Answer:
The magnitude of minimum potential difference is 1800 V
Explanation:
Given:
Electric field [tex]E = 3 \times 10^{6} \frac{V}{m}[/tex]
Gap between electrodes [tex]d = 0.060 \times 10^{-2}[/tex] m
For finding the minimum potential difference,
[tex]\Delta V = E \times d[/tex]
[tex]\Delta V = 3 \times 10^{6} \times 0.060 \times 10^{-2}[/tex]
[tex]\Delta V = 1800[/tex] V
Therefore, the magnitude of minimum potential difference is 1800 V
