Part
1 answer:
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Answer:
a) E = 4.5*10⁴ V/m
b) C= 17.7 nF
c) Q = 159. 3 nC
Explanation:
a)
- By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:
b)
- For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:
From (1), we know that V = E*d, but at the same time, applying Gauss'
Law at a closed surface half within the plate, half outside it , it can be
showed than E= σ/ε₀, so finally we get:
c)
- From (3) we can solve for Q as follows: