Answer:
The electric field value is 240 N/C
Explanation:
Given that,
Distance = 5.0 mm
Potential difference = 1.2 V
We need to calculate the electric field value
Using formula of potential difference
[tex]\Delta V=Ed[/tex]
[tex]E=\dfrac{\Delta V}{d}[/tex]
Where, E = electric field
V = potential difference
d = distance
Put the value into the formula
[tex]E=\dfrac{1.2}{5.0\times10^{-3}}[/tex]
[tex]E= 240\ N/C[/tex]
Hence, The electric field value is 240 N/C
