Four \(0.8 \mathrm{nC}\) point charges are located in free space at the corners of a square \(4 \mathrm{~cm}\) on a side. \((a)\) Find the total potential energy stored. \((b) \mathrm{A}\) fifth \(0.8 \mathrm{nC}\) charge is installed at the center of the square. Again find the total stored energy.

In the given exercise, we have four point charges each of \(0.8 \mathrm{nC}\) located at the corners of a square with side \(4 \mathrm{~cm}\) and a fifth point charge of \(0.8 \mathrm{nC}\) located at the center of the square in the second part. First, let's convert the charges and distances to SI units: \(q = 0.8 \times 10^{-9} \mathrm{C}\) \(a = 4 \times 10^{-2} \mathrm{m}\).

In first situation, we have 4 potential energy contributions between the charges placed at the corners of the square: - Between charges at top-left and top-right corners - Between charges at top-left and bottom-left corners - Between charges at top-right and bottom-right corners - Between charges at bottom-left and bottom-right corners Since all charges are equal and at equal distances from each other, the potential energy between each pair of charges at the corners can be calculated as: \(U_1 = k \frac{q^2}{a}\) The total potential energy in this situation is the sum of the potential energy contributions between the pairs: \(U_{total(a)} = 4U_1 = 4k \frac{q^2}{a}\)

In this situation, we have a fifth charge at the center of the square. Let's consider an additional charge \(q_5 = 0.8 \times 10^{-9} \mathrm{C}\). The distance from the center to the corner charges can be found using the Pythagorean theorem: \(r = \frac{a}{\sqrt{2}}\) We need to calculate the potential energy between this fifth charge and each of the four corner charges: \(U_2 = k \frac{qq_5}{r}\) The total potential energy stored in this situation is the sum of potential energy contributions from step 2 and the potential energy between the central charge and the corner charges: \(U_{total(b)} = U_{total(a)} + 4U_2 = 4k \frac{q^2}{a} + 4k \frac{qq_5}{r}\)

Now, let's substitute the values and calculate the total potential energy stored in both situations. Applying the values \(q = 0.8 \times 10^{-9} \mathrm{C}\) and \(a = 4 \times 10^{-2} \mathrm{m}\) and using the Coulomb's constant \(k = 8.9875 \times 10^9 \mathrm{Nm^2C^{-2}}\), we have: For the first situation (a): \(U_{total(a)} = 4k \frac{q^2}{a} = 4 (8.9875 \times 10^9) \frac{(0.8 \times 10^{-9})^2}{4 \times 10^{-2}} \approx 51.48 \times 10^{-9} \mathrm{J}\) For the second situation (b): \(U_{total(b)} = 4k \frac{q^2}{a} + 4k \frac{qq_5}{r} = 51.48 \times 10^{-9} + 4(8.9875 \times 10^9) \frac{(0.8 \times 10^{-9})(0.8 \times 10^{-9})}{\frac{a}{\sqrt{2}}} = 51.48 \times 10^{-9} + 4(8.9875 \times 10^9) \frac{(0.8 \times 10^{-9})^2}{(4 \times 10^{-2})/\sqrt{2}} \approx 102.96 \times 10^{-9} \mathrm{J}\) So, the total stored energy for cases \((a)\) and \((b)\) are approximately \(51.48 \times 10^{-9} \mathrm{J}\) and \(102.96 \times 10^{-9} \mathrm{J}\), respectively.