Question

How strong is the electric field between the plates of a 0.70-μFμF air-gap capacitor if they are 2.0 mmmm apart and each has a charge of 77 μCμC ?

Answer 1

The value of the electric charge in terms of capacitance and the potential difference is:

Q=CV

V=Q/C

The formula for the electric field from the potential difference for this particular problem is:

V=Ed

here

C=0.70 μF=0.70×10⁻⁶ F

Q= C=77 μC=77×10⁻⁶ C

d=2mm = 0.002 m

Therefore potential V = 77×10⁻⁶/ 0.70×10⁻⁶ = 110 V

Therefore,

E= V/d =110/.002 = 5.5×10⁴V/m

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