A $10$-m-long potentiometer wire of uniform resistance is connected to a $6$-V battery (and no other resistance in the driver loop). What is the **potential gradient** along the wire?

**Definition.** Potential gradient $k = \dfrac{V_{\text{wire}}}{L}$, the volts dropped per unit length. **Here.** The entire $6$ V drops across the $10$ m wire (no external series resistance), so $k = \dfrac{6}{10} = 0.6$ V/m. **General formula** (with a series resistance $R_s$ in the driver loop): $k = \dfrac{V}{(R_w + R_s)} \cdot \dfrac{R_w}{L}$ — when $R_s = 0$, this collapses to $V/L$. Option D ($6$ V/m) confuses *total voltage* with *gradient*.