Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. Calculate the potential energy of two singly charged nuclei separated by 1.00 x 10-12 m by finding the voltage of one at that distance and multiplying by the charge of the other.

the Potential Energy is 2.304 × 10⁻¹⁶ J

Explanation:  

Given the data in the data in the question;

The expression for the electric potential energy between the charges can be expressed as follows;

PE = qV equ 1

where q is the charge and V is the electric potential

Also the formula for electric potential due to point a point in a field is;

V = kq /  r equ 2

where k is the electrostatic constant and r is the distance form the charged particle

input equation 2 into 1

PE = q × kq /  r

PE = kq²/r equ 3

so we substitute into equation 3; 1.00×10⁻¹² for r, 9.00×10⁹ for k( constant ) and 1.60×10⁻¹⁹ for q( charge )

PE = ((9.00×10⁹) (1.60×10⁻¹⁹)²) / 1.00×10⁻¹²

PE = 2.304 × 10⁻²⁸ / 1.00×10⁻¹²

PE = 2.304 × 10⁻¹⁶ J

Therefore, the Potential Energy is 2.304 × 10⁻¹⁶ J

v=ir = 3x15 =45 v

dang that balls went far