i.
Φ=∫EdA=qϵ0=4πkq=4⋅227⋅8.99⋅109⋅1⋅10−6=113017(V⋅m)\Phi=\int EdA=\frac{q}{\epsilon_0}=4\pi kq=4\cdot\frac{22}{7}\cdot8.99\cdot10^9\cdot1\cdot10^{-6}=113017(V\cdot m)Φ=∫EdA=ϵ0q=4πkq=4⋅722⋅8.99⋅109⋅1⋅10−6=113017(V⋅m)
ii.
Φ=qϵ0→ϵ0=qΦ=1⋅10−6113017=8.85⋅10−12(F/m)\Phi=\frac{q}{\epsilon_0}\to \epsilon_0=\frac{q}{\Phi}=\frac{1\cdot10^{-6}}{113017}=8.85\cdot10^{-12}(F/m)Φ=ϵ0q→ϵ0=Φq=1130171⋅10−6=8.85⋅10−12(F/m)
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments