Q.To increase the sensitivity of a potentiometer (A) the cross-section area of its wire will have to be increased
(B) current in its wire will have to be decreased
(C) current in its wire will have to be increased
(D) length of its wire will have to be increased

Potential Gradient: The "Slope" of Voltage Along a Wire

Imagine you're walking down a hill. The steeper the hill, the faster your potential energy drops with every step you take. Now replace "height" with "electric potential" and "walking along the hill" with "moving along a wire." That's the core idea of a potential gradient.

The Intuition

In a potentiometer, we have a long uniform wire with a battery connected across its ends. The battery creates a potential difference $V$ between the two ends of the wire. As you move from one end to the other, the potential doesn't jump — it falls smoothly and steadily along the length.

If the wire is $L$ metres long and the total potential drop is $V$ volts, then every metre of wire you travel, the potential drops by a fixed amount. That fixed drop per metre is the potential gradient.

The Precise Definition

The potential gradient $k$ is defined as the fall of potential per unit length of the potentiometer wire.

where:

• $V$ is the total potential difference across the ends of the wire (in volts)
• $L$ is the total length of the wire (in metres)
• $k$ is the potential gradient (in volts per metre, V/m)

Why It Matters: Sensitivity

The potential gradient directly determines how sensitive your potentiometer is. A smaller $k$ means the potential changes very slowly along the wire — so a small unknown voltage will produce a measurable balancing length. A larger $k$ means the potential drops quickly, and you might not be able to measure small voltages accurately.

You can reduce $k$ by:

• Decreasing the voltage $V$ across the wire (using a smaller driving cell or a rheostat)
• Increasing the length $L$ of the wire

The Formula in Action

Suppose a 10 m long potentiometer wire has a 2 V battery across it. Then:

$k = \frac{2 \text{ V}}{10 \text{ m}} = 0.2 \text{ V/m}$ …