Answer
$E= 1.6 \times 10^{-19}V/m$
Work Step by Step
Given:
$\theta _{1}=90^{\circ}$
$\theta _{2}=0^{\circ}$
$p=6.2\times 10^{-30} cm=6.2 \times 10^{-32} m $
$U=1\times 10^{-21} J$
We know that:
$U_{minimum}=-pE$
The increment in P.E. is;
$\Delta U=U_{\theta _{1}}-U_{\theta _{2}}$
$\Delta U=-pEcos90^{\circ}-(-pEcos0^{\circ})$
$\Delta U=pE$
or
$E=\frac{\Delta U}{p}=\frac{1\times 10^{-21}}{6.2 \times 10^{-32} }$
Hence, $E= 1.6 \times 10^{-19}V/m$.
