# 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?

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

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 1μm from a hydrogen nucleus?

${\displaystyle E_{electric}={\frac {Q}{4\pi \epsilon _{0}r^{2}}}={\frac {1.6\times 10^{-19}}{4\pi \times 8.85\times 10^{-12}\times (10^{-6})^{2}}}=1440{\mbox{ Vm}}^{-1}}$

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:

${\displaystyle {\frac {2}{4\pi \epsilon _{0}(1+r)^{2}}}={\frac {1}{4\pi \epsilon _{0}r^{2}}}}$

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

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

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

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

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

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