6.2.3, 6.2.4(c)Particles and Medical PhysicsCore

Uniform Fields and Spherical Capacitance

A uniform electric field gives a charged particle constant acceleration. The same definitions of potential and capacitance connect a parallel-plate capacitor to an isolated conducting sphere.

40 min 6.2 Electric Fields

What you'll be able to do

Calculate acceleration and deflection between plates
Use parallel-plate permittivity equations
Derive the capacitance of an isolated sphere

Charged-particle motion between plates

A potential difference V between wide parallel plates separated by d gives an approximately uniform field E=V/d away from the edges. A particle experiences force qE, so acceleration is qE/m. Positive charges accelerate along the field; electrons accelerate against it.

A particle entering horizontally has unchanged horizontal speed if there is no horizontal force. Its residence time is plate length divided by horizontal speed. Use vertical constant-acceleration motion to find deflection. Beyond the plates the particle continues along its exit velocity rather than abruptly returning to horizontal motion.

E=\frac Vd,\quad a=\frac{qE}{m},\quad t=\frac L{v_x},\quad y=\tfrac12a t^2

The deflection expression assumes zero initial transverse velocity.

Parallel-plate capacitance

For plates of overlapping area A with a small gap d, capacitance is εA/d. Permittivity ε=ε₀εr includes the material’s dimensionless relative permittivity. Doubling area doubles C; doubling separation halves C if the dielectric remains the same.

OCR requires using relative permittivity, but does not require a microscopic explanation of why it exceeds one. Keep that scope separate from optional discussion of polarisation. Use metres squared for area: converting cm² requires a factor of 10⁻⁴, not 10⁻².

C=\frac{\varepsilon_0\varepsilon_r A}{d}

An isolated sphere

Outside a uniformly charged conducting sphere, potential is the same as that of a point charge at its centre. Its surface potential relative to infinity is Q/(4πε₀R). Substitute into C=Q/V: the Q cancels and gives C=4πε₀R.

This result concerns an isolated sphere with the reference at infinity, not two concentric spheres or a sphere next to an earthed plate. A larger sphere stores more charge for the same surface potential. The capacitance is a property of the geometry and surroundings rather than how much charge happens to be present.

V=\frac Q{4\pi\varepsilon_0R}\quad\Rightarrow\quad C=\frac QV=4\pi\varepsilon_0R

Work through the reasoning

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.

Example 1An electron enters 0.040 m long plates horizontally at

2.0\times10^7

m s⁻¹. Field magnitude is 1000 V m⁻¹. Find deflection magnitude using e=

1.60\times10^{-19}

C and m=

9.11\times10^{-31}

kg.

t=0.040/(2.0\times10^7)=2.0\times10^{-9}

|a|=eE/m=1.76\times10^{14}

m s⁻².

|y|=\tfrac12(1.76\times10^{14})(2.0\times10^{-9})^2=3.51\times10^{-4}

AnswerAbout 0.35 mm, opposite to the field direction.

Example 2Find the capacitance of an isolated sphere of radius 0.10 m using ε₀=

8.85\times10^{-12}

F m⁻¹.

C=4\pi\varepsilon_0R

C=4\pi(8.85\times10^{-12})(0.10)=1.11\times10^{-11}

AnswerAbout 11 pF.

Equation recap

E=\frac Vd,\quad a=\frac{qE}{m},\quad t=\frac L{v_x},\quad y=\tfrac12a t^2

The deflection expression assumes zero initial transverse velocity.

C=\frac{\varepsilon_0\varepsilon_r A}{d}

Equation recap

V=\frac Q{4\pi\varepsilon_0R}\quad\Rightarrow\quad C=\frac QV=4\pi\varepsilon_0R

Equation recap

Common mistakes to avoid

Using plate length as plate separation in E=V/d.

Separation sets field strength; length sets time in the field.

Treating the isolated-sphere formula as valid near any other conductor.

Its derivation assumes the isolated geometry and infinity reference.

Key takeaways

Uniform fields give constant charged-particle acceleration.
Parallel-plate C depends on area, spacing and permittivity.
An isolated sphere has C=4πε₀R.

Test yourself

Answer a few questions, then review anything you missed.

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