asked 151k views
4 votes
A Gaussian surface in the form of a hemisphere of radius R = 3.04 cm lies in a uniform electric field of magnitude E = 1.64 N/C. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through (a) the base and (b) the curved portion of the surface?

asked
User Dennism
by
7.8k points

1 Answer

1 vote

Answer:

Part a)


\phi = -4.76 * 10^(-3) Nm^2/C

Part b)


\phi_(curved) = 4.76 * 10^(-3) Nm^2/C

Step-by-step explanation:

Part a)

Electric flux entering into the base

so it is given as


\phi = E.A


\phi = - EA


\phi = -(1.64)(\pi r^2)


\phi = -(1.64)(\pi* 0.0304^2)


\phi = -4.76 * 10^(-3) Nm^2/C

Part b)

Now since we know that there is no enclosed charge in the hemisphere

so net flux must be zero


\phi_(curved) + \phi_(flat) = 0


\phi_(curved) - 4.76 * 10^(-3) = 0


\phi_(curved) = 4.76 * 10^(-3) Nm^2/C

answered
User Tj Gienger
by
7.7k points
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