Tamil Nadu Board of Secondary EducationHSC Science Class 12

Calculate the electric field due to a dipole on its axial line and equatorial plane. - Physics

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Long Answer

Calculate the electric field due to a dipole on its axial line and equatorial plane.

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Solution

Case (i):

Electric field due to an electric dipole at points on the axial line.

  1. Consider an electric dipole placed on the x-axis A point C is located at a distance of r from the midpoint of the dipole along the axial line.

    Electric field of the dipole along the axial line
  2. The electric field at a point C due to +q is
    `vec"E"_+ = 1/(4 pi ε_0) "q"/("r - a")^2` (along BC)
  3. Since the electric dipole moment vector p is from -q to +q and is directed along BC, the above equation is rewritten as
    `vec"E"_+ = 1/(4 pi ε_0) "q"/("r - a")^2 hat"P"`   ....(1)
  4. 4. The electric field at a point C due to -q is
    `vec"E"_- = -1/(4 pi ε_0) "q"/("r + a")^2 hat"P"`   ....(2)
  5. Since +q is located closer to the point C than -q, `vec"E"_+` is stronger than `vec"E"_-`.
    Therefore, the length of the `vec"E"_+` vector is drawn larger than that of `vec"E"_-` vector.
    The total electric field at point C is calculated using the superposition principle of the electric field.
    `vec"E"_"tot" = vec"E"_+ + vec"E"_-`
    `= 1/(4pi ε_0) "q"/("r - a")^2 hat"p" - 1/(4pi ε_0) "q"/("r + a")^2 hat"p"`
    `vec"E"_"tot" = "q"/(4pi ε_0) (1/("r - a")^2 - 1/("r + a")^2)hat"p"`  .....(3)
    `vec"E"_"tot" = "q"/(4pi ε_0) ("4ra"/("r"^2 - "a"^2)^2) hat"p"`    .....(4)
    r >> a
    hence (r2 - a2)2 ≈ r4
    Substitute in eqn (4)
    `vec"E"_"tot" = 1/(4pi ε_0)("4aq"/"r"^3)hat "p"`
    `2"aq"hat"p" = vec"p"`
    `vec"E"_"tot" = 1/(4pi ε_0) (2vac"p")/"r"^3`
  6. The direction of is shown in Figure

    Total electric field of the dipole on the axial line

Case (ii)

  1. Electric field due to an electric dipole at a point on the equatorial plane.
  2. Consider a point C at a distance r from the midpoint 0 of the dipole on the equatorial plane.
  3. Since point C is equidistant from +q and -q and are the same.
  4. The direction of `vec"E"_+` is along BC and the direction of `vec"E"_-` is along CA.
    `vec"E"_+` and `vec"E"_-` are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.
  5. The perpendicular components `|vec"E"|` sinθ and `|vec"E"_-|` sinθ are oppositely directed and cancel each other.
    `"E"_"tot" = - |"E"_+| cos theta hat "p" - |vec"E"_-| cos theta hat "p"`  .....(6)
  6. The magnitudes `vec"E"_+ and vec"E"_-` are the same and are given by
  7. By substituting equation (7) into equation (6) we get
    `vec"E"_"tot" = - 1/(4piε_0) (2"q" cos theta)/("r"^2 + "a"^2) hat"p"`
    since cos θ = `"a"/sqrt("r"^2 + "a"^2) hat`
    =`- 1/(4piε_0) (2"qa")/("r"^2 +"a"^2)^(3/2) hat"p"` 
    since `vec"p" = "2qa"hat"p"`   ....(8)
    `vec"E"_"tot" = - 1/(4piε_0) vec"p"/("r"^2 +"a"^2)^(3/2)`

    `|"E"_+| = |vec"E"_-| = 1/(4piε_0) "q"/("r"^2 + "a"^2)`   .....(7)
  8. At very large distances (r»a), the equation (8) becomes
    `vec"E"_"tot" = - 1/(4piε_0) vec"p"/"r"^3` (r >> a)
Concept: Electric Dipole and Its Properties
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Chapter 1: Electrostatics - Evaluation [Page 74]

APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 12th Physics Volume 1 and 2 Answers Guide
Chapter 1 Electrostatics
Evaluation | Q III. 4. | Page 74
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