Respuesta :
Answer:
The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]
Explanation:
Given that,
Distance [tex]d=4.30\times10^{-11}\ m[/tex]
suppose, Two particles with charges +6 e and -10 e are initially very far apart
We need to calculate the change in the electric potential energy
Using formula of energy
[tex]\text{electric potential energy}=\text{final electric potential energy-initial electric potential energy}[/tex]
[tex]EPE=EPE_{f}-EPE_{i}[/tex]
Here, initial electric potential energy= 0
final electric potential energy [tex]EPE_{f}=\dfrac{kq_{1}q_{2}}{r_{2}^2}[/tex]
Put the value into the formula
[tex]EPE=\dfrac{kq_{1}q_{2}}{r_{2}^2}+0[/tex]
Put the value into the formula
[tex]EPE=\dfrac{9\times10^{9}\times6\times1.6\times10^{-19}\times(-10\times1.6\times10^{-19})}{(4.30\times10^{-11})}[/tex]
[tex]EPE=-3.2\times10^{-16}\ J[/tex]
Hence, The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]
This is an incomplete question, here is a complete question.
Two particles with charges +6 e⁻ and -10 e⁻ are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 4.30 × 10⁻¹¹ m apart. What is EPE(final) - EPE(initial), which is the change in the electric potential energy?
Answer : The change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]
Explanation : Given,
Formula used for electric potential energy of the two charges when they are separated is:
[tex]EPE=\frac{1}{4\pi \epsilon_0}\times {\frac{q_1\times q_2}{r^2}[/tex]
[tex]EPE=\frac{k\times q_1\times q_2}{r^2}[/tex]
where,
EPE = electric potential energy
k = [tex]\frac{1}{4\pi \epsilon_0}=8.99\times 10^9[/tex]
[tex]q_1[/tex] = charge on 1st particle = +6 e⁻ = [tex]6\times 10^{-19}C[/tex]
[tex]q_2[/tex] = charge on 2nd particle = -10 e⁻ = [tex]-10\times 10^{-19}C[/tex]
r = distance between two charges = [tex]4.30\times 10^{-11}m[/tex]
Now put all the given values in the above formula, we get:
[tex]EPE=\frac{(8.99\times 10^9)\times (6\times 10^{-19})\times (-10\times 10^{-19})}{(4.30\times 10^{-11})^2}[/tex]
[tex]EPE=-2.92\times 10^{-6}J[/tex]
Initially EPE = 0 J
Thus, [tex]EPE_{final}-EPE_{initial}=-2.92\times 10^{-6}J[/tex]
The positive sign indicate the attractive force and negative sign indicate the repulsive force.
Thus, the change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]
