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 ?

Electric field, E = 936.19 N/C

Explanation:

It is given that,

Charge 1,

Charge 2,

Distance between them, d = 3 mm = 0.003 m

Torque,

Angle between electric field and line connecting the charge,

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

p is the dipole moment,

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.

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explanation: