Respuesta :
Answer:
a
The radius is [tex]r_1 = 0.315m[/tex]
b
The total charge is [tex]Q= 2.912*10^{-8}C[/tex]
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 [tex]E= 2641.3 V/m[/tex]
e
The velocity is [tex]v= 1.76 *10^{14} m/s[/tex]
Explanation:
From the question we are told that
The potential is [tex]V_1 = 832 V[/tex]
The radial distance from the sphere is [tex]d = 21.0cm = \frac{21}{100} = 0.21m[/tex]
The potential at the radial distance is [tex]V_2 = 499V[/tex]
The potential at the surface of the sphere is mathematically represented as
[tex]V = \frac{kQ}{r}[/tex]
[tex]Vr = kQ[/tex]
Where kQ is a constant what this means that the the charge Q and the coulomb constant do not change
This means that
[tex]V_1 r_1 = V_2 r_2[/tex]
Where [tex]r_1[/tex] is the radius of the sphere
and [tex]r_2[/tex] is the distance from that point where the second potential was measured to the center of the sphere which is mathematically represented as
[tex]r_2 = r_1 + d[/tex]
Substituting this into the equation
[tex]v_1 r_1 = V_2 (r_1 +d)[/tex]
Now substituting value
[tex]832 * r_1 = 499 * (r_1 + 0.21)[/tex]
[tex]832r_1 - 499r_1 = 104.79[/tex]
[tex]333r_1 = 104.79[/tex]
[tex]r_1 = \frac{104.79}{333}[/tex]
[tex]r_1 = 0.315m[/tex]
From the equation above
[tex]V = \frac{kQ}{r_1}[/tex]
making Q the subject
[tex]Q = \frac{V r_1 }{k}[/tex]
k has a values of [tex]k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}[/tex]
Substituting into the equation
[tex]Q =\frac{832 * 0.315}{9*10^9}[/tex]
[tex]Q= 2.912*10^{-8}C[/tex]
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
[tex]E = \frac{kQ}{r_1^2}[/tex]
Substituting values
[tex]E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}[/tex]
[tex]E= 2641.3 V/m[/tex]
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
[tex]EPE = V * e[/tex]
Where is e is an electron
And Kinetic energy is mathematically represented as
[tex]KE = \frac{1}{2} m v^2[/tex]
From the statement above
[tex]V_2 e = V_1 e + \frac{mv^2}{2}[/tex]
But from the question we can deduce that the potential at the surface is zero
So the equation becomes
[tex]V_2 e = \frac{mv^2}{2}[/tex]
The charge an electron has a value [tex]e = 1.602*10^{-19}C[/tex]
And the mass of an electron is [tex]m = 9.109 *10^{-31}kg[/tex]
Making v the subject
[tex]v = \sqrt{\frac{2V_2 e}{m} }[/tex]
Substituting value
[tex]v = \sqrt{\frac{2 * 499 * 1.602 *10^{-19}}{9.109*10^{-31}} }[/tex]
[tex]v= 1.76 *10^{14} m/s[/tex]
