Answer:
The potential across the plates is 5.25 V
Explanation:
Let us assume that the Electric field is E; Force is F; Unit charge is q, the Electric potential is V and distance is d.
Here E = 150 v/m and d = 3.5 cm = 0.035 m
Now we know that,
Electric Field E [tex]= \frac{Force}{Unit Charge} = \frac{F}{q}[/tex] ----- [1]
[tex]Work W = Force \times distance = (F \times d)[/tex] ----- [2]
[tex]Electric potential V = \frac{work}{unit charge} = \frac{W}{q}[/tex]
So, [tex]V = \frac{W}{q} = {F\times d}{q}[/tex]
[tex]=>\frac{F}{q}\times d = (E \times d) = (150\times0.035) V = 5.25V[/tex]
Thus, The potential across the plates is 5.25 V
