Question

Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m from the center of the shells. The outer surface of the larger shell has aradius of 3.75 m. If the inner shell contains an excess charge of -5.30 μC, find the amount of chargeon the outer surface of the larger shell.

Answer 1

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

 The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

$E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

Where,

$\epsilon_0 =$ permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

$E= 49.000N/C$

$r= 4.1m$

$\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}$

Replacing we have,

$E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

$49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}$

$q= 9.16*10^{-5} C$

$q= 91.6\mu C$

Therefore the amoun of charge on the outer surface of the larger shell is $91.6 \mu C$

Answer 2

The amount of charge on the outer surface of the larger shell is:

• q= 91.6uC

What is an Electric Field?

This refers to the vector field which contains a unit charge which acts on a test charge and is measured in Coulomb.

Mathematically,

eo= permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

E= 49N/C

r= 4.1m

e0=8.854 x 10 -12^2N^-1m^-2

Therefore the amount of charge on the outer surface of the larger shell is

• q= 91.6uC

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