Question

Two charges $+3.2\times {10}^{-19}\text{C}$ and $-3.2\times {10}^{-19}\text{C}$ separated by $2r=2.4\mathring{A}$ forms an electric dipole. It is placed in a uniform electric field of $4\times {10}^5\text{V}/\text{m}$. The work done by electric force to rotate the electric dipole from the equilibrium position to ${180}^{\circ }$ is

• A. $-6\times {10}^{-23}\text{J}$
• B. $6\times {10}^{-23}\text{J}$
• C. $3\times {10}^{-23}\text{J}$
• D. Zero

Answer 1

The correct option is A $-6\times {10}^{-23}\text{J}$

Given:

$2r=2.4\mathring{A}=2.4\times {10}^{-10}\text{m}$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->

$E=4\times {10}^5\text{V}/\text{m}$

$q=3.2\times {10}^{-19}\text{C}$

We know, $p=q\left(2r\right)$

Also, work done by electric force,

$W=pE(\cos {\theta }_2-\cos {\theta }_1)...\left(1\right)$

${\theta }_1=0^{\circ };\cos 0^{\circ }=1{\theta }_2={180}^{\circ };\cos {180}^{\circ }=-1$

From $\left(1\right)$, we get,

$W=q\left(2r\right)E(\cos {180}^{\circ }-\cos 0^{\circ })$

$=3.2\times {10}^{-19}\times 2.4\times {10}^{-10}\times 4\times {10}^5\times (-1-1)$

$\Rightarrow W\approx -6\times {10}^{-23}\text{J}$