Revise Electric fields and field lines for Cambridge 9702 (18.1)

An electric field is a specific region in space around any charged object where another charged object would experience an electrostatic force. Think of it as the 'sphere of influence' of a charge. This force acts without direct contact, meaning it's a non-contact force, much like gravity or magnetism.

To quantify how strong an electric field is at any point, we use electric field strength, denoted by the symbol $E$. It's defined as the force experienced per unit positive test charge placed at that point. This 'test charge' is theoretical; it's considered so small that it doesn't significantly alter the field it's testing.

$E = F/Q$

When two point charges interact, the electrostatic force between them is described by Coulomb's Law. This fundamental law tells us that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance separating them.

$F = \frac{1}{4\pi\epsilon_0} \frac{Q_1Q_2}{r^2}$

By combining the definition of electric field strength ($E=F/Q$) with Coulomb's Law, we can determine the electric field strength generated by a single point charge $Q$ at a distance $r$. This formula is crucial for understanding fields around individual charges.

$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$

Electric field lines are a powerful visual tool to represent the invisible electric field. These imaginary lines show the direction and strength of the field at every point, making complex field patterns easier to understand.

Understanding the patterns for common charge arrangements is key. These are often tested in exams.

When multiple charges are present, the total electric field at any point is the vector sum of the electric fields that each charge would create individually at that point. This is known as the principle of superposition. For two fields $E_1$ and $E_2$ at a point, the resultant field is $E_{net} = E_1 + E_2$. You must use vector addition (e.g., resolving components or using the parallelogram/triangle law) to find the resultant.

A uniform electric field is one where the electric field strength is constant in both magnitude and direction throughout a region. This type of field is typically found between two large, parallel, oppositely charged plates. The force on a charge within this field is constant ($F=Eq$), leading to constant acceleration. This creates a situation analogous to projectile motion under gravity.

$E = \frac{V}{d}$

$F = Eq$

When a charged particle enters a uniform electric field, it experiences a constant electrostatic force. If the particle's initial velocity is perpendicular to the field, its path will be a parabola. The motion can be analysed by considering two components: a constant velocity component parallel to the plates (no force in this direction) and a constantly accelerated component perpendicular to the plates (due to the constant electric force, $F=Eq$ and $F=ma$).