Chapter 6: Q13P (page 335)
over the surface of a sphere of radius and center at the origin.
Short Answer
The solution derived is .
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Chapter 6: Q13P (page 335)
over the surface of a sphere of radius and center at the origin.
The solution derived is .
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where and is the entire surface of the tin can bounded by the cylinder
role="math" localid="1657353627256"
role="math" localid="1657353639412"
role="math" localid="1657353647648"(a) Given , sketch on one graph the curves. Ifis the electrostatic potential, the curvesconst. are equipotential, and the electric field is given by. Ifis temperature, the curves= const. are isothermals andis the temperature gradient; heat flows in the direction.
(b) Find and draw on your sketch the vectorsat the points,,. Then, remembering thatis perpendicular to= const., sketch, without computation, several curves along which heat would flow [see (a)].
The force on a charge moving with velocity in a magnetic field B iswe can write B aswhere A (called the vector potential) is a vector function of x,y,z,t . If the position vectorof the charge is a function of time, show that
Thus show that
over the entire surface of the volume in the first octant bounded byand the coordinate planes, where
Show that where U is a vector function of and .
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