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bulgar [2K]
3 years ago
13

A charge Q is uniformly distributed along the x axis from x = a to x = b. If Q = 45

Physics
1 answer:
luda_lava [24]3 years ago
3 0

Answer:

The potential at the point 8 meters is approximately 48.98  Joules/C

Explanation:

Notice that we need to find the potential at a point aligned to the rod's axis (see attached figure), that is located 6 meters from one end of the rod. Notice as well, that the length (L) of the rod is 5 meters.

Since we have a uniformly distributed charge, the charge density per unit of length is defined as:

\lambda=\frac{Q}{L} \\\lambda=\frac{45}{5}\,\frac{10^{-9}\,C}{m} \\\lambda=9\,\frac{10^{-9}\,C}{m}

In order to find the contribution of each little segment dx of charge

dq=\lambda\,dx

to the potential at the requested point, we need to perform an integral:

V=\int\limits^{11}_{6} {} \, dV \\V=\int\limits^{11}_{6} {} \, k\,\frac{dq}{x}  \\V=\lambda \,*k\,\int\limits^{11}_{6} {} \, \frac{dx}{x}  \\V=8.98*9  \, \,ln(\frac{11}{6}) \,\frac{J}{C} \\V=80.82\,* \,0.606 \,\frac{J}{C}\\V=48.98 \frac{J}{C}

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50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui
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Answer:

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Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

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