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The electric field between two parallel plates has a magnitude of 1250 N/C. The positive plate is 0.05 m away from the negative plate. The electric potential difference between the plates, rounded to the tenths place, is V.

Respuesta :

meerkat18
In electronics, electric field is simply defined as the electric force per unit charge, or Newton/Coulomb. When you multiply the electric field with the distance between the charged parallel plates, it will yield the voltage. Thus,

E = V*d
1250 N/C = V*0.05 m
V = 25,000 volts

Answer : The electric potential difference between the plates is 63 volts.

Explanation :

Given that,

Electric field between two parallel plates, E = 1250 N/C

Separation between plates,  d = 0.05 m

The relation between the electric field and electric potential is given by :

[tex]E=\dfrac{V}{d}[/tex]

So, [tex]V=E\times d[/tex]

[tex]V=1250\ N/C\times 0.05\ m[/tex]

[tex]V=62.5\ V[/tex]

or

V = 63 Volts

Hence, this is the required solution.

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