Two equal and opposite charges of 2× 10-10C are placed at a distance of 1 cm forming a dipole and are placed in an electric field of 2× 105 N/C. The maximum torque on dipole is:

Question

Two equal and opposite charges of 2× 10-10C are placed at a distance of 1 cm forming a dipole and are placed in an electric field of 2× 105 N/C. The maximum torque on dipole is: (A)  2√2× 10-6Nm  (B)  8× 108Nm  (C)  4× 10-9Nm  (D)  4× 10-7Nm

Tardigrade Admin · Accepted Answer

From the formula for torque vecτ = vecp× vecE=pE sin θ For maximum value of torque sin θ =90o=1 Hence, vecτ max =pEq.2l=(2× 10-10× 1× 10-2) × (2× 105 N/C) =4× 10-7 Nm

Q. Two equal and opposite charges of $2 \times 10^{- 10} C$ are placed at a distance of 1 cm forming a dipole and are placed in an electric field of $2 \times 10^{5}$ N/C. The maximum torque on dipole is:

$22 \times 10^{- 6} N m$

$8 \times 10^{8} N m$

$4 \times 10^{- 9} N m$

$4 \times 10^{- 7} N m$

From the formula for torque $τ = p \times E = pE sin θ$ For maximum value of torque $sin θ = 90^{o} = 1$ Hence, $τ_{m a x} = pEq .2 l = (2 \times 10^{- 10} \times 1 \times 10^{- 2})$ $\times (2 \times 10^{5} N / C)$ $= 4 \times 10^{- 7} N m$

Q. Two equal and opposite charges of 2×10−10C are placed at a distance of 1 cm forming a dipole and are placed in an electric field of 2×105 N/C. The maximum torque on dipole is:

22​×10−6Nm

8×108Nm

4×10−9Nm

4×10−7Nm