Question

Potential difference at the ends of a conductor of length \(l\) is \(V\). Diameter of the conductor is \(d\). What will be the effect on the electric field inside the conductor and its resistance, if (i) \(V\) is halved? (ii) \(l\) is doubled? (iii) \(d\) is doubled?

Hint:

Use E = V/l (independent of diameter) and R = 4 rho l/(pi d^2) (independent of V). Then scale each quantity for the change asked.

Answer 1

Step 1: Write the two governing formulae.
The electric field inside a uniform conductor carrying a potential difference \(V\) over length \(l\) is
\[ E = \frac{V}{l} \]
The resistance of the conductor of length \(l\), cross-sectional area \(A\) and resistivity \(\rho\) is
\[ R = \frac{\rho\, l}{A} \]
For a circular cross-section of diameter \(d\), \(A = \dfrac{\pi d^2}{4}\), so
\[ R = \frac{\rho\, l}{\pi d^2/4} = \frac{4\rho\, l}{\pi d^2} \]
Note that \(R\) depends only on the material (\(\rho\)) and geometry (\(l, d\)); it does not depend on \(V\). The field \(E\) depends on \(V\) and \(l\) but not on \(d\).

Step 2: Case (i) V is halved (l and d unchanged).
\[ E' = \frac{V/2}{l} = \frac{1}{2}\cdot\frac{V}{l} = \frac{E}{2} \]
Resistance formula has no \(V\), so \(R\) is unchanged.
Result: electric field becomes half; resistance unchanged.

Step 3: Case (ii) l is doubled (V and d unchanged).
\[ E' = \frac{V}{2l} = \frac{1}{2}\cdot\frac{V}{l} = \frac{E}{2} \]
\[ R' = \frac{4\rho\,(2l)}{\pi d^2} = 2\cdot\frac{4\rho\, l}{\pi d^2} = 2R \]
Result: electric field becomes half; resistance becomes double.

Step 4: Case (iii) d is doubled (V and l unchanged).
\(E = V/l\) contains no \(d\), so the field is unchanged.
\[ R' = \frac{4\rho\, l}{\pi (2d)^2} = \frac{4\rho\, l}{\pi\cdot 4 d^2} = \frac{1}{4}\cdot\frac{4\rho\, l}{\pi d^2} = \frac{R}{4} \]
Result: electric field unchanged; resistance becomes one-fourth.

Step 5: Summary.
\[\boxed{\begin{array}{l}(i)\ E\to E/2,\ R\ \text{unchanged}\\ (ii)\ E\to E/2,\ R\to 2R\\ (iii)\ E\ \text{unchanged},\ R\to R/4\end{array}}\]