How much energy is need to place four positive charges, each of magnitude +5.0 mC, at the vertices of a square of side 2.5 cm?

Didn't you ask this once already?

Same answer applies.

nope

To calculate the energy needed to place the charges at the vertices of a square, you can use the formula for the potential energy of a system of charges.

The formula for the potential energy (U) of a system of charges is given by:

U = k * q1 * q2 / r

where:
- U is the potential energy
- k is the electrostatic constant (k = 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

In this case, we have four charges of magnitude +5.0 mC (mC = 10^-3 C), and the distance between the charges is the side length of the square, which is 2.5 cm (cm = 10^-2 m).

Since the charges are positive, the potential energy of the system will be positive.

Let's calculate the potential energy for one pair of charges and then multiply it by 6 to account for all the pairs of charges:

1. Calculate the potential energy for one pair of charges:
U = k * q1 * q2 / r
= (9 x 10^9 N m^2/C^2) * (5.0 x 10^-3 C) * (5.0 x 10^-3 C) / (2.5 x 10^-2 m)
= 450 N m

2. Multiply the potential energy by 6 to calculate the total energy for all the pairs of charges:
Total energy = 6 * (450 N m)
= 2700 N m

Therefore, the energy needed to place the four positive charges at the vertices of the square is 2700 N m.