Question

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to produce an electric field of 1500 N/C just outside the surface of the sphere?

Answer 1

To produce an electric field of 1500 N/C just outside the surface of the plastic sphere, approximately 337.5 nC of excess charge must be distributed uniformly within its volume.

Step-by-step explanation:

To produce an electric field just outside the surface of the sphere, the excess charge on the surface of the sphere should be uniformly distributed within its volume. The electric field just outside a uniformly charged sphere is given by:

E = k * q / r2

Where E is the electric field, k is the Coulomb constant
`(9 * 109 N m2/C^2)` q is the charge, and r is the distance from the center of the sphere to the point outside the surface. In this case, E is given as 1500 N/C and r is half the diameter of the sphere (12.5 cm).

Substituting the values into the equation, we can solve for q:

`1500 N/C = (9 * 109 N m2/C2) * q / (0.125 m)^2`

Solving for q, we find that the excess charge on the surface of the sphere should be approximately
`3.375 x 10-7 C,` or about 337.5 nC.

Answer 2

Required charge
`q=2.6* 10^(9)C`.

`n=1.622* 10^(10)\ electrons`

Step-by-step explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

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

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere =
`(25.0)/(2)=12.5cm=0.125m`

thus, according to the given data

`1500N/C=(9* 10^(9)N* q)/((0.125m)^2)`

or

`q=(0.125^2* 1500)/(9* 10^(9))`

or

Required charge
`q=2.6* 10^(9)C`.

Now,

the number of electrons (n) required will be

`n=(required\ charge)/(charge\ of\ electron)`

or

`n=(2.6* 10^(-9))/(1.602* 10^(-19))`

or

`n=1.622* 10^(10)\ electrons`