Question

The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer 1

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

$E = \frac{T}{P*sin(\theta)}= \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C$

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

Answer 2

The magnitude of electric field is 825.64 N/C.

• The torque exerted on the dipole is  given that,

                 $T = 7.3 *10^{-9} Nm$

• Dipole moment, $P=q*d$

      where q is charge and d is distance between charges.

           $P=4.10*10^{-9}*3.6*10^{-3} \\\\P=1.476*10^{-11} Nm$

The relation between torque(T) , electric field(E) and dipole moment(P) is,

                 $PEsin\theta=T\\\\E=\frac{T}{Psin\theta} =\frac{7.3*10^{-9} }{1.476*10^{-11} *sin36.8} \\\\E=825.64N/C$

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