# Electric Fields Strength and Coulomb's Law

- Created by: Tree
- Created on: 07-12-11 23:12

**Electric Fields and Coulomb's Law**

**Definition of Electric field:**

*A region around a charged object where another object will experience a force.*

Magnetic field around a point charge - arrows indicate the direction of the **force experienced by a positive charge** placed within the field.

**Electric Field Strength Equation**

*Electric Field Strength (NC^-1) = Force on Charge, F (N) / Size of Charge, Q (C)*

*E = F/Q*

**Worked Example: Finding Force experienced within electric field**

*What force would a helium nucleus be exposed to if placed in a field of 6NC^-1?*

**Solution**

Note that the question is referring to only the nucleus which means that you should not consider the charge of the orbiting electrons.

The helium nucleus may also be referred to as an alpha particle

*The helium nucleus has 2 protons and 2 neutrons*

E=F/Q, therefore F=EQ

F = 6 (NC-1) x 2(1.9EXP-19) (C)

F = 1.92EXP-18N

= 2EXP-18N

**Charged Particles in Electric Fields**

A charged particle in an electric field experiences a force. It therefore accelerates, allowing it to gain kinetic energy.

The force it experiences is equal to electric field strength x charge on particle

**F=EQ**

Take the distance over which the particle moves to be x. The work done (energy transferred) is equal to the force x distance over which it acts

**W = F x**

In electrical terms, work done is equal to change in potential difference x charge on particle

**W = VQ**

So both equations can be made equal to one another:

**Fx = VQ**

Rearranging this formula gives:

**F/Q = V/X**

Since F.Q = E, V/X…

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