Question

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of −2.96 μC. Find the electric field at the following radii from the center of this charge configuration. (a) r = 1.00 cm magnitude:_________ N/C

direction:____________
(b) r = 3.00 cm magnitude:____________ N/C
direction:_________________
(c) r = 4.50 cm magnitude:_____________ N/C direction:______________
(d) r = 7.00 cm magnitude:_____________ N/C
direction:______________

Answer 1

Step-by-step explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

`E=(kq)/(r^2)`

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

`E=(kq)/(r^2)`

Put the value into the formula

`E=(9*10^(9)*6.88*10^(-6))/((3.00*10^(-2))^2)`

`E=6.88*10^(7)\ N/C`

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

`E=(kq)/(r^2)`

Put the value into the formula

`E=(9*10^(9)*(-2.96*10^(-6)))/((7.00*10^(-2))^2)`

`E=5.43*10^(6)\ N/C`

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.