A uniformly charged thin rod of length L and positive charge Q lies along the x-axis with its left end at the origin as shown in
Figure 1.
a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq
b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.
a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq
b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.
1 answer:
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Answer:
Explanation:
height, h = 125 miles = 125 x 1609 m = 201125 m = 0.2 x 10^6 m
Rdaius, R = 2.44 x 10^6 m
Mass, M = 3.3 x 10^23 kg
Let vo be the orbital speed.
The formula for the orbital speed is given by
where, M is the mass of mercury, R be the radius of mercury
vo = 2887.47 m/s
Time period, T = 2π(R+h) / vo
T = 2 x 3.14 x 2.64 x 10^6 / 2887.47
T = 5741.77 seconds
T = 1.6 hours