Question

For a dipole having magnitude of charge q = 4x

10-6 C and distance between the charges d =

0.02 m. If this dipole is placed in a electric field

E = 2 x 105 N/C. Then maximum torque on the

dipole will be

1.6 x 10-² Nm

3.2 x 10-2 Nm

2.4 x 10-2 Nm

4.8 x 10-2 Nm​

Answer 1

Answer:

1.6 x 10^(-2) Nm

Explanation:

Dipole moment(p) = 2lq = dq

Dipole moment = (0.02)(4 x 10^(-6)) C m

Dipole moment = 8 x 10^(-8) C m

Electric field(E) = 2 x 10^(5) N/C

Using torque = p x E (vector product)

=> torque = pEsin∅

For torque to be maximum, sin∅ must be maximum i.e. 1.

=> max. torque = pE(1)

=> max. torque = 8 x 10^(-8) × 2 x 10^(5)

=> max. torque = 1.6 x 10^(-2) Nm

Answer 2

Given that , Charge on dipole , q = 4 × 10⁻⁶ , Distance between the charges , d = 0.02 m & Electrical feild Intensity , E = 2 × 10⁵ N/C .

Exigency To Find : The maximum torque on the dipole will be ??

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Calculating Dipole moment ( p ) :

As , We know that ,

• Dipole Moment is equal to :

$\qquad \dag\:\:\bigg\lgroup \pmb{\sf Dipole \: Moment \:(\:p\:)\: = \: 2lq \:\:or\:\: dq \:\:C.m \: }\bigg\rgroup$

Where ,

• d is the Distance between the charge & ,
• q is the charge on dipole.

$\qquad \dashrightarrow \sf Dipole \: Moment \:(\:p\:)\: = \: dq \:\:C.m \:\\\\ \qquad \dashrightarrow \sf Dipole \: Moment \:(\:p\:)\: = \: \bigg( 0.02 \:\{ 4 \times 10^{-6}\:\}\bigg ) \:C.m \:\\\\\qquad \dashrightarrow \sf Dipole \: Moment \:(\:p\:)\: = \: \bigg( 8 \times 10^{-8} \bigg) \:C.m \:\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\sf{\sf Dipole \: Moment \:(\:p\:)\: = \: 8 \times 10^{-8} \:C.m }}}}\:\:\bigstar\:$

Now ,

• Using Torque :

$\qquad \dag\:\:\bigg\lgroup \pmb{\sf Torque \:\: = \: p \:\times E \: ( vector \:product\:) \:\:N.m \: }\bigg\rgroup$

Where ,

• p is a dipole moment &,
• E is the Electrical feild Intensity.

$\qquad \dashrightarrow \sf Torque \:\: = \: p \:\times E \: ( vector \:product\:) \:\:N.m\\\\ \qquad \dashrightarrow \sf Torque \:\: = \: p \:\times E \: \times sin \:\theta \:\:N.m$

• We have to find maximum torque so sin θ must be maximum , i.e. 1

$\qquad \dashrightarrow \sf Torque \:\: = \: p \:\times E \: \times sin \:\theta \:\:N.m\\\\ \qquad \dashrightarrow \sf Maximum_{(Torque)}\:\: = \: p \:\times E \: \times 1 \:\:N.m\\\\ \qquad \dashrightarrow \sf Maximum_{(Torque)}\:\: = \:\bigg( \{ 4 \times 10^{-8} \} \:\times \{ 2 \times 10^5\: \} \bigg)\times 1 \:\:N.m\\\\ \qquad \dashrightarrow \sf Maximum_{(Torque)}\:\: = \: \{ 4 \times 10^{-8} \} \:\times \{ 2 \times 10^5\: \} \:\:N.m\\\\ \qquad \dashrightarrow \sf Maximum_{(Torque)}\:\: = \: 1.6 \times 10^{-2} \: \:\:N.m\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\frak{ Maximum_{(Torque)}\:\: = \: 1.6 \times 10^{-2} \:\: N.m }}}}\:\:\bigstar\:$

$\qquad \therefore \:\underline {\sf Hence, \:\:The\:\:Maximum \:\:Torque \:\:will \:\: be \:\pmb{\bf \:Option\:\:A\:)\: 1.6 \times 10^{-2} \:\: N.m }\:.}$