Question

1. How much electrical energy is needed to bring two charges Q1= 5.5 x 10-7C and Q2= 1.7 x 10-6 C from infinity to where they are 1 meter apart.

Answer 1

Answer:

The electrical energy needed is 8.415*10⁻³ N

Explanation:

Energy is the ability of a body to make changes or work.

By separating or joining two electric charges a distance (for example, a radius r) within their electric fields, you are taking away or giving the electric charges energetic potentials, relative to each other. By releasing these charges, they will attract or repel each other, releasing that acquired electrical energy.

In other words, electric potential energy is linked to the particular configuration of a conglomerate of point charges in a defined system.

That is, it calculates the capacity of an electrical system to carry out a task based exclusively on its position or configuration. So, it is a kind of energy stored in the system, or the amount of energy that it is capable of delivering.

Thus, a charge will exert a force on any other charge and the potential energy is the result of the set of charges.

The electric potential energy that has a point charge q in the presence of another point charge Q that are separated by a certain distance r is:

$Ep=K*\frac{Q1*Q2}{r^{2} }$

where:

• Ep is the electric potential energy. It is measured in Newton (N).
• Q1 and Q2 are the values ​​of the two point charges. They are measured in Coulombs (C).
• r is the value of the distance that separates them. It is measured in meters (m).
• K is the constant of Coulomb's law. For vacuum its value is approximately 9*10⁹ N*m²/C²

In this case:

• Q1=5.5*10⁻⁷ C
• Q2=1.7*10⁻⁶ C
• r=1 m

Replacing:

$Ep=9*10^{9}\frac{N*m^{2} }{C^{2} } *\frac{5.5*10^{-7}C *1.7*10^{-6}C }{(1m)^{2} }$

Solving:

Ep= 8.415*10⁻³ N

The electrical energy needed is 8.415*10⁻³ N