An electric dipole is placed at an angle of 300 with an electric field intensity 2×105 NC-1. It experiences a torque equal to 4 Nm. Find the length of the dipole if charge on dipole is 2 mC.

Here, θ=30°,E=2×105 NC-1τ=4 N m, q=2 mC = 2 × 10-3 C,L=?τ=pEsinθ=(qL)Esinθ∴L=τEqsinθ=42×105×0.002×12 ⇒L = 0.02 m = 2 cm

An electric dipole is placed at an angle of 300 with an electric field intensity 2×105 NC-1. It experiences a torque equal to 4 Nm. Find the length of the dipole if charge on dipole is 2 mC.

Here, θ=30°,E=2×105 NC-1τ=4 N m, q=2 mC = 2 × 10-3 C,L=?τ=pEsinθ=(qL)Esinθ∴L=τEqsinθ=42×105×0.002×12 ⇒L = 0.02 m = 2 cm

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An electric dipole is placed at an angle of $30^{0}$ with an electric field intensity $2 \times 10^{5} {NC}^{- 1}$ . It experiences a torque equal to $4 Nm$ . Find the length of the dipole if charge on dipole is 2 mC.

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8 cm

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4 cm

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Detailed Solution

$Here, θ = 30 °, E = 2 \times 10^{5} {NC}^{- 1}$

$\begin{array}{l} τ = 4 N m, q = 2 mC = 2 \times 10^{- 3} C, \\ L = ? \end{array}$

$τ = pEsinθ = (qL) Esinθ$

$∴ L = \frac{τ}{Eqsinθ} = \frac{4}{2 \times 10^{5} \times 0.002 \times \frac{1}{2}} \Rightarrow L = 0.02 m = 2 cm$

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An electric dipole is placed at an angle of 300 with an electric field intensity 2×105 NC-1. It experiences a torque equal to 4 Nm. Find the length of the dipole if charge on dipole is 2 mC.

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An electric dipole is placed at an angle of $30^{0}$ with an electric field intensity $2 \times 10^{5} {NC}^{- 1}$ . It experiences a torque equal to $4 Nm$ . Find the length of the dipole if charge on dipole is 2 mC.