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Fields - Equipotentials and Field Lines

Grade 12IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Field lines represent the direction and magnitude of a field. For gravitational fields, they always point toward the mass. For electric fields, they point away from positive charges and toward negative charges.

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Equipotential surfaces are regions where the potential VV is constant. Moving a test mass or charge along an equipotential surface requires zero work (W=0W = 0) because ΔV=0\Delta V = 0.

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Field lines and equipotential surfaces are always mutually perpendicular at every point in space.

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The spacing between equipotential lines indicates field strength: the closer the lines, the steeper the potential gradient and the stronger the field strength (EE or gg).

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In a uniform field, such as between parallel plates, equipotentials are parallel, equally spaced planes, and field lines are parallel, equally spaced straight lines.

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For a point mass or charge, equipotentials are concentric spheres centered on the source. The potential VV decreases (or increases) inversely with distance rr.

📐Formulae

Vg=−GMrV_g = -\frac{GM}{r}

Ve=kQrV_e = \frac{kQ}{r}

g=−ΔVgΔrg = -\frac{\Delta V_g}{\Delta r}

E=−ΔVeΔrE = -\frac{\Delta V_e}{\Delta r}

W=mΔVgW = m\Delta V_g

W=qΔVeW = q\Delta V_e

💡Examples

Problem 1:

An electron is moved from an equipotential of −200 V-200\text{ V} to an equipotential of −500 V-500\text{ V}. Calculate the work done on the electron in electronvolts (eV\text{eV}) and determine if the work is done by or against the field.

Solution:

ΔV=Vfinal−Vinitial=−500 V−(−200 V)=−300 V\Delta V = V_{final} - V_{initial} = -500\text{ V} - (-200\text{ V}) = -300\text{ V}. The charge of an electron is q=−1eq = -1e. Work done W=qΔV=(−1e)×(−300 V)=300 eVW = q\Delta V = (-1e) \times (-300\text{ V}) = 300\text{ eV}.

Explanation:

Since the work done is positive (300 eV300\text{ eV}), work must be done against the electrostatic field to move the negative electron toward a more negative potential.

Problem 2:

In a uniform electric field, two equipotential lines representing 100 V100\text{ V} and 150 V150\text{ V} are separated by a distance of 2.5 cm2.5\text{ cm}. Calculate the electric field strength EE.

Solution:

E=∣ΔVΔr∣=150 V−100 V0.025 m=500.025=2000 V m−1E = |\frac{\Delta V}{\Delta r}| = \frac{150\text{ V} - 100\text{ V}}{0.025\text{ m}} = \frac{50}{0.025} = 2000\text{ V m}^{-1}.

Explanation:

The electric field strength is the negative gradient of the potential. In a uniform field, EE is constant and is calculated by dividing the potential difference by the perpendicular distance between the equipotentials.