Electric Fields

• Electric field strength is the force per unit charge
• Units NC⁻¹ or Vm⁻¹
• It is a vector quantity 
• The direction of the field is the direction a positive charge would move in
• Density of field lines indicates field strength (closer together = stronger field)

The diagram above shows a uniform electric field between two parallel plates. The straight field lines are equal distances apart. Equipotential lines are also equally spaced and perpendicular to the field lines.

Potential difference is the energy transferred per unit charge so, a volt is a joule per coulomb (JC⁻¹).

To move a charge q from X to Y, the force on the charge due to the electric field F = E x Q (where E is the electric field strength) must be equal in magnitude to the work done in moving the charge.

Work done = F x d.
Work done = EQ x d 

The work done is also equal to the energy gained by the charge moving through a potential difference, work done = QV. 

EQd = QV 

Ed = V 

E = V / d 

This shows that electric field strength can also be measured in Vm ⁻¹. 

The diagram below shows a charged particle (e.g an electron) suspended in a vertical electric field. The charged plates have a distance d between them and a potential difference of V. Here, the gravitational field strength would be equal in magnitude but opposite in direction to the force on the charge due to the electric field. 

F_E = F_g

Ee = m_e/g (F = EQ, in this case Q = charge on electron = e) 

g/E = e/m_e (e/m_e is the specific charge of the electron) 

By substituting in E = V/d, we get e/m_e = gd/V 

Here we have an electron being accelerated in a uniform electric field. The electron has an initial velocity u and mass m_e. The force on the electron can be found using F = EQ and E = V/d. It will travel in a straight line as the as the acceleration due to the electric field is constant.

A charged sphere is suspended in a uniform electric field:

F_E = force due to the electric field
For a charged sphere, charge may be considered to be at the centre.

We can resolve vertically and horizontally to form an equation:
R(H): Tsinθ = F_E = qV/d
R(V): Tcosθ = F_g = mg

Tsinθ/Tcosθ = F_E/F_g
tanθ = qV/mgd

The diagram below shows the plan view of an electron entering a uniform electric field along the centre line with initial velocity u.

In the x-direction, the electron experiences no force so, Vₓ is constant = u. Using the suvat equation: s = ut+ 1/2 at², in the x-direction, x = ut.

In the y-direction, F = QV/d = eV/d (e is the charge on the electron).

As F = ma (Newton’s Second Law), a = eV/mₑd
(mₑ is the mass of the electron).

Since the force is constant and directed towards the positive plate, the electron will undergo uniform acceleration. Due to the constant acceleration in the vertical direction, and no forces acting in the horizontal direction, the electron will follow a parabolic path as it travels between the plates.

s = ut+ 1/2 at²
y = eVt²/2mₑd

Subbing in x = ut:

y = eVx²/2mₑdu²

Once the electron clears the plates and leaves the electric field, it no longer experiences any force due to the field. Since no forces are acting on the electron, it will continue moving in a straight line at a constant velocity (Newton’s First Law of Motion).

The velocity will have a horizontal and vertical component. The horizontal component will be the same as when it entered the field and the vertical component will have increased due to the acceleration in the field.

Coulomb’s Law is very similar to Newton’s Law of Gravitation which may make it easier to remember.

The force between 2 point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. It is a force of attraction between charges of opposite sign and a force of repulsion if the charges are of the same sign.

Air can be treated as a vacuum when calculating force between charges.

To determine the ratio of the electromagnetic force to the gravitational force between two protons at any separation, we can divide one of the equations by the other (we want our answer to be in the form F_E : F_g).

e is the charge on a proton = 1.60 x 10⁻¹⁹
mₚ is the mass of a proton = 1.67 x 10⁻²⁷

r² cancels out and we can start rearranging and substituting in:

(9.58 x 10⁷)² x 1/(4π x 8.85 x 10⁻¹² x 6.67 x 10⁻¹¹) : 1

1.24 x 10³⁶ : 1

The electromagnetic force is roughly 10³⁶ times greater than the gravitational force. F_E >> F_g

The diagram below shows a square with sides of length a. There are four charged points on each corner of the square. You can imagine the point P is a particle with charge +q.

In moving a charge q from an equipotential V to another at V + ΔV, the work done is qΔV and is independent of the path taken.

Absolute electric potential: the work done per unit positive charge in moving it to a point from infinity.

TO BE CONTINUED