Physics, 15.04.2020 02:00 edwinadidas9938

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.

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Answer from: nyny679

The radius is r_1 = 0.315m

The total charge is $Q= 2.912*10^{-8}C$

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

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

The velocity is $v= 1.76 *10^{14} m/s$

Explanation:

From the question we are told that

The potential is V_1 = 832 V

The radial distance from the sphere is $d = 21.0cm = \frac{21}{100} = 0.21m$

The potential at the radial distance is V_2 = 499V

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

$V = \frac{kQ}{r}$

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

$r_1 = \frac{104.79}{333}$

r_1 = 0.315m

From the equation above

$V = \frac{kQ}{r_1}$

making Q the subject

$Q = \frac{V r_1 }{k}$

k has a values of $k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}$

Substituting into the equation

$Q =\frac{832 * 0.315}{9*10^9}$

$Q= 2.912*10^{-8}C$

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

$E = \frac{kQ}{r_1^2}$

Substituting values

$E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}$

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

$KE = \frac{1}{2} m v^2$

From the statement above

$V_2 e = V_1 e + \frac{mv^2}{2}$

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

So the equation becomes

$V_2 e = \frac{mv^2}{2}$

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

$v = \sqrt{\frac{2V_2 e}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 499 * 1.602 *10^{-19}}{9.109*10^{-31}} }$

$v= 1.76 *10^{14} m/s$

Answer from: Quest

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Answer from: Quest

explanation:

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