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Sagot :
Answer with Explanation:
a. Option d is true.
a negatively charged plane parallel to the end faces of the cylinder
b. Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,[tex]\phi=-2 EA=-2E(\pi r^2)[/tex]
[tex]\phi=-2\times 300\times (3.14\times (0.66)^2)[/tex]
[tex]\phi=-820.67 Nm^2/C[/tex]
c.
Net charge,[tex]Q=\epsilon_0\times \phi[/tex]
Where
[tex]\epsilon_0=8.85\times 10^{-12}[/tex]
[tex]Q=-820.67\times 8.85\times 10^{-12}[/tex]
[tex]Q=-7.26\times 10^{-9} C[/tex]
[tex]Q=-7.26nC[/tex]
Where [tex]1nC=10^{-9}C[/tex]
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