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Fields - Electric Fields

Grade 11IBPhysics

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

🔑Concepts

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Electric charge (qq or QQ) is a fundamental property of matter, measured in Coulombs (CC). The elementary charge is e=1.60×10−19 Ce = 1.60 \times 10^{-19} \text{ C}.

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Coulomb's Law states that the electrostatic force FF between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance rr between them.

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The Electric Field Strength (EE) at a point is defined as the electrostatic force per unit positive test charge acting on a small stationary charge at that point (E=FqE = \frac{F}{q}). It is a vector quantity.

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Electric field lines represent the direction of the force on a positive test charge. They originate from positive charges and terminate on negative charges.

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For a point charge, the electric field follows an inverse square law: E∝1r2E \propto \frac{1}{r^2}.

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A uniform electric field is created between two parallel conducting plates with a potential difference VV and separation dd, where the field strength is constant throughout.

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The permittivity of free space, ϵ0\epsilon_0, is approximately 8.85×10−12 F m−18.85 \times 10^{-12} \text{ F m}^{-1}.

📐Formulae

F=kq1q2r2F = k \frac{q_1 q_2}{r^2}

k=14πϵ0≈8.99×109 N m2 C−2k = \frac{1}{4\pi\epsilon_0} \approx 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}

E=FqE = \frac{F}{q}

E=kQr2E = \frac{kQ}{r^2}

E=VdE = \frac{V}{d}

💡Examples

Problem 1:

Calculate the magnitude of the electric field strength at a distance of r=0.20 mr = 0.20 \text{ m} from a point charge of Q=+5.0μCQ = +5.0 \mu\text{C} in a vacuum.

Solution:

Using the formula E=kQr2E = \frac{kQ}{r^2}: E=(8.99×109 N m2 C−2)×(5.0×10−6 C)(0.20 m)2E = \frac{(8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}) \times (5.0 \times 10^{-6} \text{ C})}{(0.20 \text{ m})^2} E=449500.04E = \frac{44950}{0.04} E=1.12×106 N C−1E = 1.12 \times 10^6 \text{ N C}^{-1}

Explanation:

The electric field strength is calculated by applying Coulomb's constant kk, the charge magnitude QQ in Coulombs, and the square of the distance rr. The direction would be radially outwards from the positive charge.

Problem 2:

An electron (charge e=−1.6×10−19 Ce = -1.6 \times 10^{-19} \text{ C}) is placed in a uniform electric field between two parallel plates separated by 0.05 m0.05 \text{ m} with a potential difference of 200 V200 \text{ V}. Determine the magnitude of the force acting on the electron.

Solution:

First, find the electric field strength EE: E=Vd=200 V0.05 m=4000 V m−1E = \frac{V}{d} = \frac{200 \text{ V}}{0.05 \text{ m}} = 4000 \text{ V m}^{-1} Then, calculate the force FF: F=Eq=(4000 V m−1)×(1.6×10−19 C)F = Eq = (4000 \text{ V m}^{-1}) \times (1.6 \times 10^{-19} \text{ C}) F=6.4×10−16 NF = 6.4 \times 10^{-16} \text{ N}

Explanation:

In a uniform field, EE is constant. The force on a charge is the product of the field strength and the magnitude of the charge.

Electric Fields - Revision Notes & Key Formulas | IB Grade 11 Physics