Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. the electric repulsion force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart. what is the magnitude of the electric force between two protons separated by 2.00 ✕ 10-15 m

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

.

Now, making the substitutions

,

,

,

we can find:

.

.

Not so big for everyday life, but enormous for subatomic particles.

The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton

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)

r = distance between charges (m)

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

Let's tackle the problem now !

Given:

distance between protons = d = 2 × 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

Unknown:

electric force = F = ?

Solution:

Grade: High School

Subject: Physics

Chapter: Static Electricity

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