Question

A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 21.0 cm from this surface, the potential is 499 V. Calculate the radius of the sphere. Submit Answer Tries 0/32 Determine the total charge on the sphere. Submit Answer Tries 0/100 What is the electric potential inside the sphere at a radius of 4.0 cm? Submit Answer Tries 0/32 Calculate the magnitude of the electric field at the surface of the sphere. Submit Answer Tries 0/32 If an electron starts from rest at a distance of 21.0 cm from the surface of the sphere, calculate the electron's speed when it reaches the sphere's surface.

Answer 1

a

The radius is
`r_1 = 0.315m`

b

The total charge is
`Q= 2.912*10^(-8)C`

c

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

d

The magnitude of the electric field is
`E= 2641.3 V/m`

e

The velocity is
`v= 1.76 *10^(14) m/s`

Step-by-step explanation:

From the question we are told that

The potential is
`V_1 = 832 V`

The radial distance from the sphere is
`d = 21.0cm = (21)/(100) = 0.21m`

The potential at the radial distance is
`V_2 = 499V`

The potential at the surface of the sphere is mathematically represented as

`V = (kQ)/(r)`

`Vr = kQ`

Where kQ is a constant what this means that the the charge Q and the coulomb constant do not change

This means that

`V_1 r_1 = V_2 r_2`

Where
`r_1` is the radius of the sphere

and
`r_2` is the distance from that point where the second potential was measured to the center of the sphere which is mathematically represented as

`r_2 = r_1 + d`

Substituting this into the equation

`v_1 r_1 = V_2 (r_1 +d)`

Now substituting value

`832 * r_1 = 499 * (r_1 + 0.21)`

`832r_1 - 499r_1 = 104.79`

`333r_1 = 104.79`

`r_1 = (104.79)/(333)`

`r_1 = 0.315m`

From the equation above

`V = (kQ)/(r_1)`

making Q the subject

`Q = (V r_1 )/(k)`

k has a values of
`k = 9*10^9 \ kg\cdot m^3 \cdot s^(-4) \cdot A^(-2)`

Substituting into the equation

`Q =(832 * 0.315)/(9*10^9)`

`Q= 2.912*10^(-8)C`

According to Gauss law the electric field from outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

`E = (kQ)/(r_1^2)`

Substituting values

`E = (9*10^(9) * 2.912*10^(-8))/(0.315^2)`

`E= 2641.3 V/m`

According the the law of energy conservation

The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

`EPE = V * e`

Where is e is an electron

And Kinetic energy is mathematically represented as

`KE = (1)/(2) m v^2`

From the statement above

`V_2 e = V_1 e + (mv^2)/(2)`

But from the question we can deduce that the potential at the surface is zero

So the equation becomes

`V_2 e = (mv^2)/(2)`

The charge an electron has a value
`e = 1.602*10^(-19)C`

And the mass of an electron is
`m = 9.109 *10^(-31)kg`

Making v the subject

`v = \sqrt{(2V_2 e)/(m) }`

Substituting value

`v = \sqrt{(2 * 499 * 1.602 *10^(-19))/(9.109*10^(-31)) }`

`v= 1.76 *10^(14) m/s`