# Electric Field Strength ## Defining Electric Field Strength - An electric field is a region where charged particles experience forces - To quantify how strong or weak these forces are, we use **electric field strength** - Electric field strength at a point is defined as: **The electrostatic force per unit positive charge acting on the charge at that point** - The equation for electric field strength is: $$E = \frac{F}{Q}$$ - Where: - $E$ = electric field strength (N C$^{-1}$) - $F$ = electrostatic force on the charge (N) - $Q$ = charge (C)

Electric field strength E = F / Q, where F is the force on a charge Q.

- Using a **positive test charge** in this definition is crucial because it determines the direction of the electric field - Electric field strength is a **vector quantity** - Its direction is always: - **Away** from a positive charge - **Towards** a negative charge - Think of electric field strength like water pressure in pipes: higher pressure pushes water harder, just as stronger fields push charges with greater force ## Electric Field Strength in a Uniform Field ### The Equation - Between two charged parallel plates, the electric field is **uniform** - The magnitude of the electric field strength in this uniform field is defined as: $$E = \frac{V}{d}$$ - Where: - $E$ = electric field strength (V m$^{-1}$) - $V$ = potential difference between the plates (V) - $d$ = separation between the plates (m)

Electric field strength has alternative units: V m⁻¹ or N C⁻¹ (both are equivalent).

### What This Equation Tells Us - **The greater the voltage** between the plates, the **stronger the field** - **The greater the separation** between the plates, the **weaker the field** - Imagine stretching a rubber band: the more you stretch it (greater separation), the weaker the tension becomes The graphs below illustrate these two key relationships: **Effect of plate separation on field strength (at constant voltage of 1000 V):**

Notice the **inverse relationship**: doubling the separation halves the field strength. **Effect of voltage on field strength (at constant separation of 0.02 m):**

Notice the **direct proportionality**: doubling the voltage doubles the field strength. ### Important Restrictions - This equation $E = \frac{V}{d}$ **cannot** be used for: - Point charges (which produce radial fields, not uniform fields) - Spherical conductors - Any non-uniform field configuration - The direction of the electric field is from: - The plate connected to the **positive terminal** → towards the plate connected to the **negative terminal**

If one parallel plate is earthed, it has a voltage of 0 V

## Deriving the Uniform Field Equation ### Step-by-Step Derivation **Step 1: Work done and potential difference** - When two points in an electric field have different potentials, there is a potential difference $\Delta V$ between them - Moving a charge $Q$ across this potential difference requires **work to be done** - Potential difference is defined as the energy transferred per unit charge: $$\Delta V = \frac{W}{Q}$$ - Therefore, the work done in transferring charge $Q$ is: $$W = \Delta V \times Q$$ **Step 2: Work done in moving between plates** - When charge $Q$ moves from one plate to the other through distance $d$, the work done is also: $$W = F \times d$$ - Where: - $W$ = work done (J) - $F$ = force (N) - $d$ = distance (m) **Step 3: Equating the two expressions** - Since both expressions equal $W$: $$F \times d = \Delta V \times Q$$ **Step 4: Rearranging** - Dividing both sides by $Q$ and $d$: $$\frac{F}{Q} = \frac{\Delta V}{d}$$ **Step 5: Final result** - Since electric field strength $E = \frac{F}{Q}$: $$E = \frac{F}{Q} = \frac{\Delta V}{d}$$ - This shows that electric field strength in a uniform field equals the potential difference per unit separation ## Worked Example **Question:** A dust particle carrying a charge of $4.2 \times 10^{-18}$ C is placed in an electric field. The electric field strength is $6.0 \times 10^{4}$ N C$^{-1}$. Calculate the force experienced by the particle. **Answer:** **Step 1: Write down the equation for electric field strength** $$E = \frac{F}{Q}$$ **Step 2: Rearrange for force $F$** $$F = E \times Q$$ **Step 3: Substitute in the values** $$F = (6.0 \times 10^{4}) \times (4.2 \times 10^{-18})$$ $$F = 2.52 \times 10^{-13} \text{ N}$$ **Step 4: State the final answer** The force acting on the dust particle is $2.5 \times 10^{-13}$ N (2 s.f.) ## Worked Example: Parallel Plates **Question:** Two parallel metal plates are separated by 4.2 cm and have a potential difference of 950 V. Calculate the magnitude of the electric force acting on a stationary electron between the plates. Charge on an electron = $1.60 \times 10^{-19}$ C **Answer:** **Step 1: Write down the known values** - Potential difference, $V$ = 950 V - Distance between plates, $d$ = 4.2 cm = $4.2 \times 10^{-2}$ m - Charge, $Q = 1.60 \times 10^{-19}$ C **Step 2: Calculate the electric field strength between the parallel plates** $$E = \frac{V}{d} = \frac{950}{4.2 \times 10^{-2}} = 22619 \text{ V m}^{-1}$$ **Step 3: Calculate the electric force on the electron** $$F = E \times Q$$ $$F = 22619 \times (1.60 \times 10^{-19})$$ $$F = 3.62 \times 10^{-15} \text{ N}$$ **Step 4: State the final answer** The magnitude of the electric force acting on the electron is $3.6 \times 10^{-15}$ N (2 s.f.)