**Electrostatic Force and Electric Fields – Learn**

Electric field strength allows us to quantitatively look at electric fields. Mathematically, the electric field, , is given by:

where;

= the electric field strength in

= the force on the charged particle in

= the charge on the object experiencing the force in

The previous equation is also commonly arranged to give:

The acceleration of a charged particle in an electric field can also be determined by using Newton’s 2^{nd} Law:

where;

= the acceleration of the charged particle in

= the mass of the accelerating particle in

**Coulomb’s Law**

Coulomb’s Law is used to determine the force between two charged particles:

where:

= the force on each charged object (in N)

and = the charges on the two points (in C)

= the distance between each charged point (in m)

= the permittivity of free space, which is equal to 8.854 × 10^{−12} A^{2}S^{4}Kg^{−1}m^{−3} in air or in a vacuum.

is known as **Coulomb’s constant, k**, and is equal to 8.9875 × 10^{9 }Nm^{2}C^{−2}. This is usually rounded off to 9.0 × 10^{9 }Nm^{2}C^{−2} for calculations.

**Coulomb’s Law can now become:**

where k = 9.0 × 10^{9 }Nm^{2}C^{−2}

**Example 1:**

Calculate the uniform electric field that will cause a force of 2.4 × 10^{−}^{22} N on an electron (q_{e} = −1.602 × 10^{−}^{19} C)

**Answer:**

using:

, or

in the opposite direction to the force

**Example 2:**

Calculate the acceleration of the electron in example 1 (mass of an electron = 9.1 × 10^{−}^{31} Kg):

**Answer:**

using:

and rearranging to give:

**Example 3:**

Calculate the force between 2 protons that are separated by 5mm (q_{p} = 1.602 × 10^{−19} C):

**Answer:**

using:

(repulsion)