Question

Charged particles q1=− 4.80 nC and q2=+ 4.80 nC are separated by distance 3.00 mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.4° with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 8.00×10^−9 N⋅m ?

Answer 1

Answer:

Electric field, E = 936.19 N/C

Explanation:

It is given that,

Charge 1, $q_1=-4.8\ nC=-4.8\times 10^{-9}\ C$

Charge 2, $q_2=+4.8\ nC=+4.8\times 10^{-9}\ C$

Distance between them, d = 3 mm = 0.003 m

Torque, $\tau=8\times 10^{-9}\ N-m$

Angle between electric field and line connecting the charge, $\theta=36.4^{\circ}$

We need to find the torque exerted on the dipole. The torque experienced by the dipole in the electric field is given by :

$\tau=pE\ sin\theta$

p is the dipole moment, $p=qd$

$\tau=qdE\ sin\theta$

$E=\dfrac{\tau}{qd\ sin\theta}$

$E=\dfrac{8\times 10^{-9}}{4.8\times 10^{-9}\times 0.003\ sin(36.4)}$

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.