Two 5.0-cm-diameter conducting spheres are 8.0 m apart, and each carries 0.12 mC. Determine (a) the potential on each sphere, (b

Explanation: Two spheres are positioned 10m apart, each with a charge of 0.12mC, which is equivalent to 0.12×10^-3C. The diameter is given as 5cm or 0.05m, hence the radius is calculated by dividing the diameter by 2, yielding r = 0.025m. The potential V is expressed as V = kq/r, with k being 9×10^9Nm²/C². a. The potential at the surface of each sphere will be identical due to their equal charges. At the sphere surface, where r = 0.025m, we find V = 9E9×0.12E-3/0.025, simplifying to V = 4.32E7Volts or 4.32×10^7Volts. b. The electric field strength at the surface of each sphere remains the same since both hold identical charges. The electric field is calculated as E = kq/r². Thus, at the surface: E = 9E9×0.12E-3/0.025², which results in E = 1.728×10^8N/C. c. To determine the potential midway between the two spheres, we need to note that they are 10m apart, meaning the mid-point is 5m away. Adding the sphere radius of 0.025m gives a total distance of 5.025m from the center. Thus, the potential from the first sphere is V1=9E9×0.12E-3/5.025, leading to V1=2.149×10^5 Volts. Since the second sphere's contribution is identical, the overall potential becomes V1 + V2 = 2.149E5 + 2.149E5, which totals V = 4.3E5Volts. d. The potential difference between the spheres at any given point matches the potential difference calculated in part c. Therefore, the final potential difference is V = 4.3×10^5Volts.