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Point charges q1=− 4.40 nC and q2=+ 4.40 nC are separated by distance 3.10 mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle of 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 ?

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Answer:

988.39 N/C

Step-by-step explanation:

First, we start by finding the magnitude of the electric dipole moment, as it's going to be needed

p = 4.40*10^-9 * 3.10*10^-3

p = 1.364*10^-11 Cm

The charges are in a uniform electric field E whose direction makes an angle of 36.4° with the line connecting the charges.

Magnitude of torque exerted on the dipole,

t = 8.0×10−9 N.m

pEsinO = t, making E subject of formula

E = t / psinO

E = 8.0*10^-9/1.364*10^-11 (sin36.4)

E = 8.0*10^-9/1.364*10^-11 * (0.5934)

E = 8.0*10^-9 / 8.094*10^-12

E = 988.39 N/C

the magnitude of this field is E = 988.39 N/C

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