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Arada [10]
2 years ago
8

M A spherical conductor has a radius of 14.0cm and a charge of 26.0σC . Calculate the electric field and the electric potential

at (a) r=10.0cm
Physics
1 answer:
Yuri [45]2 years ago
4 0

The Electric field is E=23400 V/m.

What is electric potential ?

The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential). More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion.

The electric field of a spherical conductor is given by:

{\displaystyle E={\frac {Q}{4\pi \varepsilon _{0}\,r^{2}}}

Here 'E' is the electric field and 'Q' is the electric charge.

therefore putting the values we get,

E=23400 V/m

To learn more about electric field click on the link below:

brainly.com/question/11509296

#SPJ4

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Tonya is thinking about the topic presented in the text, "Do opposites really attract?" Which of her thoughts is an example of c
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3 years ago
Consider a block on frictionless ice. Starting from rest, the block travels a distance din
sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{F}{m}

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

\displaystyle d = vo.t+\frac{at^2}{2}

If the block starts from rest, vo=0:

\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

\displaystyle d = \frac{\frac{F}{m}t^2}{2}

Simplifying:

\displaystyle d = \frac{Ft^2}{2m}

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

\displaystyle a'=\frac{4F}{m}

And the distance is now:

\displaystyle d' = \frac{4Ft^2}{2m}

Dividing d'/d:

\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}

Simplifying:

\displaystyle \frac{d' }{d}=4

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2 years ago
How many centimeters are in 3.50 feet?
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4 0
3 years ago
The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU
Ipatiy [6.2K]

Answer:

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The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

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Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

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