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) An infinite straight line has uniform charge density per unit length 2,
given in C/m, is inserted into a conducting sphere of radius R with its
center lies on the line (Fig.6). The electric flux through the sphere is
(a) λR/&;
(b) 0;
(c) 22R/&o;
(d) λR/2πε. ·
o
Fig.6 conducting sphere
2
line
Chapter 22 Solutions
Physics for Scientists and Engineers with Modern Physics
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