A electric dipole is placed at an angle of 300 with an electric field intensity 2×105N C−1 . It experiences a torque equal to 8 N m . The charge on the dipole, if the dipole length is 2 cm , is

Here, θ=30º,E=2×105NC−1 τ=8Nm,l=2cm=0.02m,q=?τ=PEsinθ=(ql)Esinθ∴q=τElsinθ=82×105×0.02×12=82×103=4×10−3C=4mC .

A electric dipole is placed at an angle of 300 with an electric field intensity 2×105N C−1 . It experiences a torque equal to 8 N m . The charge on the dipole, if the dipole length is 2 cm , is

Here, θ=30º,E=2×105NC−1 τ=8Nm,l=2cm=0.02m,q=?τ=PEsinθ=(ql)Esinθ∴q=τElsinθ=82×105×0.02×12=82×103=4×10−3C=4mC .

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A electric dipole is placed at an angle of $30^{0}$ with an electric field intensity $2 \times 10^{5} N C^{- 1}$ . It experiences a torque equal to $8 N m$ . The charge on the dipole, if the dipole length is $2 c m$ , is

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$8 m c$

$7 μ C$

$4 m c$

$5 m c$

answer is B.

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Detailed Solution

Here, $θ = 30 º, E = 2 \times 10^{5} N C^{- 1}$

$τ = 8 N m, l = 2 c m = 0.02 m, q = ?$

$τ = P E \sin θ = (q l) E \sin θ$

$∴ q = \frac{τ}{E l \sin θ} = \frac{8}{2 \times 10^{5} \times 0.02 \times \frac{1}{2}}$

$= \frac{8}{2 \times 10^{3}} = 4 \times 10^{- 3} C = 4 m C$ .

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