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Chapter 23: Problem 15

Using Gauss's Law and the relation between electric potential and electric field, show that the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere and equal to the total charge of the sphere. What is the potential at the surface of the sphere? How does the potential change if the charge distribution is not uniform but has spherical (radial) symmetry?

Short Answer

Expert verified
Question: Prove that the electric potential outside a uniformly charged sphere is the same as a point charge at the center of the sphere, holding the total charge of the sphere. Short Answer: The electric potential outside a uniformly charged sphere is given by V = (1 / (4 * π * ε₀)) * (Q / r), where Q is the total charge, r is the distance from the center of the sphere, and ε₀ is the vacuum permittivity. This expression is the same as the electric potential of a point charge at the center of the sphere. Therefore, the electric potential outside a uniformly charged sphere is the same as a point charge at the center of the sphere, holding the total charge of the sphere.

Step by step solution

01

Find the electric field of a uniformly charged sphere

Using Gauss's Law, which states that the total electric flux through a closed surface is equal to the charge enclosed divided by the dielectric constant, we find the electric field outside a uniformly charged sphere. For a sphere with radius R and total charge Q, the electric field E outside the sphere can be determined using the following equation: E = (1 / (4 * \pi * \epsilon_0)) * (Q / r^2) Here, E is the electric field, \epsilon_0 is the vacuum permittivity, r is the distance from the center of the sphere to the point where the electric field is being measured, and Q is the total charge of the sphere.
02

Find the electric potential outside the sphere

To find the electric potential outside the sphere, we use the relation between electric potential V and the electric field E: V = - \int E dr For our case, we'll find the potential outside the sphere: V = - \int (1 / (4 * \pi * \epsilon_0)) * (Q / r^2) dr Integrating the electric field equation from infinity to r gives: V = (1 / (4 * \pi * \epsilon_0)) * (Q / r)
03

Compare the potential to that of a point charge

The electric potential of a point charge Q is also given by: V = (1 / (4 * \pi * \epsilon_0)) * (Q / r) We can see that the electric potential outside a uniformly charged sphere is equal to the potential of a point charge placed at the center of the sphere.
04

Calculate the potential at the sphere's surface

To find the potential at the surface of the sphere, we replace r with R, the radius of the sphere: V_surface = (1 / (4 * \pi * \epsilon_0)) * (Q / R)
05

Discuss the potential change for a non-uniform charge distribution with spherical symmetry

If the charge distribution is not uniform but has spherical (radial) symmetry, the electric field outside the sphere will still follow the same formula: E = (1 / (4 * \pi * \epsilon_0)) * (Q / r^2) The electric potential formula outside the sphere remains the same: V = (1 / (4 * \pi * \epsilon_0)) * (Q / r) However, the potential at the surface and the electric field inside the sphere will be different and would depend on the specific charge distribution. In conclusion, the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere, equal to the total charge of the sphere. The potential at the surface of the sphere is given by V_surface = (1 / (4 * \pi * \epsilon_0)) * (Q / R). If the charge distribution is not uniform but has spherical symmetry, the potential outside the sphere remains the same, but the potential at the surface and the electric field inside the sphere will be different.

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Most popular questions from this chapter

The electron beam emitted by an electron gun is controlled (steered) with two sets of parallel conducting plates: a horizontal set to control the vertical motion of the beam, and a vertical set to control the horizontal motion of the beam. The beam is emitted with an initial velocity of \(2.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\). The width of the plates is \(d=5.00 \mathrm{~cm}\), the separation between the plates is \(D=4.00 \mathrm{~cm},\) and the distance between the edge of the plates and a target screen is \(L=40.0 \mathrm{~cm}\) In the absence of any applied voltage, the electron beam hits the origin of the \(x y\) -coordinate system on the observation screen. What voltages need to be applied to the two sets of plates for the electron beam to hit a target placed on the observation screen at coordinates \((x, y)=(0 \mathrm{~cm}, 8.00 \mathrm{~cm}) ?\)

Each of the following pairs of charges are separated by a distance \(d\). Which pair has the highest potential energy? a) \(+5 \mathrm{C}\) and \(+3 \mathrm{C}\) d) \(-5 \mathrm{C}\) and \(+3 \mathrm{C}\) b) \(+5 \mathrm{C}\) and \(-3 \mathrm{C}\) e) All pairs have the \(\begin{array}{ll}\text { c) }-5 \mathrm{C} \text { and }-3 \mathrm{C} & \text { same potential energy. }\end{array}\)

An electron is accelerated from rest through a potential difference of \(370 \mathrm{~V}\). What is its final speed?

A point charge \(Q\) is placed a distance \(R\) from the center of a conducting sphere of radius \(a,\) with \(R>a\) (the point charge is outside the sphere). The sphere is grounded, that is, connected to a distant, unlimited source and/or sink of charge at zero potential. (Neither the distant ground nor the connection directly affects the electric field in the vicinity of the charge and sphere.) As a result, the sphere acquires a charge opposite in sign to \(Q\), and the point charge experiences an attractive force toward the sphere. a) Remarkably, the electric field outside the sphere is the same as would be produced by the point charge \(Q\) plus an imaginary mirror-image point charge \(q\), with magnitude and location that make the set of points corresponding to the surface of the sphere an equipotential of potential zero. That is, the imaginary point charge produces the same field contribution outside the sphere as the actual surface charge on the sphere. Calculate the value and location of \(q\). (Hint: By symmetry, \(q\) must lie somewhere on the axis that passes through the center of the sphere and the location of \(Q .)\) b) Calculate the force exerted on point charge \(Q\) and directed toward the sphere, in terms of the original quantities \(Q, R,\) and \(a\) c) Determine the actual nonuniform surface charge distribution on the conducting sphere.

What would be the consequence of setting the potential at \(+100 \mathrm{~V}\) at infinity, rather than taking it to be zero there? a) Nothing; the field and the potential would have the same values at every finite point. b) The electric potential would become infinite at every finite point, and the electric field could not be defined. c) The electric potential everywhere would be \(100 \mathrm{~V}\) higher, and the electric field would be the same. d) It would depend on the situation. For example, the potential due to a positive point charge would drop off more slowly with distance, so the magnitude of the electric field would be less.
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