recall and use E = ∆V / ∆d to calculate the field strength of the uniform field between charged parallel plates

Uniform Electric Fields (Cambridge International AS & A Level Physics 9702 – Topic 18)

Learning Objectives

• Recall and use the relationship \(E = \dfrac{\Delta V}{\Delta d}\) to calculate the magnitude of a uniform electric field between charged parallel plates.
• Explain the vector nature of the electric field and interpret field‑line diagrams.
• Apply the gradient relation \(\mathbf{E}= -\nabla V\) to uniform fields and to equipotential surfaces.
• Connect the uniform‑field result to electric flux density, the capacitance of a parallel‑plate capacitor and to experimental field‑mapping techniques.

Key Concepts

1. Definition of the Electric Field (Syllabus 18.1)

The electric field at a point is defined as the force \(\mathbf{F}\) experienced by a small positive test charge \(q\) placed at that point, divided by the magnitude of the charge:

\[

\mathbf{E}= \frac{\mathbf{F}}{q}\qquad\text{(units N C}^{-1}\text{)}

\]

Because \(\mathbf{F}\) is a vector, \(\mathbf{E}\) is also a vector. Field‑line diagrams visualise the direction (tangent to the line) and the relative strength (density of lines).

Typical field‑line sketch

2. Uniform Field Between Parallel Plates (Syllabus 18.2)

Derivation of \(E = \dfrac{\Delta V}{\Delta d}\)

1. Two large, parallel conducting plates are separated by a distance \(d\) and carry equal and opposite charges \(\pm Q\). The field between them is essentially uniform and directed from the positive to the negative plate.
2. Move a test charge \(q\) from the positive plate to the negative plate. The work done on the charge is

   \[

   W = q\,\Delta V .

   \]

3. Work is also the scalar product of force and displacement:

   \[

   W = \mathbf{F}\!\cdot\!\Delta\mathbf{d}= q\mathbf{E}\!\cdot\!\Delta\mathbf{d}= qE\,\Delta d

   \]

   (because \(\mathbf{E}\) and \(\Delta\mathbf{d}\) are parallel in a uniform field).

4. Equating the two expressions and cancelling \(q\) gives the magnitude

   \[

   E = \frac{\Delta V}{\Delta d}.

   \]

5. In vector form the direction is included:

   \[

   \boxed{\displaystyle \mathbf{E}= -\frac{\Delta V}{\Delta d}\,\hat{n}}

   \]

   where \(\hat{n}\) is a unit vector pointing from the negative to the positive plate (the opposite of the field direction).

Limitations – Edge Effects

• The uniform‑field approximation is valid only in the central region of the plates.
• Near the edges the field lines diverge (fringing); the simple relation \(E=\Delta V/d\) no longer gives the exact field strength.
• In A‑Level questions you are usually told to “neglect edge effects”.

Link to Electric Flux Density

For a parallel‑plate arrangement in free space:

\[

\mathbf{D}= \varepsilon_0\mathbf{E}= \sigma\,\hat{n},

\qquad

\sigma = \varepsilon0E = \varepsilon0\frac{\Delta V}{\Delta d}.

\]

This expression is useful when moving on to the capacitance formula (Topic 19).

Step‑by‑Step Procedure for Using \(E = \Delta V/\Delta d\)

1. Confirm uniformity: Are the plates large compared with the separation? Are edge effects ignored?
2. Obtain \(\Delta V\): Use the given potential difference or measure it with a voltmeter. Keep the sign if the direction of \(\mathbf{E}\) is required.
3. Obtain \(\Delta d\): Measure the separation of the points where the potentials are taken; convert to metres before substituting.
4. Calculate magnitude: \(\displaystyle E = \frac{|\Delta V|}{\Delta d}\) (units V m⁻¹ = N C⁻¹).
5. Assign direction: From the higher‑potential plate toward the lower‑potential plate (or use \(\mathbf{E}= -\nabla V\)).

3. Electric Field of a Point Charge (Syllabus 18.4)

\[

\boxed{\displaystyle \mathbf{E}= \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}\,\hat{r}}

\]

Field lines are radial; the magnitude follows the inverse‑square law. Superposition allows the construction of more complex fields, including the parallel‑plate case as the limit of many closely spaced charges.

4. Gradient Relation \(\mathbf{E}= -\nabla V\) (Syllabus 18.5)

In general:

\[

\mathbf{E}= -\nabla V .

\]

For a one‑dimensional uniform field (between plates) this reduces to

\[

E = -\frac{dV}{dx}\;\Longrightarrow\;E = \frac{\Delta V}{\Delta d},

\]

the minus sign indicating that the field points toward decreasing potential.

Equipotential surfaces are always perpendicular to \(\mathbf{E}\); between parallel plates they are planar and parallel to the plates.

5. Experimental Mapping of a Uniform Field (AO3)

1. Connect a pair of large parallel plates to a variable DC supply.
2. Place a high‑impedance probe (or a non‑conducting rod with a calibrated voltmeter) at several equally spaced positions between the plates.
3. Record the potential \(V\) at each position \(x\).
4. Plot \(V\) versus \(x\); the slope of the straight‑line fit equals \(-E\) because \(V = V_0 - Ex\).
5. Compare the experimental value of \(E\) with the theoretical \(E = \Delta V/d\) and comment on any discrepancy caused by edge effects.

6. Example Calculation

Problem: Two parallel plates are 5.0 cm apart and have a potential difference of 250 V. Find the magnitude and direction of the uniform electric field.

Solution:

• Convert distance: \(\Delta d = 5.0\ \text{cm}=0.050\ \text{m}\).
• Magnitude: \(\displaystyle E = \frac{250\ \text{V}}{0.050\ \text{m}} = 5.0\times10^{3}\ \text{V m}^{-1}\) (or N C⁻¹).
• Direction: From the positively charged plate toward the negatively charged plate.

Thus \(\mathbf{E}= 5.0\times10^{3}\ \text{V m}^{-1}\,\hat{n}\) where \(\hat{n}\) points from + to –.

7. Link to Capacitance (Topic 19)

For a parallel‑plate capacitor of area \(A\) and separation \(d\):

\[

C = \frac{Q}{\Delta V}= \frac{\varepsilon_0 A}{d}.

\]

Using \(E = \Delta V/d\) we can write \(Q = \varepsilon_0 A E\), showing how the uniform‑field result underpins the capacitance formula.

8. Common Mistakes to Avoid

• Forgetting to convert \(\Delta d\) to metres before using the formula.
• Using the absolute value of \(\Delta V\) without stating the field direction; remember \(\mathbf{E}\) points from high to low potential.
• Applying \(E = \Delta V/\Delta d\) to non‑uniform fields (e.g. near plate edges or around point charges).
• Mixing units: V m⁻¹ and N C⁻¹ are equivalent, but the numerical value must be consistent with the chosen system.
• Ignoring edge effects when the plate size is comparable to the separation; this leads to systematic error.

9. Practice Questions

1. A pair of parallel plates are 12 mm apart and have a potential difference of 480 V. Calculate the electric field strength.
2. If the uniform electric field between two plates is \(2.5\times10^{4}\ \text{V m}^{-1}\) and the separation is 0.020 m, what is the potential difference between the plates?
3. A proton (charge \(+e\)) enters a uniform field of \(1.0\times10^{3}\ \text{N C}^{-1}\) that extends over a distance of 0.10 m. What is the change in the proton’s electric potential energy?
4. Two large parallel plates of area \(0.10\ \text{m}^{2}\) are 3 mm apart and carry a surface charge density of \(5.0\times10^{-6}\ \text{C m}^{-2}\).

   1. Find the magnitude of the electric field between the plates.
   2. Determine the potential difference.

5. Sketch a field‑line diagram for a positive point charge and label the direction of the electric field and the equipotential surfaces.

Answers to Practice Questions

10. Summary

• The electric field is a vector defined by \(\mathbf{E}= \mathbf{F}/q\); its magnitude in a uniform region is \(E = \Delta V/\Delta d\).
• Direction: from higher to lower electric potential (\(\mathbf{E}= -\nabla V\)).
• For parallel plates the uniform‑field approximation holds away from edges; edge effects cause deviations.
• Relations to other topics: \(\mathbf{D}= \varepsilon0\mathbf{E}\), \(C = \varepsilon0A/d\), and the inverse‑square law for point charges.
• Experimental skill: map the field by measuring potential at several points and using the slope of a \(V\)‑vs‑\(x\) graph to obtain \(E\).
• Always keep units consistent (V for \(\Delta V\), m for \(\Delta d\)) and include the sign when the direction of the field is required.