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Likurg_2 [28]
1 year ago
5

two charges experience exert a force of 1n on each other when they are 1m apart. what force will these charges experience if the

y are placed 2m apart?
Physics
1 answer:
Kisachek [45]1 year ago
8 0

The force experienced by the charges when they are placed 2 m apart will be 0.25 N.

The force between two charges is given by Coulomb’s Law.

According to this law, the force between two charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges.

The above statement is represented by the following equation,

F= K Q1 x Q2/r^2

Here, F = Force between the two charges

K = Constant

Q1 and Q2 =  Magnitude of the two charges

r = Separation between the two charges

According to this equation, the force is inversely proportional to the square of the distance between two charges.

Initially, when the two charges are 1 m apart, the force is 1 N.

When the distance between the two charges is doubled,

then according to Coulomb’s Law, the force should decrease by 4 times.

Hence, if the charges are placed 2 meters apart,  the force becomes 1/4= 0.25 N.

To know more about "Coulomb's Law", refer to the following link:

brainly.com/question/9261306?referrer=searchResults

#SPJ4

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ni is the index of refraction of the medium you want, water in this case;

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Substituting and calculating for water (nwater = 1.3):

\frac{1}{f} = (1.5 - 1.3)(\frac{1}{10} - \frac{1}{15})

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For air (nair = 1):

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f = \frac{30}{0.5} = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

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protons in an atomic nucleus are typically 10-15 m apart. what is the electric force of repulsion between nuclear protons?
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Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

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The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

F = k(q1 q2)/ r^2

Distance between protons = d = 10⁻¹⁵ m

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Here q1=q2

electric force = F =230N

Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.

2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.

Find more about electric force of repulsion between nuclear protons

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