A-level Physics (Advancing Physics)/Electric Field/Worked Solutions

'''1. Two metal plates are connected to a 9V battery with negligible internal resistance. If the plates are 10 cm apart, what is the electric field at either of the plates?'''

$$E_{electric} = \frac{V_{electric}}{d} = \frac{9}{0.1} = 90\mbox{ Vm}^{-1}$$

2. What is the electric field at the midpoint between the plates?

The whole point of a uniform electric field is that the field does not change anywhere between the plates - at the midpoint, as anywhere, it is 90 NC−1.

3. The charge on an electron is -1.6 x 10−19 C. What is the electric field 1&mu;m from a hydrogen nucleus?

$$E_{electric} = \frac{Q}{4\pi\epsilon_0r^2} = \frac{1.6 \times 10^{-19}}{4\pi \times 8.85 \times 10^{-12} \times (10^{-6})^2} = 1440\mbox{ Vm}^{-1}$$

4. What is the direction of this field?

The hydrogen nucleus has a positive charge, so the electric field goes away from the nucleus (by convention).

'''5. A 2C charge is placed 1m from a -1C charge. At what point will the electric field be 0?'''

Define the distance r as shown:



$$\frac{2}{4\pi\epsilon_0(1 + r)^2} = \frac{1}{4\pi\epsilon_0r^2}$$

$$\frac{2}{(1 + r)^2} = \frac{1}{r^2}$$

$$\frac{(1 + r)^2}{2} = r^2$$

$$(1 + r)^2 = 2r^2$$

$$1 + r = r\sqrt{2}$$

$$r(\sqrt{2} - 1) = 1$$

$$r = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \sqrt{2} + 1 \approx 2.41\mbox{ m}$$