A solid conducting sphere (radius = 5.0 cm) has a charge of 0.25 nC distributed uniformly on its surface. If point A is located at the center of the sphere and point B is 15 cm from the center, what is the magnitude of the electric potential difference between these two points in volt ?
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For a solid conducting sphere, charge resides on the surface. The electric field inside (r < R) is zero, so the potential is constant and equal to the surface potential. Outside (r > R), it behaves like a point charge.
The potential at the center (point A) is the same as on the surface: VA = kQ/R.
The potential at an external point B (r = 15 cm) is VB = kQ/r.
The potential difference is VA - VB = kQ(1/R - 1/r).
Given: Q = 0.25 × 10-9 C, R = 0.05 m, r = 0.15 m, k = 9 × 109 N m²/C².
Calculation: VA - VB = (9e9)(0.25e-9)(1/0.05 - 1/0.15) = 2.25(20 - 6.666...) = 2.25 × 13.333... = 30 V.
Final Answer: 30