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    Current Subject
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    Applied Physics
    GE-169
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    Topics
    1. Electric Force and Its Applications2. Conservation of Charge3. Charge Quantization4. Electric Fields Due to Point Charge and Lines of Force5. Electric Fields: Ring of Charge and Disk of Charge6. A Point Charge in an Electric Field7. Dipole in an Electric Field8. Flux of a Vector Field9. Flux of an Electric Field10. Gauss’ Law and Its Applications11. Spherically Symmetric Charge Distribution12. Charge Isolated Conductor13. Electric Potential Energy14. Electric Potentials and Related Problems15. Calculating Potential from the Field16. Potential Due to Point and Continuous Charge Distribution17. Potential Due to a Dipole18. Equipotential Surfaces19. Calculating the Field from the Potential20. Electric Current and Current Density21. Resistance, Resistivity, and Conductivity22. Ohm's Law and Its Applications23. The Hall Effect24. Magnetic Force on a Current25. The Biot-Savart Law26. Line of Magnetic Field (B)27. Two Parallel Conductors28. Ampere's Law29. Solenoids and Toroids30. Faraday's Experiments and Law of Induction31. Lenz's Law32. Motional EMF33. Induced Electric Fields34. The Basic Equations of Electromagnetism35. Induced Magnetic Fields36. The Displacement Current37. Reflection and Refraction of Light Waves38. Total Internal Reflection39. Two Source Interference40. Double-Slit Interference and Related Problems41. Interference from Thin Films42. Diffraction and Wave Theory43. Single-Slit Diffraction and Related Problems44. Polarization of Electromagnetic Waves45. Polarizing Sheets and Related Problems
    GE-169›Calculating the Field from the Potential
    Applied PhysicsTopic 19 of 45

    Calculating the Field from the Potential

    10 minread
    1,749words
    Intermediatelevel

    Calculating the Electric Field from the Electric Potential

    In electrostatics, the electric field E⃗\vec{E}E and electric potential VVV are closely related. The electric field is the negative gradient of the electric potential. This relationship allows us to calculate the electric field if we know the electric potential, and vice versa.

    1. General Relationship Between Electric Field and Potential

    The electric field E⃗\vec{E}E at a point is the negative gradient of the electric potential VVV. In mathematical terms, this is:

    E⃗=−∇V\vec{E} = - \nabla VE=−∇V

    This equation expresses that the electric field points in the direction of greatest decrease of the potential, and the magnitude of the field is proportional to how rapidly the potential changes with distance.

    Components of the Gradient in Cartesian Coordinates:

    If the potential VVV is a function of the spatial coordinates xxx, yyy, and zzz, the gradient in Cartesian coordinates is given by:

    ∇V=i^∂V∂x+j^∂V∂y+k^∂V∂z\nabla V = \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z}∇V=i^∂x∂V​+j^​∂y∂V​+k^∂z∂V​

    Therefore, the electric field components are:

    E⃗=−i^∂V∂x−j^∂V∂y−k^∂V∂z\vec{E} = - \hat{i} \frac{\partial V}{\partial x} - \hat{j} \frac{\partial V}{\partial y} - \hat{k} \frac{\partial V}{\partial z}E=−i^∂x∂V​−j^​∂y∂V​−k^∂z∂V​

    In Spherical Coordinates:

    In spherical coordinates (rrr, θ\thetaθ, ϕ\phiϕ), where the potential VVV depends on the radial distance rrr, the polar angle θ\thetaθ, and the azimuthal angle ϕ\phiϕ, the gradient is:

    ∇V=r^∂V∂r+θ^1r∂V∂θ+ϕ^1rsin⁡θ∂V∂ϕ\nabla V = \hat{r} \frac{\partial V}{\partial r} + \hat{\theta} \frac{1}{r} \frac{\partial V}{\partial \theta} + \hat{\phi} \frac{1}{r \sin \theta} \frac{\partial V}{\partial \phi}∇V=r^∂r∂V​+θ^r1​∂θ∂V​+ϕ^​rsinθ1​∂ϕ∂V​

    So, the electric field in spherical coordinates is:

    E⃗=−r^∂V∂r−θ^1r∂V∂θ−ϕ^1rsin⁡θ∂V∂ϕ\vec{E} = - \hat{r} \frac{\partial V}{\partial r} - \hat{\theta} \frac{1}{r} \frac{\partial V}{\partial \theta} - \hat{\phi} \frac{1}{r \sin \theta} \frac{\partial V}{\partial \phi}E=−r^∂r∂V​−θ^r1​∂θ∂V​−ϕ^​rsinθ1​∂ϕ∂V​

    2. Calculating Electric Field from Potential: Examples

    Now, let's look at how to calculate the electric field from the potential in specific situations:

    a. Electric Field Due to a Point Charge

    The potential due to a point charge QQQ at a distance rrr from the charge is:

    V(r)=14πϵ0QrV(r) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}V(r)=4πϵ0​1​rQ​

    To find the electric field, we take the negative gradient of the potential. In spherical coordinates, the gradient of V(r)V(r)V(r) with respect to rrr is:

    ∂V∂r=−14πϵ0Qr2\frac{\partial V}{\partial r} = - \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}∂r∂V​=−4πϵ0​1​r2Q​

    So, the electric field E⃗\vec{E}E is:

    E⃗=−r^(−14πϵ0Qr2)\vec{E} = - \hat{r} \left( - \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \right)E=−r^(−4πϵ0​1​r2Q​)

    Thus, the electric field due to a point charge is:

    E⃗=14πϵ0Qr2r^\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \hat{r}E=4πϵ0​1​r2Q​r^

    This is Coulomb's law, which states that the electric field due to a point charge is radially outward (for a positive charge) or inward (for a negative charge) and decreases with the square of the distance.

    b. Electric Field Due to a Uniformly Charged Infinite Line

    For a uniformly charged infinite line with linear charge density λ\lambdaλ, the potential at a distance rrr from the line is:

    V(r)=λ2πϵ0ln⁡(1r)V(r) = \frac{\lambda}{2 \pi \epsilon_0} \ln \left( \frac{1}{r} \right)V(r)=2πϵ0​λ​ln(r1​)

    To find the electric field, we take the gradient of V(r)V(r)V(r) with respect to rrr:

    ∂V∂r=−λ2πϵ0r\frac{\partial V}{\partial r} = - \frac{\lambda}{2 \pi \epsilon_0 r}∂r∂V​=−2πϵ0​rλ​

    The electric field is then:

    E⃗=−r^(−λ2πϵ0r)\vec{E} = - \hat{r} \left( - \frac{\lambda}{2 \pi \epsilon_0 r} \right)E=−r^(−2πϵ0​rλ​)

    Thus, the electric field due to an infinite line of charge is:

    E⃗=λ2πϵ0rr^\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r}E=2πϵ0​rλ​r^

    This electric field points radially outward from the line if λ>0\lambda > 0λ>0 and inward if λ<0\lambda < 0λ<0, and it decreases with distance from the line.

    c. Electric Field Due to a Uniformly Charged Disk

    The potential at a point along the axis of a uniformly charged disk with surface charge density σ\sigmaσ and radius RRR is:

    V(z)=σ2ϵ0(R2+z2−z)V(z) = \frac{\sigma}{2 \epsilon_0} \left( \sqrt{R^2 + z^2} - z \right)V(z)=2ϵ0​σ​(R2+z2​−z)

    To find the electric field along the axis, we take the derivative of the potential V(z)V(z)V(z) with respect to zzz:

    dVdz=σ2ϵ0(zR2+z2−1)\frac{dV}{dz} = \frac{\sigma}{2 \epsilon_0} \left( \frac{z}{\sqrt{R^2 + z^2}} - 1 \right)dzdV​=2ϵ0​σ​(R2+z2​z​−1)

    So, the electric field along the axis of the disk is:

    E(z)=−dVdz=σ2ϵ0(1−zR2+z2)E(z) = - \frac{dV}{dz} = \frac{\sigma}{2 \epsilon_0} \left( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right)E(z)=−dzdV​=2ϵ0​σ​(1−R2+z2​z​)

    This shows that the electric field on the axis of a uniformly charged disk decreases as zzz increases and approaches a value similar to the field of a point charge at large zzz.


    3. Electric Field from Potential for More Complex Distributions

    For more complex charge distributions, the general process for calculating the electric field from the potential involves:

    1. Writing down the expression for the potential: This could be due to a point charge, a charged sphere, a charged line, or any other configuration of charges.
    2. Taking the gradient: You compute the gradient of the potential to find the electric field.
    3. Interpreting the result: The resulting electric field may be in vector form or expressed in terms of radial, angular, or Cartesian components.

    For example, in the case of a spherically symmetric charge distribution, you would calculate the electric potential for that distribution, and the electric field can be found by taking the derivative of that potential with respect to rrr.


    4. Summary

    • The electric field E⃗\vec{E}E can be calculated from the electric potential VVV using the relationship: E⃗=−∇V\vec{E} = - \nabla VE=−∇V
    • The gradient operation tells us how the potential changes in space, and the electric field points in the direction of greatest decrease of the potential.
    • In Cartesian coordinates, the electric field components are found by taking partial derivatives of the potential with respect to xxx, yyy, and zzz.
    • In spherical coordinates, the gradient is expressed in terms of rrr, θ\thetaθ, and ϕ\phiϕ, and the electric field is calculated accordingly.
    • Examples: For a point charge, the electric field is Qr2\frac{Q}{r^2}r2Q​; for an infinite line charge, it is λr\frac{\lambda}{r}rλ​; and for a uniformly charged disk, it can be computed along the axis using the derivative of the potential.

    This relationship between electric field and potential is a fundamental concept in electrostatics and is widely used in solving problems related to electric fields and potentials.

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    Electric Current and Current Density

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