Hello there, Goodmorning and God Bless...Here is your answer to your following question:
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Answer:
Explanation:
The capacitance of the parallel-plate capacitor is given by:
where
is the vacuum permittivity
is the area of the plates
is the separation between the plates
Substituting,
The energy stored in the capacitor is given by
Since we know the energy
we can re-arrange the formula to find the charge, Q:
Answer: 1.55 x 10⁴ Nm²c^-1
Explanation: The electric flux, electric field intensity and area are related by the formulae below.
Φ= EAcosθ,
Where Φ= electric flux (Nm²c^-1)
E =electric field intensity (N/m²)
A = Area (m²)
θ= this is angle between the planar area and the magnetic flux
For our question E=3.80KN/c= 3800 N/c
A= 0.700 x 0.350= 0.245m²
θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).
Hence Φ= 3800 x 0.245 x cos(0)
= 3800 x 0.245 x 1 (value of cos 0° =1)
= 1.55 x 10⁴ Nm²c^-1
Thus the electric field is 1.55 x 10⁴ Nm²c^-1
