A hemisphere of radius R is placed in a charge-free region of space where a uniform electric field exists of magnitude E directed perpendicular to the hemisphere’s circular base (Fig. 22–50). ( a ) Using the definition of Φ E through an “open” surface, calculate (via explicit integration) the electric flux through the hemisphere. [ Hint : In Fig. 22–50 you can see that, on the surface of a sphere, the infinitesimal area located between the angles θ and θ + dθ is dA = (2 πR sin θ )( R dθ ) = 2 πR 2 sin θ dθ. ] ( b ) Choose an appropriate gaussian surface and use Gauss’s law to much more easily obtain the same result for the electric flux through the hemisphere. FIGURE 22–50 Problem 66.
A hemisphere of radius R is placed in a charge-free region of space where a uniform electric field exists of magnitude E directed perpendicular to the hemisphere’s circular base (Fig. 22–50). ( a ) Using the definition of Φ E through an “open” surface, calculate (via explicit integration) the electric flux through the hemisphere. [ Hint : In Fig. 22–50 you can see that, on the surface of a sphere, the infinitesimal area located between the angles θ and θ + dθ is dA = (2 πR sin θ )( R dθ ) = 2 πR 2 sin θ dθ. ] ( b ) Choose an appropriate gaussian surface and use Gauss’s law to much more easily obtain the same result for the electric flux through the hemisphere. FIGURE 22–50 Problem 66.
A hemisphere of radius R is placed in a charge-free region of space where a uniform electric field exists of magnitude E directed perpendicular to the hemisphere’s circular base (Fig. 22–50). (a) Using the definition of ΦE through an “open” surface, calculate (via explicit integration) the electric flux through the hemisphere. [Hint: In Fig. 22–50 you can see that, on the surface of a sphere, the infinitesimal area located between the angles θ and θ + dθ is dA = (2πR sin θ)(R dθ) = 2πR2sin θ dθ.] (b) Choose an appropriate gaussian surface and use Gauss’s law to much more easily obtain the same result for the electric flux through the hemisphere.
(3) A circular surface with a radius 0.057 m is exposed to a uniform external electric field of magnitude 1.44 x 104
N/C. The magnitude of electric flux through the surface is 78 Nm²/C. What is the angle between the direction of the
electric field and the normal to the surface?
A very thin filament of uniform linear charge density "A" is located on the x-axis from
x=0 to x=a.
Prove that the components of the electric field at a point P on the y-axis, located at the
distance "y" from the origin are:Ex = -k^(1/y-1/√/y² + a²) i, Ey = kha/y√/y² + a²)]
6 In Fig. 22-27, two identical circu-
lar nonconducting rings are centered
on the same line with their planes
perpendicular to the line. Each ring
has charge that is uniformly distrib-
uted along its circumference. The
rings each produce electric fields at points along the line. For three
situations, the charges on rings A and B are, respectively, (1) qo and
9o, (2) -90 and -90, and (3) - and qo. Rank the situations
according to the magnitude of the net electric field at (a) point P1
midway between the rings, (b) point P, at the center of ring B, and
(c) point P3 to the right of ring B. greatest first.
P,
P3
Ring A
Ring B
Figure 22-27 Question 6.
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