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Chapter 22: Problem 34

A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials and (b) the electric field strengths at their surfaces.

Short Answer

Expert verified
The potentials at the surfaces of two spheres are the same, and the electric field strength at the surface of the larger sphere is a third of that at the surface of the smaller sphere.

Step by step solution

01

Find Potential for Each Sphere

The potential \(V\) at the surface of a sphere can be found using the formula \(V = K_e \frac{q}{r}\), where \(K_e\) is Coulomb's constant, \(q\) is the charge on the sphere, and \(r\) is the radius of the sphere. For the smaller sphere, let its charge be \(q\) and radius be \(r\). Thus, its potential \(V_1\) is \(K_e \frac{q}{r}\). For the larger sphere, its charge is three times the smaller sphere (\(3q\)), and its radius is three times the smaller sphere (\(3r\)). The potential \(V_2\) at its surface is \(K_e \frac{3q}{3r} = K_e \frac{q}{r}\), which is equal to \(V_1\). So, the potentials at the surfaces of both spheres are the same.
02

Find Electric Field Strength for Each Sphere

The electric field \(E\) at the surface of a sphere can be found using the formula \(E = K_e \frac{q}{r^2}\), where \(K_e\) is Coulomb's constant, \(q\) is the charge on the sphere, and \(r\) is the radius of the sphere. For the smaller sphere, the electric field \(E_1\) is \(K_e \frac{q}{r^2}\). For the larger sphere, its charge is three times the smaller sphere (\(3q\)), and its radius is three times the smaller sphere (\(3r\)). The electric field \(E_2\) at its surface is \(K_e \frac{3q}{(3r)^2} = K_e \frac{q}{3r^2}\), which is one third of \(E_1\). So, the electric field strength at the surface of the larger sphere is one third of that on the surface of the smaller sphere.

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Most popular questions from this chapter

Can equipotential surfaces intersect? Explain.

You're sizing a new electric transmission line, and you can save money with thinner wire. The potential difference between the line and the ground, \(60 \mathrm{m}\) below, is \(115 \mathrm{kV}\). The field at the wire surface cannot exceed \(25 \%\) of the 3 -MV/m breakdown field in air. Neglecting charges in the ground itself, what minimum wire diameter do you specify? (Hint. You'll have to do a numerical calculation.)

A 5.0 -g object carries \(3.8 \mu \mathrm{C}\). It acquires speed \(v\) when accelerated from rest through a potential difference \(V\). If a 2.0 -g object acquires twice the speed under the same circumstances, what's its charge?

A conducting sphere \(5.0 \mathrm{cm}\) in radius carries \(60 \mathrm{nC} .\) It's surrounded by a concentric spherical conducting shell of radius \(15 \mathrm{cm}\) carrying \(-60 \mathrm{nC}\). (a) Find the potential at the sphere's surface, taking \(V=0\) at infinity. (b) Repeat for the case when the shell carries \(+60 \mathrm{nC}\)

An electric field is given by \(\vec{E}=E_{0} \hat{\jmath},\) where \(E_{0}\) is a constant. Find the potential as a function of position, taking \(V=0\) at \(y=0\)
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