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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 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) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
In a region where there is an electric field, the electric forces do $+8.0 \times 10^{-19} \mathrm{J}\( of work on an electron as it moves from point \)X$ to point \(Y\). (a) Which point, \(X\) or \(Y\), is at a higher potential? (b) What is the potential difference, \(V_{Y}-V_{X},\) between point \(Y\) and point \(X ?\)
An axon has the outer part of its membrane positively charged and the inner part negatively charged. The membrane has a thickness of \(4.4 \mathrm{nm}\) and a dielectric constant \(\kappa=5 .\) If we model the axon as a parallel plate capacitor whose area is \(5 \mu \mathrm{m}^{2},\) what is its capacitance?
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
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