Respuesta :
Answer:
Electric field, E = 936.19 N/C
Explanation:
It is given that,
Charge 1, [tex]q_1=-4.8\ nC=-4.8\times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=+4.8\ nC=+4.8\times 10^{-9}\ C[/tex]
Distance between them, d = 3 mm = 0.003 m
Torque, [tex]\tau=8\times 10^{-9}\ N-m[/tex]
Angle between electric field and line connecting the charge, [tex]\theta=36.4^{\circ}[/tex]
We need to find the torque exerted on the dipole. The torque experienced by the dipole in the electric field is given by :
[tex]\tau=pE\ sin\theta[/tex]
p is the dipole moment, [tex]p=qd[/tex]
[tex]\tau=qdE\ sin\theta[/tex]
[tex]E=\dfrac{\tau}{qd\ sin\theta}[/tex]
[tex]E=\dfrac{8\times 10^{-9}}{4.8\times 10^{-9}\times 0.003\ sin(36.4)}[/tex]
E = 936.19 N/C
So, the magnitude of electric field on the dipole is 936.19 N/C. Hence, this is the required solution.
