Four identical charges (+2.0ìC each)

are brought in from infinity and fixed to a straight line. The charges are located 0.40 meters
apart. Determine the electric potential energy of this group.

thank you! It seems to all make sense. I'm guessing you meant V2=V3 though? Due to symmetry that is.

To calculate the electric potential energy of this group, we can use the equation:

U = k * (Q1 * Q2) / r

Where:
U is the electric potential energy
k is the Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2)
Q1 and Q2 are the charges
r is the distance between the charges

In this case, since all four charges are identical (+2.0ìC each), we can simplify the equation by plugging in the values Q1 = Q2 = 2.0ìC and r = 0.40 meters:

U = (8.99 x 10^9 N m^2/C^2) * (2.0ìC * 2.0ìC) / 0.40 meters

U = (8.99 x 10^9 N m^2/C^2) * (4.0ìC^2) / 0.40 meters

U = (8.99 x 10^9 N m^2/C^2) * (4.0 x 10^-12 C^2) / 0.40 meters

U = (8.99 x 4.0 x 10^-12) N m^2 / 0.40 meters

U = (35.96 x 10^-12) N m^2 / 0.40 meters

U = 89.9 x 10^-12 N m^2 / 0.40 meters

U ≈ 2.248 x 10^-10 N m

Therefore, the electric potential energy of this group is approximately 2.248 x 10^-10 N m.

To determine the electric potential energy of the group of charges, we need to use the formula:

Electric Potential Energy (U) = k * (q1 * q2) / r

Where:
- k is the electrostatic constant, approximately equal to 9 x 10^9 N m^2 / C^2
- q1 and q2 are the magnitudes of the charges
- r is the distance between the charges

Given that the charges are identical and each charge has a magnitude of +2.0 μC, the equation becomes:

U = k * (q * q) / r

Now we can substitute the values into the equation:

U = (9 x 10^9 N m^2 / C^2) * (2.0 μC * 2.0 μC) / 0.40 m

To solve this equation, we first need to convert the charge magnitude from microcoulombs (μC) to coulombs (C).

1 μC = 1 x 10^-6 C

So, 2.0 μC = 2.0 x 10^-6 C

Plugging in this value:

U = (9 x 10^9 N m^2 / C^2) * (2.0 x 10^-6 C * 2.0 x 10^-6 C) / 0.40 m

U = (9 x 10^9 N m^2 / C^2) * (4.0 x 10^-12 C^2) / 0.40 m

Now we can calculate the electric potential energy using a calculator:

U = (9 x 10^9 N m^2 / C^2) * (4.0 x 10^-12 C^2) / 0.40 m

U ≈ 9.0 x 10^-3 J

Therefore, the electric potential energy of the group of charges is approximately 9.0 x 10^-3 joules.

label the points 1,2,3,4

V1=V2+V3+V4
V1=kq/x+kq/2x + kq/3x
V1=kq/6x(6+3+2)=11kq/6x
where x is .4m
Now, due to symettry, V4=V1

V2=V1+V3+V4=kq/x+kq/x+kq/2x=kq/6x (15)
and due to symettry, V3=V3

Electric PE= sum of V * q

Electric PE=kq^2/6x (15+15+11+11)
= 7kq^2/x

check my thinking.