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For both a conducting sphere and an insulating sphere, the external field depends only on the total charge, whether it is distributed on the surface (conducting sphere) or throughout the interior (insulating sphere).

>Model

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Iframe

>Top

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Concept

>Top

In the case of a spherical Gaussian surface, the electric field ($\vec{E}$) is constant in the direction of the versor normal to the section ($\hat{n}$). Therefore, using the charge ($Q$), the electric field constant ($\epsilon_0$), and the dielectric constant ($\epsilon$), it can be calculated by integrating over the surface where the electric field is constant ($dS$):

Since the surface area of the surface of a sphere ($S$) is equal to the pi ($\pi$) and the disc radius ($r$), we have:

what is shown in the graph

Outside the sphere, the electric field, sphere, outer ($E_e$) with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the distance between charges ($r$) is equal to:

While in the case of an insulating sphere, the electric field, sphere, interior ($E_i$) with the sphere radius ($R$) is:

If the sphere is conductive, the charges will distribute over the surface, and the electric field, sphere, interior ($E_i$) will be zero.

ID:(11839, 0)

Concept

>Top

As the potential difference is the reference electrical, sphere, outer ($\varphi_e$) with the electric field, sphere, outer ($E_e$), the electric field, sphere, interior ($E_i$), the sphere radius ($R$), and the radius ($r$), we get:

Given that the electric field, sphere, outer ($E_e$) with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the distance between charges ($r$) is equal to:

and that the electric field, sphere, interior ($E_i$) with the internal radius ($r_i$) is equal to:

In spherical coordinates, we have:

$\varphi_e = -\displaystyle\int_0^R du \displaystyle\frac{ Q u }{4 \pi \epsilon_0 \epsilon R ^3 } -\displaystyle\int_R^r du \displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon u ^2 }= -\displaystyle\frac{ 1 }{ 4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r }$

Therefore, the reference electrical, sphere, outer ($\varphi_e$) results in:

As illustrated in the following graph:

the field at two points must have the same energy. Therefore, the variables the charge ($Q$), the particle mass ($m$), the speed 1 ($v_1$), the speed 2 ($v_2$), and the electric potential 1 ($\varphi_1$) according to the equation:



and the electric potential 2 ($\varphi_2$), according to the equation:



must satisfy the following relationship:

ID:(11846, 0)

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Parameters

$\epsilon$

epsilon

Dielectric constant

-

$\epsilon_0$

epsilon_0

Electric field constant

C^2/m^2N

$m$

m

Particle mass

kg

$\pi$

pi

Pi

rad

$R$

R

Sphere radius

m

Variables

$Q$

Q

Charge

C

$E_i$

E_i

Electric field, sphere, interior

V/m

$E_e$

E_e

Electric field, sphere, outer

V/m

$\varphi_1$

phi_1

Electric potential 1

V

$\varphi_2$

phi_2

Electric potential 2

V

$r_1$

r_1

Radius 1

m

$r_2$

r_2

Radius 2

m

$v_1$

v_1

Speed 1

m/s

$v_2$

v_2

Speed 2

m/s

$q$

q

Test charge

C

Calculations

• Alternative method
• Traditional method
• Strategy
• 2D
• 3D

Equation

Number

First, select the equation: to , then, select the variable: to

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Calculations

Symbol

Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable Given Calculate Target : Equation To be used

Equations

ID:(15807, 0)

Equation

>Top, >Model

The electric field, sphere, interior ($E_i$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the sphere radius ($R$) and the distance between charges ($r$) is equal to:

$E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r_1 }{ R ^3 }$

$E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$

Conditions

Variables

$Q$

Charge

$C$

5459

$\epsilon$

Dielectric constant

$-$

5463

$r$

$r_1$

Radius 1

$m$

10390

$\epsilon_0$

Electric field constant

8.854187e-12

$C^2/m^2N$

5462

$E_i$

Electric field, sphere, interior

$V/m$

8530

$\pi$

Pi

3.1415927

$rad$

5057

$R$

Sphere radius

$m$

8541

Relation

Development

Considering a spherical Gaussian surface, the electric field is constant. Therefore, the electric eield ($E$) is equal to the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the surface of the conductor ($S$) as shown by:

$E S = \displaystyle\frac{ Q }{ \epsilon_0 \epsilon }$

Since the surface area of the surface of a sphere ($S$) is equal to the pi ($\pi$) and the disc radius ($r$), we have:

$S = 4 \pi r ^2$

The charge enclosed by the Gaussian surface, with the encapsulated charge on Gauss surface ($q$), the sphere radius ($R$), and the distance between charges ($r$), is given by:

$q =\displaystyle\frac{ r ^3}{ R ^3} Q$

Therefore, the electric field, sphere, interior ($E_i$) results in:

$E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$

ID:(11447, 0)

Equation

>Top, >Model

The electric field, sphere, outer ($E_e$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$) and the distance between charges ($r$) is equal to:

$E_e=\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r_2 ^2 }$

$E_e=\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r ^2 }$

Conditions

Variables

$Q$

Charge

$C$

5459

$\epsilon$

Dielectric constant

$-$

5463

$r$

$r_2$

Radius 2

$m$

10391

$\epsilon_0$

Electric field constant

8.854187e-12

$C^2/m^2N$

5462

$E_e$

Electric field, sphere, outer

$V/m$

8529

$\pi$

Pi

3.1415927

$rad$

5057

Relation

Development

In the case of a spherical Gaussian surface, the electric field is constant, so the electric eield ($E$) can be calculated using the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the surface of the conductor ($S$), resulting in:

$E S = \displaystyle\frac{ Q }{ \epsilon_0 \epsilon }$

Given that the surface of a sphere ($S$) is equal to the pi ($\pi$) and the disc radius ($r$), we obtain:

$S = 4 \pi r ^2$

Finally, the electric field, sphere, outer ($E_e$) together with the distance between charges ($r$) is equal to:

$E_e=\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r ^2 }$

ID:(11446, 0)

Equation

>Top, >Model

The reference electrical, insulating sphere, inner ($\varphi_i$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the distance between charges ($r$) and the sphere radius ($R$) is equal to:

$\varphi_1 = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r_1 ^2 }{ R ^3 }$

$\varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$

Conditions

Variables

$Q$

Charge

$C$

5459

$\epsilon$

Dielectric constant

$-$

5463

$r$

$r_1$

Radius 1

$m$

10390

$\epsilon_0$

Electric field constant

8.854187e-12

$C^2/m^2N$

5462

$\pi$

Pi

3.1415927

$rad$

5057

$\varphi_i$

$\varphi_1$

Electric potential 1

$V$

10392

$R$

Sphere radius

$m$

8541

Relation

Development

As the potential difference is the reference electrical, insulating sphere, inner ($\varphi_i$) with the electric field, sphere, interior ($E_i$) and the radius ($r$), we get:

$\varphi_i = -\displaystyle\int_0^ r du\, E_i$

Given that the electric field, sphere, interior ($E_i$) with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the sphere radius ($R$), and the distance between charges ($r$) is equal to:

$E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r_1 }{ R ^3 }$

In spherical coordinates, this is:

$\varphi_i = -\displaystyle\int_0^{r} du \displaystyle\frac{ Q u }{4 \pi \epsilon_0 \epsilon R ^3 }= -\displaystyle\frac{ Q }{ 8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$

Therefore, the reference electrical, insulating sphere, inner ($\varphi_i$) with the distance between charges ($r$) results in:

$\varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$

ID:(11583, 0)

Equation

>Top, >Model

The reference electrical, sphere, outer ($\varphi_e$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$) and the distance between charges ($r$) is equal to:

$\varphi_2 = -\displaystyle\frac{ 1 }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r_2 }$

$\varphi_e = -\displaystyle\frac{ 1 }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r }$

Conditions

Variables

$Q$

Charge

$C$

5459

$\epsilon$

Dielectric constant

$-$

5463

$r$

$r_2$

Radius 2

$m$

10391

$\epsilon_0$

Electric field constant

8.854187e-12

$C^2/m^2N$

5462

$\pi$

Pi

3.1415927

$rad$

5057

$\varphi_e$

$\varphi_2$

Electric potential 2

$V$

10393

Relation

Development

As the potential difference is the reference electrical, sphere, outer ($\varphi_e$) with the electric field, sphere, outer ($E_e$), the electric field, sphere, interior ($E_i$), the sphere radius ($R$), and the radius ($r$), we get:

$\varphi_e = - \displaystyle\int_0^ R du\, E_i - \displaystyle\int_ R ^ r du\, E_e$

Given that the electric field, sphere, outer ($E_e$) with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the distance between charges ($r$) is equal to:

$E_e=\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r_2 ^2 }$

and that the electric field, sphere, interior ($E_i$) with the internal radius ($r_i$) is equal to:

$E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r_1 }{ R ^3 }$

In spherical coordinates, we have:

$\varphi_e = -\displaystyle\int_0^R du \displaystyle\frac{ Q u }{4 \pi \epsilon_0 \epsilon R ^3 } -\displaystyle\int_R^r du \displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon u ^2 }= -\displaystyle\frac{ 1 }{ 4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r }$

Therefore, the reference electrical, sphere, outer ($\varphi_e$) results in:

$\varphi_e = -\displaystyle\frac{ 1 }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r }$

ID:(11584, 0)

Equation

>Top, >Model

Electric potentials, which represent potential energy per unit of charge, influence how the velocity of a particle varies. Consequently, due to the conservation of energy between two points, it follows that in the presence of variables the charge ($q$), the particle mass ($m$), the speed 1 ($v_1$), the speed 2 ($v_2$), the electric potential 1 ($\varphi_1$), and the electric potential 2 ($\varphi_2$), the following relationship must be satisfied:

$\displaystyle\frac{1}{2} m v_1 ^2 + q \varphi_1 = \displaystyle\frac{1}{2} m v_2 ^2 + q \varphi_2$

Conditions

Variables

$\varphi_1$

Electric potential 1

$V$

10392

$\varphi_2$

Electric potential 2

$V$

10393

$m$

Particle mass

$kg$

5516

$v_1$

Speed 1

$m/s$

8562

$v_2$

Speed 2

$m/s$

8563

$q$

Test charge

$C$

8746

Relation

Development

ID:(11596, 0)