Question

A potential difference V is applied across a conductor of length L and diameter D. How are the electric field E and the resistance R of conductor affected, when
(a) V is halved

(b) L is halved

(c) D is doubled
Justify your answer in each case.

Answer 1

Electric Field, $E=\frac{\text{potential difference, V}}{\text{length of the conductor, L}}$.

Hence, the electric field is directly proportional to potential difference and inversely proportional to the length of the conductor.

And, Resistance, $R=\frac{\text{Resistivity}\times \text{length of the conductor, L}}{\text{area of the cross section, A}}$

Hence, the resistance is directly proportional to the length of the conductor and inversely proportional to the area f cross section.

And, area of cross section can be written as $A=\frac{\pi D^2}{4}$

Hence area of cross section is directly proportional to the square of the diameter of the conductor.

Case 1: when $V$ is halved

As $V$ and $E$ are directly proportional to each other, $E$ is halved.

On halving the voltage, the current will also get halved, and by Ohm's law $V=IR$, such that $R$ will remain unchanged.

Case 2: $L$ is halved

As $L$ and $E$ are inversely proportional to each other, $E$ will get doubled.

And as $L$ and $R$ are directly proportional to each other, $R$ will get halved.

Case 3: $D$ is doubled

As $E$ is independent of the area of cross-section, so $E$ will remain unchanged.

Resistance is inversely proportional to area of cross-section and area of cross-section is directly proportional to the square of the diameter of the conductor, hence resistance is inversely proportional to square of the diameter of the conductor.

hence when $D$ is doubled, area of cross section will become 4 times more, and resistance will become ${\left(\frac{1}{4}\right)}^{th}$ times the original.