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Chapter 23: Problem 60

A solid metal ball with a radius of \(3.00 \mathrm{~m}\) has a charge of \(4.00 \mathrm{mC}\). If the electric potential is zero far away from the ball, what is the electric potential at each of the following positions? a) at \(r=0 \mathrm{~m},\) the center of the ball b) at \(r=3.00 \mathrm{~m},\) on the surface of the ball c) at \(r=5.00 \mathrm{~m}\)

Short Answer

Expert verified
Answer: The electric potentials at the given positions are: a) At the center of the ball, the electric potential is 0 V. b) On the surface of the ball, the electric potential is 1.199 x 10^7 V. c) At a distance of 5.00 m from the center of the ball, the electric potential is 7.190 x 10^6 V.

Step by step solution

01

Understand the concept of electric potential

Electric potential is the amount of work needed to move a test charge from a reference point to a specific point in an electric field without accelerating the test charge. The reference point can be chosen arbitrarily, and in this case, it is far away from the ball, where the electric potential is zero. The electric potential at any point is given by the formula \(V = \frac{kQ}{r}\), where \(V\) is the electric potential, \(k\) is the electrostatic constant (\(8.9875 \times 10^9 \mathrm{Nm^2/C^2}\)), \(Q\) is the charge of the source, and \(r\) is the distance between the point and the source.
02

Calculate the electric potential at \(r=0\mathrm{~m}\) (center of the ball)

At the center of the ball, \(r=0\mathrm{~m}\). Since the electric potential is the amount of work needed to move a test charge from a reference point to this point, and since the center of the ball is at zero distance from the source of charge, the electric field is uniform and doesn't change with the movement of a test charge. Therefore, the electric potential at the center of the ball, \(V_{center}\), is the same as that far away from the ball, which is zero. So, \(V_{center} = 0\mathrm{V}\).
03

Calculate the electric potential at \(r=3.00\mathrm{~m}\) (surface of the ball)

At the surface of the ball, \(r=3.00\mathrm{~m}\). We can use the formula for electric potential \(V = \frac{kQ}{r}\) to find the electric potential at the surface of the ball, \(V_{surface}\). Plugging in the values, we get \(V_{surface} = \frac{8.9875\times10^9\mathrm{Nm^2/C^2} \times 4.00\times10^{-3}\mathrm{C}}{3.00\mathrm{m}} = 1.199 \times 10^7\mathrm{V}\).
04

Calculate the electric potential at \(r=5.00\mathrm{~m}\)

At a distance of \(5.00\mathrm{~m}\) from the center of the ball, \(r=5.00\mathrm{~m}\). We can use the formula for electric potential \(V = \frac{kQ}{r}\) to find the electric potential at this point, \(V_{5m}\). Plugging in the values, we get \(V_{5m} = \frac{8.9875\times10^9\mathrm{Nm^2/C^2} \times 4.00\times10^{-3}\mathrm{C}}{5.00\mathrm{m}} = 7.190 \times 10^6\mathrm{V}\). In summary, the electric potentials at the given positions are: a) \(V_{center} = 0\mathrm{V}\) b) \(V_{surface} = 1.199 \times 10^7\mathrm{V}\) c) \(V_{5m} = 7.190 \times 10^6\mathrm{V}\)

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