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antoniya [11.8K]
4 years ago
12

How many electrons should be removed from an initially uncharged spherical conductor of radius 0.200 m to produce a potential of

5.50 kv at the surface?
Physics
2 answers:
Lubov Fominskaja [6]4 years ago
4 0

To calculate the number of electrons from spherical conductor first we use the formula,

V= k\frac{q}{r}

Here, k is a constant with a value of 8.99 \times 10^9 N.m^{2} /C^2 , q is the charge and r is the radius and its value of 0.200 m.

Substituting these value in above formula, we get  

5.50 kV =\frac{8.99 \times 10^9 N.m^{2} /C^2\times q}{0.200 m}

or                q= \frac{5.50\times10^3 V\times0.200 m}{8.99 \times 10^9 N.m^{2} /C^2}

q=1.2\times 10^{-7} C

Now number of electron,

N= \frac{1.2\times 10^{-7} C}{1.6\times 10^{-19}C/e }

N=7.5\times 10^{11} electrons

Hence, the number of electrons to be removed from conductor would be 7.5\times 10^{11} electrons

                     


Lena [83]4 years ago
3 0

Solution:

Potential is defined as

V = kq/r  

=> q= Vr/k  

= 5500 * 0.2 / 9 x 10^9 C  

= (110/9) x 10^-7  

If n = no. of electrons to be removed

So,  

n x 1.6 x 10^-19  

Therefore,

n=v/q

=> n = (110/14.4) x 10^12 electrons  

= 7.6382 x 10^12 electrons.  

This will leave an equal amount of positive charge in protons and produce the give potential on the surface of the sphere.

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Answer:

1) a = -1 m/s²

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Explanation:

Given,

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The time taken by the object to travel is, t = 5 s

Using the first equation of motion

                               <em>v = u + at</em>

                               a = (v - u) / t

Substituting the values

                                a = (10 - 15) / 5

                                    = -1 m/s²

The negative sign indicates the body is decelerating

The acceleration of the object is, a = -1 m/s²

The speed of the object after 2 seconds

From the above equations of motion

                                  v = 15 + (-1) 2

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Hence, the speed of the object after 2 seconds is, v = 12 m/s

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4 years ago
Work and Energy
Sergeeva-Olga [200]

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The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

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g = 9.8 m/s^2 is the acceleration of gravity

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For the first student, \Delta h = 3.5 m, so the work done is

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