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Chapter 24: Q65P (page 714)

What is the excess charge on a conducting sphere of radius r=0.15mif the potential of the sphere is1500V and at infinityV=0 ?

Short Answer

Expert verified

The excess charge on a conducting sphere is Q=2.5×10−8C.

Step by step solution

01

Given data:

The potential of the sphere, V=1500V

The permittivity of free space, ε0=8.85×10−12C2/N⋅m2

The radius of the conducting sphere, r=0.15m

02

Understanding the concept:

Use the formula for the potential inside a sphere. The electric potential due to a sphere of radius ris,

V=14πε0Qr

Hereε0is permittivity of free space,Qis the charge, and ris the radius of the sphere.

Rearrange the equation for the charge on the sphere.

Q=V(4πε0)r

03

Calculate the excess charge on a conducting sphere:

Write the equation for the charge as below.

Q=V(4πε0)r

Substitute known values in the above equation.

Q=(1500V)(4×3.14(8.85×10−12C2/N⋅m2))(0.15m)=2.5×10−8C

Hence, the excess charge on the sphere is 2.5×10−8C .

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Most popular questions from this chapter

The chocolate crumb mystery. This story begins with Problem 60 in Chapter 23. (a) From the answer to part (a) of that problem, find an expression for the electric potential as a function of the radial distance r from the center of the pipe. (The electric potential is zero on the grounded pipe wall.) (b) For the typical volume charge density ρ=-1.1×10-3C/m3 , what is the difference in the electric potential between the pipe’s center and its inside wall? (The story continues with Problem 60 in Chapter 25.)

An infinite nonconducting sheet has a surface charge density isσ=0.20μC/m2on one side. How far apart are equipotential surfaces whose potentials differ by 50 V?

Question: How much work is required to set up the arrangement of Fig. 24-52 if, q =2.30 pC, a = 64.0 cm and the particles are initially infinitely far apart and at rest?

Question: In Fig. 24-41a, a particle of elementary charge +eis initially at coordinate z = 20 nmon the dipole axis (here a zaxis) through an electric dipole, on the positive side of the dipole. (The origin of zis at the center of the dipole.) The particle is then moved along a circular path around the dipole center until it is at coordinate z = -20 nm, on the negative side of the dipole axis. Figure 24-41bgives the work done by the force moving the particle versus the angle u that locates the particle relative to the positive direction of the z-axis. The scale of the vertical axis is set byWas=4.0×10-30J.What is the magnitude of the dipole moment?

A uniform charge of 16.0μCis on a thin circular ring lying in xy plane and centered on the origin. The ring’s radius is3.00cm. If point A is at the origin and point B is on the z-axis at z=4.00cm, what isVB-VA?
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