Three point charges are on the x axis: q1 is at the origin, q2 is at x = +2.50 m, and q3 is at x = +6.00 m. find the electrostatic potential energy of this system of charges for the following charge values. (assume the potential energy is zero when the charges are very far from each ) q1 = q2 = q3 = +4.50 µc(b) q1 = q2 = +4.50 µc and q3 = -4.50 µc(c) q1 = q3 = 3.00 µc and q2 = -3.00 µc

a) U(system) = 155.346 * 10 ⁻³ J

b)  U(system) = -9.546* 10⁻³ J

c)  U(system) = - 34.24 * 10⁻³ J

Step-by-step explanation:

The  electrostatic potencial energy of a system of charges, is the algebraic sum of the potencial energy of all possible combinacion of charges taking in pairs, in other words if the system consist of (3) three charges, the total potencial energy of the system consist of, the potencial energy between

-charge q1 and q2

-charge q1 and q3

-charge q2 and q3

Then depending on the sign of the calculated quantities (which depends on the sense of the force between them) we add or subtract such quantities.

In our case we have:

a) all charges are iqual  +4.50 μC        or    + 4.50* 10⁻⁶ C

b) distance between charges q1 and q2 2.5 m   between q1 and q3  6 m and between q2 and q3  3.5 m

K = 1 ÷4πε₀   = 9 * 10⁹  Nm²/C⁻²

The potencial energy is U = K (q1*q2 )÷d

Then calculating for charges q1 and q2 we have

U₁₋₂  =   [ 9*10⁹*(+4.5)*(+4.5)*10⁻⁶ *10⁻⁶]÷2.5      [Nm²*C²/C² *m]

U₁₋₂  =  72.9 * 10⁻³   J

U₁₋₃  = [ 9*10⁹*(+4.5)*(+4.5)*10⁻⁶ *10⁻⁶]÷6        [Nm²*C²/C² *m]

U₁₋₃  = 30.375 * 10⁻³     J

U₂₋₃ =   [ 9*10⁹*(+4.5)*(+4.5)*10⁻⁶ *10⁻⁶]÷3.5        [Nm²*C²/C² *m]

U₂₋₃ =  52.071*  10⁻³ J

Note that U₁₋₂  U₂₋₃  and  U₁₋₃   are positive (meaning the system is acumulating energy)

U(system) = U₁₋₂ + U₂₋₃ + U₁₋₃

U(system) = 155.346 * 10 ⁻³ J

b) Charges q1 = q2 =  +4.50 μC and charge q3 = -4.5 μC

The change here respect case a is charge q3 is negative. There are no changes in distances therefore we can use results from case a (taken into accont that in this case the potencial energy with components in q3 are negative then we have:

U₁₋₂ = 72.9 * 10⁻³ J      U₁₋₃  =  -30.375* 10⁻³ J     U₂₋₃ = - 52.071* 10⁻³ J

Then:

U(system) = -9.546* 10⁻³ J

That means that system  lost energy, nstead of acmlate lke in the fist case

c) q1 =q3 = 3 μC     and q3 = -3 μC

Calclating as before:

U₁₋₂  =   [ 9*10⁹*(+3)*(+3)*10⁻⁶ *10⁻⁶]÷2.5      [Nm²*C²/C² *m]

U₁₋₂  = 2.4 *10⁻³ J

U₁₋₃  = [ 9*10⁹*(+3)*(-3)*10⁻⁶ *10⁻⁶]÷6

U₁₋₃  =  - 13.5 * 10⁻³ J  

U₂₋₃ =   [ 9*10⁹*(+3)*(-3)*10⁻⁶ *10⁻⁶]÷3.5  

U₂₋₃ = - 23.14 *10⁻³ J

U(system) = (2.4 - 13.5 -23.14)*10⁻³ J

U(system) = - 34.24 * 10⁻³ J