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Question: For a dipole \(q = 2 \times 10^{- 6}C\) and \(d = 0.01m\). Calculate the maximum torque for this dip...

For a dipole q=2×10−6Cq = 2 \times 10^{- 6}C and d=0.01md = 0.01m. Calculate the maximum torque for this dipole if E=5×105N/CE = 5 \times 10^{5}N/C

A

1×10−3Nm−11 \times 10^{- 3}Nm^{- 1}

B

10×10−3Nm−110 \times 10^{- 3}Nm^{- 1}

C

10×10−3Nm10 \times 10^{- 3}Nm

D

1×102Nm21 \times 10^{2}Nm^{2}

Answer

10×10−3Nm10 \times 10^{- 3}Nm

Explanation

Solution

τmax⁡\tau_{\max} =2×10−6×0.01×5×105= 2 \times 10^{- 6} \times 0.01 \times 5 \times 10^{5}

=10×10−3N−m= 10 \times 10^{- 3}N - m