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Chapter 22: Problem 50

Consider a hollow spherical conductor with total charge \(+5 e\). The outer and inner radii are \(a\) and \(b\), respectively. (a) Calculate the charge on the sphere's inner and outer surfaces if a charge of \(-3 e\) is placed at the center of the sphere. (b) What is the total net charge of the sphere?

Short Answer

Expert verified
#Answer# The charge distribution on the inner surface of the sphere is +3e, and on the outer surface is +2e. The total net charge on the sphere remains +5e.

Step by step solution

01

Apply Gauss' Law for the inner surface of the sphere

Gauss' Law states that the enclosed electric charge within a closed surface is proportional to the total electric flux through the surface. Mathematically, it's expressed as: $$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_\text{enclosed}}{\varepsilon_0}$$ When a charge of -3e is placed at the center of the sphere, the charge on the inner surface of the sphere \((\)radius \(b)\) should be equal and opposite to the charge placed at the center to maintain equilibrium. Thus, the charge on the inner surface of the shell, \(Q_1\), will be +3e.
02

Applying conservation of charge

The total charge on the sphere is given to be +5e. Since the charge on the inner surface is +3e, we can find the charge on the outer surface, \(Q_2\), using the conservation of charge principle: $$Q_\text{total} = Q_1 + Q_2$$ Plugging in the values: $$5e = 3e + Q_2$$
03

Solving for the charge on the outer surface

Now, just solve the equation for \(Q_2\): $$Q_2 = 5e - 3e = 2e$$ The charge on the outer surface of the sphere is +2e.
04

Calculate the total net charge on the sphere

As we have found the charge on both inner and outer surfaces of the sphere, calculating the total net charge on the sphere is straightforward. We simply sum the charge on both surfaces: $$Q_\text{net} = Q_1 + Q_2$$ Plugging in the values: $$Q_\text{net} = 3e + 2e = 5e$$ The total net charge on the sphere is +5e.

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Most popular questions from this chapter

A solid nonconducting sphere has a volume charge distribution given by \(\rho(r)=(\beta / r) \sin (\pi r / 2 R) .\) Find the total charge contained in the spherical volume and the electric field in the regions \(rR\). Show that the two expressions for the electric field equal each other at \(r=R\).

A solid metal sphere of radius \(8.00 \mathrm{~cm},\) with a total charge of \(10.0 \mu C\), is surrounded by a metallic shell with a radius of \(15.0 \mathrm{~cm}\) carrying a \(-5.00 \mu \mathrm{C}\) charge. The sphere and the shell are both inside a larger metallic shell of inner radius \(20.0 \mathrm{~cm}\) and outer radius \(24.0 \mathrm{~cm} .\) The sphere and the two shells are concentric. a) What is the charge on the inner wall of the larger shell? b) If the electric field outside the larger shell is zero, what is the charge on the outer wall of the shell?

There is a uniform charge distribution of \(\lambda=\) \(8.00 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\) along a thin wire of length \(L=6.00 \mathrm{~cm}\) The wire is then curved into a semicircle that is centered about the origin, so the radius of the semicircle is \(R=L / \pi .\) Find the magnitude of the electric field at the center of the semicircle.

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside the surface of the wire. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.81 \lambda\) and radius \(6.5 r ?\)

A solid nonconducting sphere of radius \(a\) has a total charge \(+Q\) uniformly distributed throughout its volume. The surface of the sphere is coated with a very thin (negligible thickness) conducting layer of gold. A total charge of \(-2 Q\) is placed on this conducting layer. Use Gauss's Law to do the following. a) Find the electric field \(E(r)\) for \(ra\) (outside the coated sphere, beyond the sphere and the gold layer). c) Sketch the graph of \(E(r)\) versus \(r\). Comment on the continuity or discontinuity of the electric field, and relate this to the surface charge distribution on the gold layer.
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