]

Determine both the magnitude and sign of the charge on the oil drop.

(b(i))[2]

The plate potentials are swapped at once, with the top plate now at $0\,\text{V}$ and the bottom plate at $+1340\,\text{V}$. As a result, the oil drop begins to travel downwards. Compare the new arrangement of electric field lines between the plates with the original one.

(b(ii))[1]

Determine the size of the resultant force acting on the oil drop.

(b(iii))[2]

Show that the oil drop’s acceleration has magnitude $20\,\text{m s}^{-2}$.

(b(iv))[2]

Take the radius of the oil drop as negligible. Use the information in (b)(iii) to calculate how long the oil drop takes to travel to the bottom metal plate from its starting point halfway between the plates.

(c)[2]

The oil drop in (b) starts moving at time $t = 0$. The distance of the oil drop from the bottom plate is $x$. On Fig. 2.2, sketch how distance $x$ varies with time $t$ as the drop moves from its initial position until it reaches the surface of the bottom plate. Numerical values of $t$ are not required.

Worked solution & mark scheme

This 14-mark question has a full step-by-step worked solution and mark scheme. One marking point: $E=(\Delta V)/(\Delta d)$

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