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Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle θ with the axis of the dipole, then potential at the point P is given by
  • a)
    V = p cos θ/4 π ∈₀ r2
  • b)
    V = p cos θ/4 π ∈₀ r
  • c)
    V = p sn θ/4 π ∈₀ r
  • d)
    V = p cos θ/4 π ∈₀ r3
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Two equal charges q of opposite sign separated by a distance 2a consti...
Θ with the axis of the dipole, then the electric field at point P is given by:

E = (1/4πε₀) * [(2p cosθ)/(r² - a²)²]

where ε₀ is the permittivity of free space.

Explanation:

An electric dipole is a system of two equal and opposite charges separated by a distance d. The dipole moment p is defined as the product of the magnitude of either charge and the distance between them, i.e. p = qd.

In this case, we have two charges q of opposite sign separated by a distance 2a, so the dipole moment is given by:

p = q(2a) = 2qa

Now, let's consider a point P at a distance r from the centre of the dipole. We want to find the electric field at this point due to the dipole.

First, let's consider the electric field due to one of the charges at P. This is given by Coulomb's law:

E₁ = (1/4πε₀) * (q/r₁²)

where r₁ is the distance between the charge and point P.

Next, let's consider the electric field due to the other charge at P. This is also given by Coulomb's law:

E₂ = (1/4πε₀) * (-q/r₂²)

where r₂ is the distance between the other charge and point P. Note that the negative sign is because the charge is negative.

Now, the total electric field at point P due to the dipole is the vector sum of E₁ and E₂. However, since the charges are equal and opposite, their magnitudes are the same and their directions are opposite. Therefore, we can write:

E = E₁ + E₂ = (1/4πε₀) * [q/r₁² - q/r₂²]

Using the geometry of the problem, we can express r₁ and r₂ in terms of r and θ. We have:

r₁ = √(r² + a² - 2ar cosθ)

r₂ = √(r² + a² + 2ar cosθ)

Substituting these expressions into the equation for E, we get:

E = (1/4πε₀) * [q/(r² + a² - 2ar cosθ) - q/(r² + a² + 2ar cosθ)]

Now, we can simplify this expression using the identity:

1/(x - y) - 1/(x + y) = 2y/(x² - y²)

Applying this identity with x = r² + a² and y = 2ar cosθ, we get:

E = (1/4πε₀) * [(2qar cosθ)/(r⁴ - a⁴ - 2a²r²cos²θ)]

Finally, we can substitute the expression for p = 2qa and simplify to get the desired result:

E = (1/4πε₀) * [(2p cosθ)/(r² - a²)²]
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Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle θ with the axis of the dipole, then potential at the point P is given bya)V = p cos θ/4 π ∈ r2b)V = p cos θ/4 π ∈ rc)V = p sn θ/4 π ∈ rd)V = p cos θ/4 π ∈ r3Correct answer is option 'A'. Can you explain this answer?
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Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle θ with the axis of the dipole, then potential at the point P is given bya)V = p cos θ/4 π ∈ r2b)V = p cos θ/4 π ∈ rc)V = p sn θ/4 π ∈ rd)V = p cos θ/4 π ∈ r3Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle θ with the axis of the dipole, then potential at the point P is given bya)V = p cos θ/4 π ∈ r2b)V = p cos θ/4 π ∈ rc)V = p sn θ/4 π ∈ rd)V = p cos θ/4 π ∈ r3Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle θ with the axis of the dipole, then potential at the point P is given bya)V = p cos θ/4 π ∈ r2b)V = p cos θ/4 π ∈ rc)V = p sn θ/4 π ∈ rd)V = p cos θ/4 π ∈ r3Correct answer is option 'A'. Can you explain this answer?.
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