Answer:
The potential difference between the plates is [tex]8.4\times10^{3}\ V[/tex]
Explanation:
Given that,
Distance = 1.4 mm
Electric field strength [tex]E= 6.0\times10^{6}\ N/C[/tex]
Let the potential difference is V.
We need to calculate the potential difference between the plates
Using formula of electric field
[tex]E=\dfrac{V}{d}[/tex]
[tex]V=Ed[/tex]
Where, V = potential
d = distance
Put the value into the formula
[tex]V=6.0\times10^{6}\times1.4\times10^{-3}[/tex]
[tex]V=8.4\times10^{3}\ V[/tex]
Hence, The potential difference between the plates is [tex]8.4\times10^{3}\ V[/tex]
