Explore a world of knowledge and get your questions answered on IDNLearn.com. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.
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]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.