Connect with us

# Electric Field

Electric field cannot be discussed without first all talked about electric force.

Electric force: According to coulomb’s law, which state that the force of attraction between two charges is directly proportional to that products of the two charges and inversely proportion to the square of distance separating them and also act along the line joining the charges.

You can as well

$that\quad is\quad F=\frac { { KQ }_{ 1 }{ Q }_{ 2 } }{ { d }^{ 2 } }$

Where k = constant of the proportionality which is give as

$\frac { 1 }{ 4\pi { \varepsilon }_{ 0 }{ d }^{ 2 } } \quad therefore\quad F\quad =\frac { { Q }_{ 1 }{ Q }_{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ d }^{ 2 } }$

Where Ɛo = relative permittivity = 8.854 x 10-12Fm-1 , k = 1/4πƐo =9 x 10 109(Nm2C2  )

Also d= distance separating the charges,  Q1 and Q2 = charges and is measured in coulomb

# The electric field

Electric field can be defined as a region where electric force is felt. Electrified can also be described in terms of line of force which represent the direction of  motion of  small positive charge  place at a point in the field assuming that the charge is so small that the field did not change by the presence of charge.

# Electric field intensity

The electric field intensity at any point can be defined as electric force per a unit charges which it exerts at that point. The unit of  electric field intensity is Newton per coulomb (NC-1). It is given as:

$E=\frac { F }{ Q } \quad but\quad F\quad =\frac { QQ }{ 4\pi { \varepsilon }_{ 0 }{ d }^{ 2 } } \quad therefore\quad E\quad =\frac { Q }{ 4\pi { \varepsilon }_{ 0 }{ d }^{ 2 } }$

You can also read Newton’s laws of motion

# Electric field flux

Assuming that the electric field intensity E, makes an angle Ѳ, with a small surface area ⊿A, then the electric field flux Ψ = EcosѲ⊿A.  But for the flux through an area perpendicular to the lines of force is E x area . Example for a sphere the flux is given as

$\Psi =E\times area\quad \frac { Q }{ 4\pi { \varepsilon }_{ 0 }{ d }^{ 2 } } \times 4{ \pi }{ d }^{ 2 },\quad therefore\quad \Psi =\frac { Q }{ { \varepsilon }_{ 0 } }$

What is work done

# Electric field potential

Electric field potential is work done in bringing a unit positive charge from infinity to that point against the action of the field. It is measure in volts or Joules per coulomb (JC-1)

The potential due to point charges is given as

$V=\frac { Q }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ } }$

and for the potential difference Vab which is the work done in taking a unit positive charge, from b to a against the force is given as

${ V }_{ ab }=\frac { Q }{ 4\pi { \varepsilon }_{ 0 } } (\frac { 1 }{ { r }_{ a } } -\frac { 1 }{ { r }_{ b } } )$

$but\quad as\quad b\rightarrow a,\quad { V }_{ a }=\frac { Q }{ 4\pi { \varepsilon }_{ 0 }{ r } }$