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    Electric Field and Potential

    Flashcards for topic Electric Field and Potential

    Intermediate52 cardsphysics

    Preview Cards

    Card 1

    Front

    Using vector calculus, establish the relationship between electric field (E⃗\vec{E}E) and electric potential (V). How can you determine the electric field if you know the potential distribution?

    Back

    The relationship is: E⃗=−∇V\vec{E} = -\nabla VE=−∇V, where ∇\nabla∇ is the gradient operator

    In Cartesian coordinates:

    • Ex=−∂V∂xE_x = -\frac{\partial V}{\partial x}Ex​=−∂x∂V​
    • Ey=−∂V∂yE_y = -\frac{\partial V}{\partial y}Ey​=−∂y∂V​
    • Ez=−∂V∂zE_z = -\frac{\partial V}{\partial z}Ez​=−∂z∂V​

    Key insights:

    • Electric field points in direction of maximum potential decrease
    • Field magnitude equals rate of potential change in that direction
    • Perpendicular to equipotential surfaces (where V is constant)
    • Units consistency: V/m = N/C
    • Zero field where potential has zero gradient (local extrema)

    Application: To find field from potential, calculate partial derivatives of V with respect to coordinates.

    Card 2

    Front

    Compare and contrast the distance dependencies of electric potential and electric field for: (1) point charge, (2) electric dipole, and (3) quadrupole. What physical insight does this reveal?

    Back

    Distance dependencies:

    1. Point charge (monopole):

      • Potential: V ∝ 1/r
      • Field: E ∝ 1/r²
    2. Electric dipole:

      • Potential: V ∝ 1/r²
      • Field: E ∝ 1/r³
    3. Quadrupole:

      • Potential: V ∝ 1/r³
      • Field: E ∝ 1/r⁴

    Physical insight:

    • Each higher-order multipole moment has faster spatial decay
    • At sufficient distance, lowest non-zero multipole dominates
    • Neutral objects with dipole moments influence surroundings less extensively than charged objects
    • Complex charge distributions behave approximately as point charges at large distances
    Card 3

    Front

    What physical principles determine the shape and direction of electric field lines, and how can you determine the relative field strength in different regions?

    Back

    Principles determining field line shape and direction:

    • Field lines start on positive charges and end on negative charges
    • Direction indicates the force experienced by a positive test charge
    • Lines never cross (would indicate multiple field directions at one point)
    • The number of lines from a charge is proportional to the charge magnitude
    • In free space, field lines extend to infinity unless terminated on opposite charges

    Field strength determination:

    • Higher density of field lines indicates stronger field
    • Field strength is proportional to the number of lines crossing a unit area perpendicular to the lines
    • Mathematically: E∝NAE \propto \frac{N}{A}E∝AN​ where N is the number of lines and A is the area
    • Field strength decreases with distance: E∝1r2E \propto \frac{1}{r^2}E∝r21​ for point charges
    • Where field lines converge, field strength increases
    • Where field lines diverge, field strength decreases

    Example: Between two opposite charges, the field lines are closest together midway between charges of equal magnitude, indicating the strongest field in that region.

    Card image
    Card 4

    Front

    What is the relationship between electric field lines and equipotential surfaces for a point charge?

    Back

    • Electric field lines and equipotential surfaces are perpendicular to each other at every point • For a point charge:

    • Electric field lines radiate directly outward (or inward) from the charge
    • Equipotential surfaces form concentric spheres around the charge • The electric field points in the direction of maximum potential decrease • No work is done moving a test charge along an equipotential surface • The closer the equipotential surfaces are to each other, the stronger the electric field

    Example: Moving a test charge in a circular path around a point charge requires zero work because you stay on the same equipotential surface.

    Card image
    Card 5

    Front

    How do small displacements in different directions relate to the electric field of a dipole, and how is this used to determine field components?

    Back

    In analyzing the electric field of a dipole, small displacements reveal the field components:

    1. Relationship between displacement and field:

      • For any small displacement dl⃗\vec{dl}dl, the potential change is dV=−E⃗⋅dl⃗dV = -\vec{E} \cdot \vec{dl}dV=−E⋅dl
      • This means El=−dVdlE_l = -\frac{dV}{dl}El​=−dldV​ for any direction lll
    2. Key displacements for a dipole:

      • Radial displacement (drdrdr): Changes only the distance rrr while keeping angle θ\thetaθ constant
      • Transverse displacement (rdθrd\thetardθ): Changes only the angle θ\thetaθ while keeping rrr constant
    3. Resulting field components:

      • Radial: Er=−∂V∂rE_r = -\frac{\partial V}{\partial r}Er​=−∂r∂V​
      • Transverse: Eθ=−1r∂V∂θE_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta}Eθ​=−r1​∂θ∂V​

    This method is particularly powerful because it allows us to derive vector field components by taking derivatives of a scalar potential function along different directions.

    Example: At a 45° angle from a dipole, both components contribute equally to the total field, which has a magnitude of p4πϵ0r35\frac{p}{4\pi\epsilon_0 r^3}\sqrt{5}4πϵ0​r3p​5​.

    Card image
    Card 6

    Front

    Derive the expression for the torque acting on an electric dipole placed in a uniform electric field.

    Back

    τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E}τ=p​×E

    Derivation:

    1. For a dipole with charges +q at B and -q at A with midpoint O:
    2. Force on +q: F⃗1=qE⃗\vec{F}_1 = q\vec{E}F1​=qE
    3. Force on -q: F⃗2=−qE⃗\vec{F}_2 = -q\vec{E}F2​=−qE
    4. Torque from +q: τ⃗1=OB⃗×qE⃗\vec{\tau}_1 = \vec{OB} \times q\vec{E}τ1​=OB×qE
    5. Torque from -q: τ⃗2=OA⃗×(−qE⃗)=q(AO⃗×E⃗)\vec{\tau}_2 = \vec{OA} \times (-q\vec{E}) = q(\vec{AO} \times \vec{E})τ2​=OA×(−qE)=q(AO×E)
    6. Net torque: τ⃗=q(OB⃗×E⃗)+q(AO⃗×E⃗)=q(OB⃗+AO⃗)×E⃗=qAB⃗×E⃗=p⃗×E⃗\vec{\tau} = q(\vec{OB} \times \vec{E}) + q(\vec{AO} \times \vec{E}) = q(\vec{OB} + \vec{AO}) \times \vec{E} = q\vec{AB} \times \vec{E} = \vec{p} \times \vec{E}τ=q(OB×E)+q(AO×E)=q(OB+AO)×E=qAB×E=p​×E

    Magnitude: τ=pEsin⁡θ\tau = pE\sin\thetaτ=pEsinθ, where θ is the angle between dipole axis and field.

    Card 7

    Front

    Given that the potential energy of an electric dipole in a uniform field is U(θ)=−p⃗⋅E⃗U(\theta) = -\vec{p}\cdot\vec{E}U(θ)=−p​⋅E, derive this expression starting from work principles.

    Back

    Derivation:

    1. The torque on a dipole is τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E}τ=p​×E with magnitude τ=pEsin⁡θ\tau = pE\sin\thetaτ=pEsinθ
    2. For a small rotation dθ, work done by the torque is dW=τdθ=pEsin⁡θ dθdW = \tau d\theta = pE\sin\theta\,d\thetadW=τdθ=pEsinθdθ
    3. This work is negative as rotation dθ opposes the torque
    4. Change in potential energy: dU=−dW=pEsin⁡θ dθdU = -dW = pE\sin\theta\,d\thetadU=−dW=pEsinθdθ
    5. For change from θ = 90° to some angle θ: U(θ)−U(90°)=∫90°θpEsin⁡θ dθ=pE[−cos⁡θ]90°θ=−pEcos⁡θU(\theta) - U(90°) = \int_{90°}^{\theta}pE\sin\theta\,d\theta = pE[-\cos\theta]_{90°}^{\theta} = -pE\cos\thetaU(θ)−U(90°)=∫90°θ​pEsinθdθ=pE[−cosθ]90°θ​=−pEcosθ
    6. If we set U(90°)=0U(90°) = 0U(90°)=0 (zero potential energy when dipole perpendicular to field): U(θ)=−pEcos⁡θ=−p⃗⋅E⃗U(\theta) = -pE\cos\theta = -\vec{p}\cdot\vec{E}U(θ)=−pEcosθ=−p​⋅E

    This shows the potential energy is minimized when the dipole aligns with the field (θ = 0°).

    Card 8

    Front

    If I have an electric dipole made of charges +q and -q separated by distance d, and I double both the charge and halve the separation distance, how does the:

    1. Dipole moment
    2. Electric field at a distant point
    3. Potential energy in a uniform field change compared to the original configuration?

    Back

    Original dipole: charges ±q separated by distance d Modified dipole: charges ±2q separated by distance d/2

    1. Dipole moment:

      • Original: porig=qdp_{orig} = qdporig​=qd
      • Modified: pmod=2q⋅d2=qdp_{mod} = 2q \cdot \frac{d}{2} = qdpmod​=2q⋅2d​=qd
      • Result: Dipole moment remains unchanged
    2. Electric field at a distant point:

      • Field is proportional to dipole moment: E∝pr3E \propto \frac{p}{r^3}E∝r3p​
      • Since p is unchanged, the electric field remains identical at all points
    3. Potential energy in uniform field:

      • Potential energy: U=−p⃗⋅E⃗U = -\vec{p}\cdot\vec{E}U=−p​⋅E
      • Since p is unchanged, the potential energy in any given field remains identical

    This demonstrates that different charge configurations can produce identical external electromagnetic effects if they have the same dipole moment.

    Card 9

    Front

    What is the expression for the torque acting on an electric dipole placed in a uniform electric field when the dipole axis makes an angle θ with the field?

    Back

    The torque acting on an electric dipole in a uniform electric field is:

    τ = p⃗ × E⃗

    Where:

    • p⃗ is the dipole moment vector
    • E⃗ is the uniform electric field vector

    The magnitude of the torque is: τ = pE sin θ

    Where:

    • p is the magnitude of the dipole moment
    • E is the magnitude of the electric field
    • θ is the angle between the dipole axis and the field

    The direction of the torque is perpendicular to the plane containing the dipole axis and the electric field.

    Card image
    Card 10

    Front

    Explain the mechanism of charge redistribution in conductors when exposed to an external electric field.

    Back

    When a conductor is exposed to an external electric field:

    1. Initial state: Free electrons are uniformly distributed throughout the conductor
    2. When field is applied: Free electrons experience a force opposite to the field direction
    3. Electron movement:
      • Electrons drift toward the side facing the positive direction of the field
      • This creates a negative charge density on that surface
      • The side facing the field's negative direction develops a positive charge density due to electron deficiency
    4. Result: Surface charge accumulation creates an internal opposing field
    5. Equilibrium: Reached when the internally generated field exactly cancels the external field

    This process demonstrates how conductors shield their interior from external electric fields.

    Card image

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