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3 years ago

The magnitude of the electric force between two protons is 2.30 10-26 n. how far apart are they?

2 answers:

7 0

The electric force between two charge objects is calculated through the Coulomb's law.
F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m.

Feliz [49]3 years ago

7 0

Answer:

0.079m

Explanation:

According to coulombs law, the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of their distance between them.

Mathematically, F = kq1q2/d² where

k is the coulombs constant

F is the force between the charges

q1 and q2 are the charges

d is the distance between the charges

Given F = 2.30×10^-26

d = ?

q1 = q2 = 1.602 x10^-19 C.

k = 9×10^9N/C²m²

Note that charge of proton is 1.602 x10^-19 C.

Substituting the values in the formula to get 'd' we have;

2.3×10^-26 = 9×10^9×(1.602×10^-19)²/d²

2.3×10^-26d² = 9×10^9×(1.602×10^-19)²

2.3×10^-26d² = 9×10^9×1.602×10^-38

2.3×10^-26d² = 1.442×10^-28

d² = 1.442×10^-28/2.3×10^-26

d² = 0.627×10^-2

d² = 0.00627

d = √0.00627

d = 0.079m

It means they are 0.079m apart

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