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sergij07 [2.7K]
3 years ago
8

Suppose the electric field between two points separated by 4 meters is 37 Volt/m. What is the electric potential (in Volt) betwe

en the two points? Use exact numbers; do not estimate.
Physics
1 answer:
o-na [289]3 years ago
3 0

Answer:

148 V

Explanation:

The electric field can be obtained using the formula;

E= V/d

where;

E is the electric field intensity = 37 volt/m

V is the potential difference (the unknown)

m is the distance of separation = 4m

V = Ed

V = 37 volt/m * 4 m

V= 148 V

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Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

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There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle \angle \mathrm{B\hat{C}D} which corresponds to this minor arc.

This angle comes can be split into two parts:

\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}.

Also,

\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}.

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\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}.

The lengths of segment DC, AC, BC can all be found:

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In the two right triangles \triangle\mathrm{DAC} and \triangle \rm BAC, the value of \angle \mathrm{B\hat{C}A} and \angle \mathrm{A\hat{C}D} can be found using the inverse cosine function:

\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}

\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}

\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}.

The length of the minor arc will be:

\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km.

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