Question

Can someone please show me how to do this question, I will attach the file

Answer 1

attached

Answer 2

For part a), assume you can imagine the electrode as a wire, use the equation for resistivity:
\[R=\rho L /A\]

All I can think is,
The length of this hemispheric 'wire' would be the radius, a.
The cross-sectional area would be the average cross-sectional area of the hemisphere (it would be \[\pi a^2\] at the base, 0 at the top) so I assume the average area would be \[\pi a^2 /2\].
From these you can answer part a).

(However, I am not happy with this working, unless I know the level of this question, e.g. a-level or uni)

Answer 3

ok, so \[r=\int\limits_{0}^{a} ( \rho/ A ) dr\]
We know the equation of the cross-sectional area of a hemisphere : \[\pi r^2\]
Therefore, integrate:
\[R=(\rho / 2\pi ) \int\limits_{0}^{a}(1/ r^2 ) dr\]

So you should get
\[R=\rho / (2 \pi a)\]