Question

An electric dipole experiences a torque of 10√3 Nm. ∅1 = 60° with a electric field of magnitude 10³ N/C
Find the charge q if it's length is 3 × 10^-2 m​

Answer 1

${\boxed{\mathcal{Electric\:Field\:and\:Torque}}}$

✔Given that

• Torque = 10√3 Nm
• Angle = 60°
• Electric field = 10³ N/C
• Length = 3 × 10^-2 m

→We know that

$\tau$ = P × E [ Cross Product of Vectors ]

→ PE Sin∅

$10 \sqrt{3} = 10 {}^{3} \times sin \: 60° × P$

[ Sin 60° = √3/2 ]

From this ,

=> P = 2 × 10^-2

•°• ${\bold{\underline{The\:Dipole\:Movement\:is}}}$ → 2 × 10^-2

Now , the dipole movement (P) is equal to the product of the charge and small distance between them. i,e

♦P = q.d

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

=> q = 2/3 = 0.66 C

•°• ${\sf{\underline{The\:magnitude \:of\:the\:Charge \:is\:0.66\:C}}}$

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Answer 2

Answer:

0.66 coloum charge!!!!!!