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Fields - Electrostatic Fields

Grade 12IBPhysics

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

πŸ”‘Concepts

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Coulomb's Law states that the electrostatic force FF between two point charges q1q_1 and q2q_2 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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Electric Field Strength (EE) at a point is defined as the electrostatic force per unit charge experienced by a small positive test charge placed at that point. It is a vector quantity with units N Cβˆ’1N\,C^{-1} or V mβˆ’1V\,m^{-1}.

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Electric Potential (VV) at a point is the work done per unit charge in bringing a small positive test charge from infinity to that point. It is a scalar quantity measured in Volts (VV), where 1 V=1 J Cβˆ’11\,V = 1\,J\,C^{-1}.

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Electric Potential Energy (EpE_p) is the energy a charge possesses due to its position in an electric field, calculated as Ep=qVE_p = qV.

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Equipotential surfaces are surfaces where the electric potential is constant. Electric field lines are always perpendicular to equipotential surfaces, and no work is done when moving a charge along such a surface (W=qΞ”V=0W = q\Delta V = 0).

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The Electric Field Strength can also be expressed as the negative gradient of the potential: E=βˆ’Ξ”VΞ”rE = -\frac{\Delta V}{\Delta r}. This indicates that the electric field points in the direction of decreasing potential.

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In a uniform electric field (such as between two parallel charged plates), the field strength is constant and given by E=VdE = \frac{V}{d}.

πŸ“Formulae

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

k=14πϡ0β‰ˆ8.99Γ—109 N m2 Cβˆ’2k = \frac{1}{4\pi\epsilon_0} \approx 8.99 \times 10^9 \, N\,m^2\,C^{-2}

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

E=kQr2E = k\frac{Q}{r^2}

V=kQrV = k\frac{Q}{r}

Ep=kq1q2rE_p = k\frac{q_1 q_2}{r}

W=qΞ”VW = q\Delta V

E=βˆ’Ξ”VΞ”rE = -\frac{\Delta V}{\Delta r}

E=VdΒ (forΒ uniformΒ fields)E = \frac{V}{d} \text{ (for uniform fields)}

πŸ’‘Examples

Problem 1:

Calculate the electric field strength at a distance of 0.20 m0.20\,m from a point charge of +4.0Γ—10βˆ’9 C+4.0 \times 10^{-9}\,C.

Solution:

E=kQr2=(8.99Γ—109)4.0Γ—10βˆ’9(0.20)2=899 N Cβˆ’1E = k\frac{Q}{r^2} = (8.99 \times 10^9) \frac{4.0 \times 10^{-9}}{(0.20)^2} = 899 \, N\,C^{-1}

Explanation:

We use the formula for the electric field strength of a point charge. Substitute the Coulomb constant kk, the charge QQ, and the square of the distance rr.

Problem 2:

An electron (q=βˆ’1.6Γ—10βˆ’19 Cq = -1.6 \times 10^{-19}\,C) is accelerated from rest through a potential difference of 500 V500\,V. Determine its final kinetic energy.

Solution:

Ξ”Ek=W=qΞ”V=(1.6Γ—10βˆ’19 C)(500 V)=8.0Γ—10βˆ’17 J\Delta E_k = W = q\Delta V = (1.6 \times 10^{-19}\,C)(500\,V) = 8.0 \times 10^{-17}\,J

Explanation:

The work done by the electric field on the electron is equal to its gain in kinetic energy. Since we are looking for the magnitude of energy, we use the magnitude of the electron's charge.

Problem 3:

Two parallel plates are separated by 0.05 m0.05\,m and connected to a 12 V12\,V battery. What is the magnitude of the electric field between the plates?

Solution:

E=Vd=120.05=240 V mβˆ’1E = \frac{V}{d} = \frac{12}{0.05} = 240 \, V\,m^{-1}

Explanation:

In a uniform electric field between parallel plates, the field strength is the potential difference divided by the separation distance.

Electrostatic Fields - Revision Notes & Key Formulas | IB Grade 12 Physics