Calculate the amount of work done in turning an electric dipole of dipole moment $3 \times 10^{- 8} C$ -m from its position of unstable equilibrium to the position of stable equilibrium in a uniform electric field of intensity $10^{3} N C^{- 1}$

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Solution

Formual,

$W = P E (1 - cos θ)$

$= 2 P E$ $∵ θ = 180 \to cos θ = - 1$

$= 2 \times 3 \times 10^{- 8} \times 10^{3}$

$= 6 \times 10^{5} J$

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Calculate the amount of work done in turning an electric dipole of dipole moment 3×10−8C-m from its position of unstable equilibrium to the position of stable equilibrium in a uniform electric field of intensity 103NC−1

Formual,W=PE(1−cosθ)=2PE ∵θ=180→cosθ=−1=2×3×10−8×103=6×105J

Calculate the amount of work done in turning an electric dipole of dipole moment $3 \times 10^{- 8} C$ -m from its position of unstable equilibrium to the position of stable equilibrium in a uniform electric field of intensity $10^{3} N C^{- 1}$

Formual,

$W = P E (1 - cos θ)$

$= 2 P E$ $∵ θ = 180 \to cos θ = - 1$

$= 2 \times 3 \times 10^{- 8} \times 10^{3}$

$= 6 \times 10^{5} J$