Question

(Last answer was wrong)

What is the electric field strength between two parallel charged plates that are 3.50 cm apart and have a potential difference of 18.0 V?

Answer 1

The electric field strength between two parallel plates is given by:

E = V/d

where E is the electric field strength, V is the potential difference between the plates, and d is the distance between the plates.

Substituting the given values, we have:

E = 18.0 V / 0.035 m

E = 514.3 N/C

Therefore, the electric field strength between the two parallel charged plates is 514.3 N/C.

Answer 2

514 V/m

Step-by-step explanation:

You want the field strength between two plates 3.50 cm apart with a potential difference of 18.0 V.

Field strength

The field strength is measured in volts per meter.

(18.0 V)/(0.035 m) ≈ 514 V/m

The electric field strength is about 514 V/m.

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Additional comment

The given numbers have 3 significant figures, so we have rounded the answer to 3 significant figures.

The units "volts per meter" are equivalent to the units "newtons per coulomb."