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e-lub [12.9K]
3 years ago
13

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.00mm away. What is

the initial acceleration of the proton after it is released?
Chemistry
1 answer:
myrzilka [38]3 years ago
3 0
<span>F = m.a
   = mass x acceleration

So,
a = F / m
Where,
F is coulomb repulsion.
k q q / r^2
with,
k = 9 x 10^9 in SI units (q in C and r in m)

Hence,
</span><span>a = 2.40 × 10^11 m/s^2 </span>
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alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

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E = k \frac{q}{r^{2} }

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r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

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we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

E = k \frac{q}{r^{2} }

q = <u>E x r²</u>

        k

q =  <u>1250 N/C x 0.205m</u>²

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q =   5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = \frac{q}{e}

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       1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

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