Question

10) The dipole moment of a dipole in a 300N/C electric field is initially perpendicular to the field, but it rotates so it is in same direction as field. If moment has magnitude of 2x10^-9 Cm. work done by field is
?????

Answer 1

Answer:

• Work Done (W) by the field is - 6 × 10⁻⁷ J

Given:

1. Electric field (E) = 300 N/C
2. Dipole moment (p) = 2 × 10⁻⁹ Cm

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\large\bigstar\;\underline{\boxed{\sf U=- pE\;\cos\theta}}$

Here,

• p Denotes Dipole moment.
• E Denotes Electric field.
• θ Denotes angle b/w them.

Now, as given, firstly the dipole is perpendicular to the electric field, so angle (θ₁) will be 90°. and now the dipole is rotated such that they are in same direction so the angle (θ₂) will be 0°.

So, let's find Change in Potential energy which will be equal to the work done by the electric field.

$\longmapsto\sf \Delta U=U_{f}-U_{i}\\\\\\\\\longmapsto\sf \Delta U=\Bigg[-pE\;\cos \theta_{2}\Bigg]-\Bigg[-pE\;\cos \theta_{1}\Bigg]\\\\\\\\\longmapsto\sf \Delta U=\Bigg[-pE\;\cos \theta_{2}\Bigg]+pE\;\cos \theta_{1}\\\\\\\\\longmapsto\sf \Delta U=pE\times \Bigg[-\cos \theta_{2}+\cos \theta_{1}\Bigg]$

• Substituting the values,

$\longmapsto\sf \Delta U=pE\times \Bigg[-\cos 0^{\circ}+\cos 90^{\circ}\Bigg]\\\\\\\\\longmapsto\sf \Delta U=pE\times \Bigg[-1+0\Bigg]\\\\\\\\\longmapsto\sf \Delta U=-pE\\\\\\\\\longmapsto\sf \Delta U=-2\times 10^{-9}\times 300\\\\\\\\\longmapsto\sf \Delta U=-6\times 10^{-9+2}\\\\\\\\\longmapsto\sf \Delta U=-6\times 10^{-7}\\\\\\\\\longmapsto\sf W=\Delta U=-6\times 10^{-7}\\\\\\\\\longmapsto\large{\underline{\boxed{\red{\sf W=-\; 6\times 10^{-7}\;J}}}}$

∴ Work Done (W) by the field is - 6 × 10⁻⁷ J.

$\rule{300}{1.5}$

Answer 2

Answer:

-6×10^-7 J

Explanation:

is the correct answer