Question

Two protons are a distance 3 10-9 m apart. What is the electric potential energy of the system consisting of the two protons

Answer 1

Answer:

The electric potential energy of the system is 7.87x10⁻²⁰ J.

Explanation:

The electric potential energy is given by:

$E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}$

Where:

q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C

r is the distance = 3x10⁻⁹ m

K: is the electrostatic constant = 9x10⁹ Nm²/C²

$E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J$

Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.

I hope it helps you!

Answer 2

The electric potential energy of the system should be 7.87x10⁻²⁰ J.

Calculation of the electric potential energy:

SInce We know that

fdr = kq1q2/r

Here

q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C

r should be the distance = 3x10⁻⁹ m

K should be the electrostatic constant = 9x10⁹ Nm²/C²

Now electric potential energy should be

= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) /  3x10⁻⁹ m

=  7.87x10⁻²⁰ J.

hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.

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