Question

Point charges q1=− 4.30 nC and q2=+ 4.30 nC are separated by distance 4.00 mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.3 ∘ 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?

Answer 1

The magnitude of the electric field is 5.70x10^4 N/C.

Step-by-step explanation:

To calculate the magnitude of the electric field, we can use the formula:

τ = pEsinθ

Where τ is the torque on the dipole, p is the dipole moment, E is the magnitude of the electric field, and θ is the angle between the direction of the electric field and the line connecting the charges.

In this case, the torque is given as 7.30x10^-9 N⋅m, the dipole moment is qd, where q is the charge magnitude (4.30 nC) and d is the separation distance (4.00 mm).

Plugging in the values, we have:

7.30x10^-9 N⋅m = (4.30x10^-9 C)(4.00x10^-3 m)Esin(36.3°)

Solving for E, we find:

E = 5.70x10^4 N/C