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?

Compute the potential energy of the configuration, assuming that the charges are brought in from infinity one at a time. The will be the energy required. Let each corner charge be q and the square side length a. brining in the first charge requires no work. The second charge is placed a distance a from the first charge. The third charge gets placed a distance a from one charge and sqrt2*a from the other charge. Then bring in the fourth charge. You get the idea.

W = k[q/a + q/a + q/(a*sqrt2) + q/a + q/a + q/(a*sqrt2)]
= (k/a)[4 + sqrt2]

k is the Coulomb's law constant. Make sure q is in coulombs and a is in meters to get the answer in joules.

To calculate the energy needed to place the charges, we can use the concept of electric potential energy.

Step 1: Calculate the electric potential energy between each pair of charges.

The electric potential energy between two charges q1 and q2 separated by a distance r is given by 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 * q2| represents the magnitude of the product of the charges
- r is the distance between the charges

For a square, each charge is at the vertices, and the distance between each pair of charges is the length of one side, which is 2.5 cm.

Step 2: Calculate the total energy needed.

Since there are four charges, we need to calculate the energy between each pair of charges and then sum them up.

Let's calculate it step by step:

1. Energy between charges at opposite vertices of the square:
U1 = (k * |q1 * q2|) / r
= (9 x 10^9 N m^2/C^2) * (5.0 mC * 5.0 mC) / (2.5 cm)
= (9 x 10^9 N m^2/C^2) * (5.0 x 10^-3 C)^2 / (2.5 x 10^-2 m)
= (9 x 10^9 N m^2/C^2) * (25 x 10^-6 C^2) / (2.5 x 10^-2)
= 9 x 25 x 10^3 / 2.5 J
= 90 x 10^3 / 2.5 J
= 36,000 J
(Numeric value: 36,000 J)

2. Energy between charges at adjacent vertices of the square:
U2 = (k * |q1 * q2|) / r
= (9 x 10^9 N m^2/C^2) * (5.0 mC * 5.0 mC) / (2.5 cm)
= (9 x 10^9 N m^2/C^2) * (25 x 10^-6 C^2) / (2.5 x 10^-2)
= 9 x 25 x 10^3 / 2.5 J
= 90 x 10^3 / 2.5 J
= 36,000 J
(Numeric value: 36,000 J)

3. Total energy needed:
Total Energy = U1 + U2 + U1 + U2
= 36,000 J + 36,000 J + 36,000 J + 36,000 J
= 144,000 J
(Numeric value: 144,000 J)

Therefore, the total energy needed to place four positive charges, each of magnitude +5.0 mC, at the vertices of a square of side 2.5 cm is 144,000 joules.