A field exist when one object is able to exert a force on another object at a distance.

Electric fields | Gravitational fields |
---|---|

An electrostatic force exists between charged objects | A gravitational force exists between objects |

Associated with positive and negative charge | A gravitational field is associated with each mass. It acts on other masses in the field |

When the two objects have the same charge, the force between them is repulsive. When two objects have opposite charge, the force between them is attractive. | Gravity is always an attractive force. |

Electric fields | Gravitational fields |
---|---|

\begin{equation} electric\ field\ strenght=\frac{force\ acting\ on\ a\ positive\ charge}{magnitude\ of\ test\ charge} \end{equation} | \begin{equation} gravitational\ field\ strength=\frac{force\ acting\ on\ test\ mass}{magnitude\ of\ test\ mass} \end{equation} |

The direction of the field is the same as the direction of the force acting on a positive charge. | Gravitational force is always attractive. |

Measured in N C^{-1} |
Measured in N kg^{-1} |

Electric fields | Gravitational fields |
---|---|

Potential difference | Potential energy |

\begin{gather} V=\frac{W}{Q}\\ V=potential\ difference,\ W=work\ done,\ Q=charge \notag \end{gather} | \begin{gather} \Delta E_{P} =mg\Delta h\\ \Delta E_{P} =change\ in\ gravitational\ potential\ energy,m=mass, \notag\\ \Delta h=change\ in\ height,\ g=gravitational\ field\ strenght\ \ \notag \end{gather} |

Measured in volts. | Applies only when g is relatively constant over the height change. |

Field lines visualise the shape of the electric field that arises from static charges.

The electric field between two parallel plates: is uniform betweeen the plates and becomes weaker at the edges (**edge effects**).

The strenght of the electric field depends on the potential difference between plates.

The lines of equal potential difference between two plates are equipotentials.

Equipotentials: are regions where charges can move without work being done, are cut by field lines at 90°, and are scalars.

**The electric potential at a point is the work done in bringing a unit positive test charge from infinity to the point.**

Gravitational fields that surround a point are radial fields.

Gravitational potential difference is the work done in moving a unit mass between two points.

\begin{gather}
\Delta V_{g} =\frac{W}{m}\\
\Delta V_{g} =change\ in\ gravitational\ potential,\ W=work\ done,\ \notag\\
m=mass\ of\ the\ object \notag
\end{gather}

Gravitational potential is defined to be zero at infinity

\begin{gather}
F\varpropto \frac{1}{r^{2}}\\
F=force,\ r=distance \notag\\
\notag\\
Therefore,\ \notag\\
\lim _{r\rightarrow \infty }\frac{1}{r^{2}} =0 \notag
\end{gather}

When the distance is very large, the gravitational force decreases to 0.

Gravitational potential at a point is defined to be work done per unit mass in moving a test mass from infinity to the point.

Gravitational & Electric Fields - video explanationBoth fields obey an inverse-square law, where the force between the objects is inversely proportional to the distance between them squared.

Electric fields | Gravitational fields |
---|---|

\begin{gather} F_{E} =+\frac{kq_{1} q_{2}}{r^{2}}\\ F_{E} \ =\ force\ between\ the\ point\ charges,\ r=distance\ between\ them \notag\\ k=\frac{1}{4\pi \epsilon _{0}} \notag\\ \epsilon _{0} =permittivity\ of\ free\ space\ or\ the\ medium\ \notag \end{gather} | \begin{gather} F_{g} =\frac{Gm_{1} m_{2}}{r^{2}}\\ F_{g} \ =\ force\ between\ the\ point\ masses,m=mass\ \notag\\ r=distance\ between\ them \notag\\ \notag \end{gather} |

The sign of the overall result indicates the direction of the force. Negative indicates attraction, positive indicates repulsion. | The value of G is a universal constant |

For a uniform field between parallel plates:

\begin{gather}
E=\frac{V}{d} =\frac{F}{Q}\\
\notag\\
E=electric\ field\ strenght,\ V=potential\ difference\ between\ plates \notag\\
d=distance,\ F=force\ that\ acts\ on\ the\ charge,\ Q=charge \notag\\
\notag
\end{gather}

Electric charge, field, and potential - Khan Academy
\begin{gather}
\frac{Q}{A} =\epsilon _{0}\frac{V}{d}\\
Q=charge,\ A=area,\ \epsilon _{0} =permittivity,\ d=distance \notag
\end{gather}

Therefore,

\begin{gather}
\frac{Q}{A} =\sigma \\
\sigma =surface\ charge\ density \notag
\end{gather}

Thus between the plates:

\begin{gather}
E=\frac{\sigma }{\epsilon _{0}}\\
E=electric\ field\ strenght,\ \sigma =surface\ charge\ density,\ \notag\\
\epsilon _{0} =permittivity \notag
\end{gather}

For surface of any conductor (singular):

\begin{gather}
E=\frac{\sigma }{2\epsilon _{0}}\\
E=electric\ field\ strenght,\ \sigma =surface\ charge\ density,\ \notag\\
\epsilon _{0} =permittivity \notag
\end{gather}

\begin{gather}
E=-potential\ gradient=-\frac{\Delta V_{e}}{\Delta x}\\
E=electric\ field\ strenght,\ \Delta V_{e} =variation\ of\ the\ potential \notag\\
\Delta x=change\ in\ distance \notag
\end{gather}

The minus sign means that the direction of the electric field is opposite the variation of potential of a positive charge.

\begin{gather}
V_{e} =\frac{Q}{4\pi \epsilon _{0} r}\\
V_{e} =electric\ potential,\ \epsilon _{0} =permittivity,\ r=distance\ from\ point\ charge \notag\\
Q=charge\ creating\ the\ field \notag
\end{gather}

If charge q is in a field that arises from another single point charge Q:

\begin{gather}
E_{ \begin{array}{l}
p\\
\end{array}} =\frac{kqQ}{r} =\frac{qQ}{4\pi \epsilon _{0} r}\\
E_{ \begin{array}{l}
p\\
\end{array}} =potential\ energy,\ r=distance\ between\ charges \notag\\
\notag
\end{gather}

**Example:**

A point charge has a magnitude of -0.48 nC. Calculate the potential 1.5m from this charge.

\begin{gather}
V_{e} =\frac{Q}{4\pi \epsilon _{0} r}\\
\notag\\
V_{e} =\frac{-0.48\ *\ 10-9}{4\pi \epsilon _{0} *\ 1.5} =2.9\ V \notag
\end{gather}

Outside a charged conducting sphere, the potential is indistinguishable from that of a point charge. This is because the field lines leave the surface of the sphere at 90°, therefore they are radial.

The surplus charge must be on the outside of the sphere because the charges will move until they are as far apart as possible, the charges will move until they are all in equilibrium (equidistant from the surface).

\begin{gather}
g=-\frac{\Delta V_{g}}{\Delta r}\\
\notag\\
g=gravitational\ field\ strenght,\ \Delta V_{g} =gravitational\ potential \notag\\
r=distance \notag
\end{gather}

For a field due to point mass M with a point mass m at a distance r from it:

\begin{gather}
V_{g} =-\frac{Gm}{r}\\
V_{g} =gravitational\ potential,\ G=gravitational\ constant \notag
\end{gather}

The potential energy of m at distance r from mass M:

\begin{gather}
E_{p} =mV_{g} =-\frac{GMm}{r}\\
\notag
\end{gather}

The forces are always attractive for gravity, therefore the potential is always negative with a maximum of zero at infinity (horizontal asymptote).

Gravitational Fields - video explanationThe inner solid sphere (inside the planet) is an Earth with a smaller radius. The gravitational field strenght of that inner sphere is given by:

\begin{gather}
g'\ =\frac{G\ *\ reduced\ mass\ of\ Earth}{radius\ of\ small\ Earth^{2}}\\
\notag
\end{gather}

The outer shell does not give rise to a gravitational field.

\begin{gather}
g'\ =\frac{4\pi G\rho r'\ }{3}\\
\rho =density\ of\ planet,\ r'\ =radius\ of\ the\ inner\ sphere \notag\\
\notag
\end{gather}

Linear orbital speed:

\begin{gather}
v_{orbit} =\sqrt{\frac{GM}{r}}\\
v_{orbit} =orbital\ velocity,\ M=mass\ of\ planet,\ r=radius\ of\ tbe\ orbit \notag\\
\notag
\end{gather}

Orbital period:

\begin{gather}
T_{orbit} =\sqrt{\frac{4\pi ^{2} r^{3}}{GM}}\\
v_{orbit} =orbital\ velocity,\ M=mass\ of\ planet,\ r=radius\ of\ tbe\ orbit \notag\\
\notag
\end{gather}

Video explanation
An orbit relatively close to the Earth's surface (100 km). The satellites often orbit over the poles. The linear speed of the satellite is about 7800 ms^{-1}.

They orbit at large distances above the Earth. The orbital period is around 24 hours, so they can be made to stay in the same area of the sky, following a figure-eight orbit.

It is a special kind of geosynchronous orbit that orbits above the Equator and will seem stationary when viewed from the surface. They are used for communication purposes and collecting images weather forecasts.

Classes of orbits - videoEscape velocity is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a massive body, that is, to achieve an infinite distance from it.

\begin{gather}
v_{e} =\sqrt{\frac{2GM}{r}}\\
r=distance\ from\ the\ center\ of\ the\ mass,\ v_{e} =escape\ velocity \notag\\
M=\ mass\ of\ the\ body\ to\ be\ escaped\ from \notag\\
\notag
\end{gather}

Escape velocity is roughly 12000 ms^{-1} for Earth.

In a magnetic field the force is always at right angles to the motion of the electron (Fleming's left hand rule) and so the resulting path of the electron is circular.

Force acting on electron:

\begin{gather}
\frac{m_{e} v^{2}}{r} =Bqv\\
\notag\\
m_{e} =mass\ of\ electron\ \left( 9.1\ x\ 10^{-31}\right) ,\ q=charge \notag\\
v=speed,\ r=radius\ of\ circular\ orbit,\ B=magnetic\ field\ strenght \notag
\end{gather}

Radius of the circle due to centripetal force:

\begin{gather}
r=\frac{m_{e} v}{Bq} =\frac{p}{Bq} =\frac{\sqrt{2m_{e} E_{k}}}{Bq}\\
\notag\\
m_{e} =mass\ of\ electron\ \left( 9.1\ x\ 10^{-31}\right) ,\ q=charge \notag\\
v=speed,\ r=radius\ of\ circular\ orbit,\ B=magnetic\ field\ strenght \notag\\
p=momentum \notag
\end{gather}

Electrons In A Uniform Magnetic Field - videoMotion in a magnetic field - video

Electron follows a parabolic path.

Electron moving in a uniform fieldCombining electric and magnetic fields at right angles to each other:

\begin{gather}
F_{E} =F_{B} \ \\
\notag\\
Thus:\ \notag\\
qE=Bqv \notag\\
q=charge,\ E=electric\ field\ strength,\ B=magnetic\ field\ strength,\ v=speed \notag\\
\notag\\
Leading\ to: \notag\\
v=\frac{E}{B} \notag
\end{gather}

There is some speed where the forces are balanced. This arrangement is a velocity selector.

Bainbridge mass spectrometer - videoNumber of correct answers: