Question

A fixed source charge particle of 4.1 nC and a second particle with a charge 2.4 nC are separated by a distance of 3.5 cm. What is the electric potential energy between the particles? And if the second particle is released will it move forward or away from the source charge

Answer 1

Answer:2.5 * 10^6 J away

Explanation:Using the equation PE = 9 * 10^9 * q1 * q2 / r,

where q is charge and r is distance. In this case, the equation would be

9*10^9{(4.1* 10^-9) * (2.4*10^9)/0.035} = 2.5 * 10^6. Both are positive, so they will be away.

Answer 2

a. 2.53 μJ b. It will move away

Step-by-step explanation:

a. What is the electric potential energy between the particles?

The electric potential energy U = kq₁q₂/r where k = 9 × 10⁹ Nm²/C², q₁ = 4.1 nC = 4.1 × 10⁻⁹ C, q₂ = 2.4 nC = 2.4 × 10⁻⁹ C and r = distance between charges = 3.5 cm = 3.5 × 10⁻² m.

Substituting the values of the variables into U, we have

U = kq₁q₂/r²

U = 9 × 10⁹ Nm²/C² × 4.1 × 10⁻⁹ C × 2.4 × 10⁻⁹ C/3.5 × 10⁻² m

U = 88.56 × 10⁻⁹ Nm²/3.5 × 10⁻² m

U = 25.3 × 10⁻⁷ Nm

U = 2.53 × 10⁻⁶ Nm

U = 2.53 × 10⁻⁶ J

U = 2.53 μJ

b. And if the second particle is released will it move forward or away from the source charge.

It will move away from the source charge since they have the same sign of charge. Since, opposite charges repel.