Question

An electric dipole consist of two opposite charges of magnitude 1/ 3 × 10^-7 c separated by 2cm dipole is placed in an external field of 3× 10^ 7N/ c what maximum torque does the electric field exert on the dipole

Answer 1

Answer:

0.02 Nm

Explanation:

Torque = P E sinx

For maximum we have to take; sinx = 1

Since it is a vector T¬ = P¬ × E¬

T = 1/3×3×0.02 ( 2cm= 0.02m)

and cancelling tenth powers

T = 0.02 Nm

Answer 2

Formula for finding TORQUE:

$\tau$ = P x E x sinΘ

and, P = 2 x d x q

• A pair of equally spaced, opposite point-charges [q] is what makes up an electric dipole.

Given in the question:

1. Magnitude of opposite charges:                   $q = \frac{1}{3} \times 10^{-7}$C
2. Magnitude of External Field:                         $E =3 \times10^{7} NC^{-1}$
3. Distance between opposite charges:          $2d = 2 \times 10^{-2} m$

Torque exerted by electric field E on dipole P can be written as:$\tau = P\times E\times \sin \Theta$

Torque will be maximum when $\sin \Theta$  will be maximum, i.e. Θ = 90°

So,

   $\tau_{max} = p \times E \times (sin\theta)_{max}$

⇒ $\tau_{max} = (2 \times d \times q) \times E \times 1$

            = $2 \times 10^{-2} Nm$

Therefore,

The external field exerts a maximum torque of $2 \times 10^{-2} Nm$ on the Dipole.

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