Question #127428

Acapacitor is to be constructed to have a capacitance of 100µF. The area\ of the plates is 6.0mx0.030m and the relative permittivity of thedielectric is7.0.Find the necessary separation of the plates and the electric fieldbstrength if a potential difference of 12V is applied across the capacitor.


Expert's answer

C=ϵϵ0Sd→d=ϵϵ0SC=7⋅8.85⋅10−12⋅(6⋅0.03)100⋅10−6=0.1⋅10−6(m)C=\frac{\epsilon \epsilon_0S}{d}\to d=\frac{\epsilon \epsilon_0S}{C}=\frac{7\cdot8.85\cdot10^{-12}\cdot(6\cdot0.03)}{100\cdot10^{-6}}=0.1\cdot10^{-6}(m)


E=U/d=12/(0.1⋅10−6=120⋅106(V/m)E=U/d=12/(0.1\cdot10^{-6}=120\cdot10^6(V/m)


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