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slamgirl [31]
2 years ago
5

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.Ex

press your answer in terms of the given quantities and appropriate constants.1. What is the electric field at the surface of the sphere?E=2.What is the electric potential at the surface of the sphere?V=
Physics
1 answer:
amid [387]2 years ago
3 0

Answer:

E =   k*Q₁/R₁² V/m

V =  k*Q₁/R₁ Volt

Explanation:

Given:

- Charge distributed on the sphere is Q₁

- The radius of sphere is R₁

- The electric potential at infinity is 0

Find:

What is the electric field at the surface of the sphere?E.

What is the electric potential at the surface of the sphere?V

Solution:

- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.

- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                                        F = k*Q₁/R₁²

- Then the electric field at that point is

                                        E =  F/1

                                        E =  k*Q₁/R₁²  V/m

- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                                        V =  k*Q₁/R₁  Volt

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Answer:

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Explanation:

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k = ΔF / Δx

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With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

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b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:

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wo = √2 * 40,000 / 50 = 40 rad/s

And the frequency:

f = wo/2π

f = 40 / 2π = 6.37 Hz

c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

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In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

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2 years ago
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