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2 years ago

What is the net charge on a sphere that has the following? (a) 5.87 106 electrons and 8.11 106 protons C

Physics

2 answers:

hammer [34]2 years ago

6 0

Answer:

3.58*10⁻¹³ C

Explanation:

When we have a net charge on a sphere, this means that there must be a difference in the number of electrons and protons on the sphere, otherwise, the sphere would be electrically neutral.

In this case, we can find this difference as follows:

Np -Ne = 8.11*10⁶ protons - 5.87*10⁶ electrons = 2.24*10⁶ protons.

The total charge carried by all these protons is just the product of the charge of only one proton (which is equal to the elementary charge e) times the number of excess protons, as follows:

Qnet = 2.24*10⁶ protons * 1.6*10⁻19 C/proton = 3.58*10⁻¹³ C

AysviL [449]2 years ago

3 0

Answer:

3.6 * 10^-13 C

Explanation:

The net charge of the sphere will be the sum of the total electron charge and total proton charge. Mathematically,

Q = Qe + Qp

TOTAL ELECTRON CHARGE:

An electron has an electronic charge of -1.6022 * 10^-19C.

Hence, the charge of 5.87 * 10^6 electrons will be:

Qe = - 1.6022 * 10^-19 * 5.87 * 10^6

Qe = - 9.405 * 10^-13 C

TOTAL PROTON CHARGE:

A proton has an electronic charge of 1.6022 * 10^-19. Hence, 8.11 * 10^6 protons will have:

Qp = 1.6022 * 10^-19 * 8.11 * 10^6

Qp = 1.3 * 10^-12 C

=> Q = (-9.405 * 10^-13) + (1.3 * 10^-12)

Q = 3.6 * 10^-13 C

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lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

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1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

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So, we have:

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So, the new force is:

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8 0

2 years ago

Read 2 more answers

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Butoxors [25]

if a volume of air is warmed it expands due to increased translational kinetic energy as it expands it will start to cool.

<h3>When does temperature increase volume?</h3>

We can then conclude that at constant pressure, temperature and volume are directly proportional: temperature increases, volume increases; decrease temperature, decrease volume.

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See more about volume at brainly.com/question/1578538

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1 year ago

An automobile rounds a curve of radius 50.0 m on a flat road.

bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve = 50m

Centripetal acceleration = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

It is given as;

a = $\frac{v^{2} }{r}$

a is the centripetal acceleration

v is the speed

r is the radius

Now insert the parameters and find v;

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v = √196 = 14m/s

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2 years ago