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.
The electric force of repulsion between two electrons is similar to the gravitational force:
where k is called Planck's constant, q is the charge on a particle, and r is the distance separating the charges.
Find the new force of repulsion between them, in terms of R:
a. when the distance between the electrons is doubled.
b, when the distance between the electrons is multiplied by 1.2.
C. when the distance between the electrons is multiplied by 1/3.
D. when the distance between the electrons is divided by 6.
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