# Electric Field Strength

## Electric Field Strength Revision

**Electric Field Strength**

Previously we have learnt how to calculate the strength of a gravitational field. In this section we look at **electric field strength **in **uniform** and **radial** **fields**.

**Electric Field Strength**

Any charged object within an **electric field** will experience a force acting upon it. If the two charges are oppositely charged , the force will be** attractive **and if the charges are like, the force will be **repulsive**.

The** electric field strength** can be defined as the amount of **electrostatic force exerted on an object per unit of charge**. In equation form we get:

E = \dfrac{F}{Q}

- E is the
**electric field strength**in volts per metre or Newtons per Coulomb\left(\text{Vm}^{-1} \: \text{or} \: \text{NC}^{-1}\right) - F is the
**electrostatic force**acting on the charged particle in Newtons \left(\text{N}\right) - Q is the
**charge**of the object in the field in Coulombs \left(\text{C}\right).

**Example: **An electron enters an electric field and experiences a force of 0.01 \: \text{N}. Given that the charge of an electron is -1.6 \times 10^{-19} \: \text{C}, calculate the electric field strength of the electric field.

**[2 marks]**

E = \dfrac{F}{Q}

E = \dfrac{0.01}{-1.6 \times 10^{-19}}

E = \boldsymbol{-6.25 \times 10^{16}} \: \textbf{Vm}^{-1} \: \text{or} \: \textbf{NC}^{-1}

**Electric Field Strength in a Uniform Field**

When connected to a **potential difference**, two parallel metal plates become oppositely charged and a **uniform electric field **forms between them:

The **electric field strength** at any point within the field will always be equal, because it is a **uniform field**.

To calculate the electric field strength between two** parallel plates** we can use the equation:

E = \dfrac{V}{d}

- E is the
**electric field strengthÂ**in volts per metre \left(\text{Vm}^{-1}\right). - V is the
**potential difference**between the plates in volts \left(\text{V}\right). - d is the
**distance**between the plates in metres \left(\text{m}\right).

**Example:** Two parallel plates are connected to a power source providing a d.c supply of 6 \: \text{V}. Calculate the electric field strength when the plates are separated by 0.1 \: \text{m}.

**[1 mark]**

E = \dfrac{V}{d}

E = \dfrac{6}{0.1}

E = 60 \: \text{Vm}^{-1}

**Moving Charges in a Uniform Electric Field**

A charged particle has its own **electric field**. If the charged particle enters a** uniform field** it will experience a** force** which will result in a change in **direction **or **velocity**. The motion of the **charged particle** depends upon its** mass**, **charge** and **velocity**.

The **charged particle** will always be** attracted to the oppositely charged plate** and **deflected away from the same charged plate**. It will move in a **parabola** as shown in the diagrams above until it collides with the plate.

To increase the amount of **deflection** we may:

**Increase the charge.****Decrease the velocity of the particle.****Decrease the mass of the particle.**

The diagram below shows a proton whose** deflection** has been increased, as a result of either increased charge, decreased velocity, or decreased mass of the moving charge.

**Work done moving a charged particle**

If we wanted to know the **work done** when moving a **charg**e between two **parallel plates**, we can derive our own equation:

Work done: W = Fd

For a moving charge: F = EQ and d = \dfrac{V}{E}

Substituting F and d into the equation for work done, we arrive at:

W = Q \Delta V

**Particle Accelerators**

A similar concept can be used for **particle accelerators **such as the one here.

A **proton** is fired through a tiny gap in the** positively charged plate **and towards a **negatively charged plate**. As the proton is positively charged, it is **repelled **away from the positive plate and **attracted **to the negative plate, causing an **acceleration**. The proton then passes through a tiny gap in the negatively charged plate at a **higher velocity** than when it entered.

**Radial Electric Fields**

A** point charge** produces a **radial electric field** like the examples below. Remember, the **electric field lines** point in opposite directions for positive and negative charges:

We can see that in the **radial** electric fields, the field lines get further apart as **distance** from the particle increases. This represents the changing** electric field strength** with distance for a radial electric field.

To calculate the electric field strength at any point in the radial electric field, the following equation may be used:

E = \dfrac{Q}{4 \pi \epsilon_0 r^2}

- Q is the
**charge**of the point charge that is causing the field in Coulombs \left(C\right). - \epsilon_0 is the
**permittivity of free space**, \left(8.85 \times 10^{-12}\right). - r is the
**distance**from the centre of the charge in metres \left(\text{m}\right).

Example: Calculate the electric field strength at a distance of 0.5 \: \text{mm} away from a proton.

**[2 marks]**

E = \dfrac{Q}{4 \pi \epsilon_0 r^2}

E = \dfrac{1.6 \times 10^{-19}}{\left(4 \pi \times 8.85 \times 10^{-12} \times \left(0.5 \times 10^{-3}\right)^2\right)}

E = 5.75 \times 10^{-3} \: \text{Vm}^{-1}

## Electric Field Strength Example Questions

**Question 1:** A proton enters an electric field and experiences a force of 0.002 \: \text{N}. Given that the charge of a proton is 1.6 \times 10^{-19} \: \text{C}, calculate the electric field strength.

**[2 marks]**

**Question 2:** Describe how a positive point charge would move through a pair of parallel plates with the positively charged plate on the left hand side, and negatively charged plate on the right hand side.

**[2 marks]**

The point charge will always be attracted to the oppositely charged plate and moves in a parabola. Therefore the **positive charge would move to the right towards the negatively charged plate** in a **parabolic motion**.

**Question 3:** Explain how oppositely charged plates could be used to accelerate a charged particle.

**[3 marks]**

A **proton is fired through a tiny gap in a positively charged plate and towards a negatively charged plate**.

As the proton is positively charged, **it is repelled away from the positive plate and attracted to the negative plate, causing an acceleration**.

The **proton then passes through a tiny gap in the negatively charged plate at a higher velocity than when it entered**.

This would also work for an electron but the polarity of the plates would need to be reversed.

**Question 4:** Calculate the electric field strength at a distance of 0.12 \: \text{mm} away from an electron.

**[2 marks]**

## Electric Field Strength Worksheet and Example Questions

### Electric Field Questions

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