Field of force

q

{\displaystyle q}

) should have units of force*
*Electric field (

E

{\displaystyle E}

) should have units of force per area (the same as stress)*
*Electric constant (

ϵ
0

{\displaystyle \epsilon\_0}

) should have units of force per area (the same as bulk modulus)*
*Magnetic field (

B

{\displaystyle B}

) should have units of force per velocity per time^2*
*Magnetic constant (

μ
0

{\displaystyle \mu\_0}

) should have units of 1/(force per time^2)*
See this for more details

In physics a **field of force** is a vector field that describes a non-contact force acting on a particle at various positions in space. Specifically, a field of force is a vector field

F
→

{\displaystyle \vec F}

, where

F
→

(

x
→

)

{\displaystyle \vec{F}(\vec{x})}

 is the force that a particle would feel if it were at the point

x
→

{\displaystyle \vec{x}}

.

## Examples

* Gravity is the force of attraction between two objects. In Newtonian gravity, a particle of mass *M* creates a gravitational field

g
→

=

−
G
M

r
2

r
^

{\displaystyle \vec{g}=\frac{-G M}{r^2}\hat{r}}

, where the radial unit vector

r
^

{\displaystyle \hat{r}}

 points away from the particle. The gravitational force experienced by a particle of light mass *m*, close to the surface of Earth is given by

F
→

=
m

g
→

{\displaystyle \vec{F} = m \vec{g}}

, where *g* is the standard gravity. A gravitational field of force is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body.
* An electric field

E
→

{\displaystyle \vec{E} }

 is a vector field. It exerts a force on a point charge *q* given by

F
→

=
q

E
→

{\displaystyle \vec{F} = q\vec{E}}

.

## Work

Work is dependent on the displacement as well as the force acting on an object. As a particle moves through a field of force along a path *C*, the work done by the force is a line integral

# W

∫
C

F
→

⋅
d

r
→

{\displaystyle W = \int\_C \vec{F} \cdot d\vec{r}}

This value is independent of the velocity/momentum that the particle travels along the path.

### Conservative field of force

For a conservative field of force, it is also independent of the path itself, depending only on the starting and ending points. Therefore, the work for an object travelling in a closed path is zero, since its starting and ending points are the same:

∮
C

F
→

⋅
d

r
→

=
0

{\displaystyle \oint\_C \vec{F} \cdot d\vec{r} = 0}

If the field is conservative, the work done can be more easily evaluated by realizing that a conservative vector field can be written as the gradient of some scalar potential function:

F
→

=
∇
ϕ

{\displaystyle \vec{F} = \nabla \phi}

The work done is then simply the difference in the value of this potential in the starting and end points of the path. If these points are given by *x* = *a* and *x* = *b*, respectively:

# W
ϕ
(
b
)
−
ϕ
(
a
)

{\displaystyle W = \phi(b) - \phi(a) }

## Similarity of electric and gravitational fields

Coulomb's law, which describes the interaction of electric charges:

F
e

=
q

(

1

4
π

r

2

1

ε
0

Q

)

r
^

=
q

E

{\displaystyle
\mathbf{F\_e}=q
\left(
\frac{1}{4 \pi \mathbf{r}^2}
\frac{1}{\varepsilon\_0} Q
\right) \mathbf{\hat{r}} =
q \mathbf{E}
}

is similar to Newton's law of universal gravitation:

F
g

=
m

(

1

4
π

r

2

1

γ
0

M

)

r
^

=
m

g

{\displaystyle
\mathbf{F\_g}=m
\left(\frac{1}{4 \pi \mathbf{r}^2}
\frac{1}{\gamma\_0} M
\right) \mathbf{\hat{r}} =
m\mathbf{g}
}

(where

r
^

=

r

r

{\displaystyle \mathbf{\hat{r}}=\mathbf{\frac{r}{|r|}}}

 and

γ
0

=

1

4
π
G

{\displaystyle \gamma\_0 =
\frac{1}{4 \pi G}}

 surface area of a sphere =

4
π

r
2

{\displaystyle 4\pi r^2}

).

**Fe** is positive when q and Q are the same sign.

**Fg** is negative when m and M are the same sign therefore gravitational mass is imaginary.

The energy in the electric field is proportional to E2 and is positive energy.

The energy in the gravitational field is proportional to g2 and is negative energy therefore the gravitational field is imaginary.

## Energy in the electric field

From Wikipedia: Electric field

The total energy per unit volume stored by the  electromagnetic field is

u

E
M

=

ε
2

E

2

1

2
μ

B

2

where *ε* is the  permittivity of the medium in which the field exists,

μ

{\displaystyle \mu}

 its  magnetic permeability, and **E** and **B** are the electric and magnetic field vectors.

As **E** and **B** fields are coupled, it would be misleading to split this expression into "electric" and "magnetic" contributions. However, in the steady-state case, the fields are no longer coupled (see Maxwell's equations). It makes sense in that case to compute the electrostatic energy per unit volume:

u

E
S

=

1
2

ε

E

2

,

{\displaystyle u\_{ES} = \frac{1}{2} \varepsilon |\mathbf{E}|^2 \, ,}

The total energy *U* stored in the electric field in a given volume *V* is therefore

U

E
S

=

1
2

ε

∫

V

E

2

d

V

,

{\displaystyle U\_{ES} = \frac{1}{2} \varepsilon \int\_{V} |\mathbf{E}|^2 \, \mathrm{d}V \, ,}

## Gravitational binding energy

From Wikipedia:Gravitational binding energy

The **gravitational binding energy** of a system is the minimum energy which must be added to it in order for the system to cease being in a gravitationally bound state. A gravitationally bound system has a lower (*i.e.*, more negative) gravitational potential energy than the sum of the energies of its parts when these are completely separated—this is what keeps the system aggregated in accordance with the minimum total potential energy principle.

For a spherical body of uniform density, the gravitational binding energy *U* is given by the formula

# U

3
G

M
2

5
R

{\displaystyle U = \frac{3GM^2}{5R}}

where *G* is the gravitational constant, *M* is the mass of the sphere, and *R* is its radius.

Assuming that the Earth is a sphere of uniform density (which it is not, but is close enough to get an order-of-magnitude estimate) with *M* = 5.97×1024 kg and *r* = 6.37×106m, then *U* = 2.24×1032 J. This is roughly equal to one week of the Sun's total energy output. It is 37.5 MJ/kg, 60% of the absolute value of the potential energy per kilogram at the surface.

The actual depth-dependence of density, inferred from seismic travel times (see Adams–Williamson equation), is given in the Preliminary Reference Earth Model (PREM). Using this, the real gravitational binding energy of Earth can be calculated numerically as *U* = 2.49×1032 J.

According to the virial theorem, the gravitational binding energy of a star is about two times its internal thermal energy in order for hydrostatic equilibrium to be maintained. As the gas in a star becomes more relativistic, the gravitational binding energy required for hydrostatic equilibrium approaches zero and the star becomes unstable (highly sensitive to perturbations), which may lead to a supernova in the case of a high-mass star due to strong radiation pressure or to a black hole in the case of a neutron star.

### Derivation for a uniform sphere

The gravitational binding energy of a sphere with radius

R

{\displaystyle R}

 is found by imagining that it is pulled apart by successively moving spherical shells to infinity, the outermost first, and finding the total energy needed for that.

Assuming a constant density

ρ

{\displaystyle \rho}

, the masses of a shell and the sphere inside it are:

m

s
h
e
l
l

=
4
π

r

2

ρ

d
r

{\displaystyle m\_\mathrm{shell}=4\pi r^{2}\rho\,dr}

      and

m

i
n
t
e
r
i
o
r

=

4
3

π

r
3

ρ

{\displaystyle m\_\mathrm{interior}=\frac{4}{3}\pi r^3 \rho}

The required energy for a shell is the negative of the gravitational potential energy:

d
U

=
−
G

m

s
h
e
l
l

m

i
n
t
e
r
i
o
r

r

{\displaystyle {\it dU}=-G\frac{m\_\mathrm{shell} m\_\mathrm{interior}}{r}}

Integrating over all shells yields:

# U
−
G

∫
0
R

(
4
π

r
2

ρ
)
(

4
3

π

r

3

ρ
)

r

d
# r
−
G

16
3

π
2

ρ
2

∫
0
R

r
4

d
# r
−
G

16
15

π

2

ρ

2

R
5

{\displaystyle U = -G\int\_0^R {\frac{(4\pi r^2\rho)(\tfrac{4}{3}\pi r^{3}\rho)}{r}} dr = -G{\frac{16}{3}}\pi^2 \rho^2 \int\_0^R {r^4} dr = -G{\frac{16}{15}}{\pi}^2{\rho}^2 R^5}

Since

ρ

{\displaystyle \rho}

 is simply equal to the mass of the whole divided by its volume for objects with uniform density, therefore

# ρ

M

4
3

π

R
3

{\displaystyle \rho=\frac{M}{\frac{4}{3}\pi R^3}}

And finally, plugging this into our result leads to

# U
−
G

16
15

π
2

R
5

(

M

4
3

π

R
3

)

2

=
−

3
G

M
2

5
R

{\displaystyle U=-G\frac{16}{15} \pi^2 R^5 \left(\frac{M}{\frac{4}{3}\pi R^3}\right)^2= -\frac{3GM^2}{5R}}

# U
−

3
G

M
2

5
R

{\displaystyle U = -\frac{3GM^2}{5R}}

## See also

* Field line
* Force