Two charges +3.2 times 10^{-19} , C and -3.2 | vedclass

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(B) The potential energy of an electric dipole in an external electric field is given by $U = -\vec{p} \cdot \vec{E} = -pE \cos 	heta$.In stable equi

Vedclass · Accepted Answer

$-3 	imes 10^{-23} \, J$

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(B) The potential energy of an electric dipole in an external electric field is given by $U = -\vec{p} \cdot \vec{E} = -pE \cos 	heta$.In stable equilibrium,the dipole moment $\vec{p}$ is aligned with the electric field $\vec{E}$,so $	heta = 0^\circ$.Given: $q = 3.2 	imes 10^{-19} \, C$,$2l = 2.4 \, \mathring{A} = 2.4 	imes 10^{-10} \, m$,and $E = 4 	imes 10^5 \, V/m$.The dipole moment $p = q 	imes 2l = (3.2 	imes 10^{-19}) 	imes (2.4 	imes 10^{-10}) = 7.68 	imes 10^{-29} \, C \cdot m$.Substituting these values into the potential energy formula:$U = -(7.68 	imes 10^{-29}) 	imes (4 	imes 10^5) 	imes \cos(0^\circ)$$U = -30.72 	imes 10^{-24} \, J = -3.072 	imes 10^{-23} \, J$.Rounding to the nearest value,we get $U \approx -3 	imes 10^{-23} \, J$.

Two charges $+3.2 \times 10^{-19} \, C$ and $-3.2 \times 10^{-19} \, C$ kept $2.4 \, \mathring{A}$ apart form a dipole. If it is kept in a uniform electric field of intensity $4 \times 10^5 \, V/m$,what will be its electrical potential energy in stable equilibrium?

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