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son4ous [18]
3 years ago
9

A 10.00-V battery is connected between two parallel metal plates 4.00 mm apart. What is the magnitude of the electric field betw

een the plates?
Physics
2 answers:
olya-2409 [2.1K]3 years ago
6 0

Answer:

2500 N/C

Explanation:

Electric Field: This can be defined as the the force per unit charge in an electric field. The S.I unit is N/C

V = E.r ....................... Equation 1

Where V = Electric potential, E = Electric Field, r = Distance between the two parallel plate,

Make E the subject of the equation,

E = V/r .................... Equation 2

Given: V = 10 V, r = 4.00 mm = (4/1000) m = 0.004 m.

Substitute into equation 2

E = 10/0.004

E = 2500 N/C

Hence the electric field = 2500 N/C

Fittoniya [83]3 years ago
4 0
<h2>Answer:</h2>

2500V/m

<h2>Explanation:</h2>

The voltage (V) across the parallel plates is the product of the electric field strength (E) and the distance between them. i.e

V = E x d            ------------------------(i)

<em><u>From the question;</u></em>

V = 10.00V

d = 4.00mm = 0.004m

<em><u>Substitute the values of V and d into equation (i) as follows;</u></em>

10.00 = E x 0.004

<em><u>Solve for E;</u></em>

E = 10.00 / 0.004

E = 2500V/m

Therefore, the magnitude of the electric field between the plates is 2500V/m.

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