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Chapter 17: Problem 22

A spherical conductor with a radius of \(75.0 \mathrm{cm}\) has an electric field of magnitude \(8.40 \times 10^{5} \mathrm{V} / \mathrm{m}\) just outside its surface. What is the electric potential just outside the surface, assuming the potential is zero far away from the conductor?

Short Answer

Expert verified
Answer: The electric potential just outside the surface of the spherical conductor is 6.30 x 10^5 V.

Step by step solution

01

Write down the given information

We are given the following information: - Radius of the spherical conductor: \(r = 75.0 \mathrm{cm} = 0.75 \mathrm{m}\) - Electric field just outside the surface: \(E = 8.40 \times 10^{5} \mathrm{V/m}\)
02

Calculate the charge of the spherical conductor

To find the electric potential, first, we need to find the charge of the conductor. The electric field due to a point charge is given by the equation: \(E = \cfrac{k \cdot |Q|}{r^2}\) where \(E\) is the electric field, \(k\) is the Coulomb's constant (\(8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\)), \(Q\) is the charge of the conductor, and \(r\) is the distance from the conductor (in this case, the radius of the sphere). Rearrange the equation to solve for the charge: \(Q = \cfrac{E \cdot r^2}{k}\) Plug in the values and calculate the charge: \(Q = \cfrac{(8.40 \times 10^{5} \mathrm{V/m}) \cdot (0.75 \mathrm{m})^2}{(8.99 \times 10^9 \mathrm{N \cdot m^2/C^2})} = 2.33 \times 10^{-4} \text{C}\)
03

Calculate the electric potential

Now that we have the charge of the conductor, we can calculate the electric potential just outside the surface. The relationship between electric field and electric potential is given by: \(V = \cfrac{E \cdot r}{r_0}\) where \(V\) is the electric potential, \(E\) is the electric field, \(r\) is the distance from the conductor, and \(r_0\) is the distance where the potential is zero (in this case, assumed to be far away from the conductor). In this particular problem, the electric potential is already calculated at the surface of the spherical conductor, so: \(V = E \cdot r\) Plug in the values: \(V = (8.40 \times 10^{5} \mathrm{V/m}) \cdot (0.75 \mathrm{m}) = 6.30 \times 10^5 \mathrm{V}\)
04

Write the final answer

The electric potential just outside the surface of the spherical conductor is: \(V = 6.30 \times 10^5 \mathrm{V}\)

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Most popular questions from this chapter

A point charge \(q=+3.0\) nC moves through a potential difference $\Delta V=V_{\mathrm{f}}-V_{\mathrm{i}}=+25 \mathrm{V} .$ What is the change in the electric potential energy?
A hydrogen atom has a single proton at its center and a single electron at a distance of approximately \(0.0529 \mathrm{nm}\) from the proton. (a) What is the electric potential energy in joules? (b) What is the significance of the sign of the answer?
An array of four charges is arranged along the \(x\) -axis at intervals of 1.0 \(\mathrm{m}\). (a) If two of the charges are \(+1.0 \mu \mathrm{C}\) and two are \(-1.0 \mu \mathrm{C},\) draw a configuration of these charges that minimizes the potential at \(x=0 .\) (b) If three of the charges are the same, $q=+1.0 \mu \mathrm{C},\( and the charge at the far right is \)-1.0 \mu \mathrm{C},$ what is the potential at the origin?
A parallel plate capacitor has a capacitance of \(2.0 \mu \mathrm{F}\) and plate separation of \(1.0 \mathrm{mm} .\) (a) How much potential difference can be placed across the capacitor before dielectric breakdown of air occurs \(\left(E_{\max }=3 \times 10^{6} \mathrm{V} / \mathrm{m}\right) ?\) (b) What is the magnitude of the greatest charge the capacitor can store before breakdown?
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) There is a charge of magnitude \(0.800 \mu\) C on each plate. (a) What is the potential difference between the plates? (b) If the plate separation is doubled, while the charge is kept constant, what will happen to the potential difference?
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