Question

The figure below shows a dipole. If the positive particle has a charge of 37.3 mC and the particles are 3.08 mm apart, what is the electric field at point A located 2.00 mm above the dipole's midpoint? (Express your answer in vector form.)

Answer 1

The electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.

Electric field of the positive particle

The electric field is calculated as follows;

E = kq/r²

where;

• r is the distance between the charges
• k is Coulomb's constant
• q is magnitude of the charge

midpoint of 3.08 m, x = 1.54 mm

r(1.54 mm, 2.00 mm)

|r| = √(1.54² + 2²)

|r| = 2.52 mm

E = (9 x 10⁹ x 37.3 x 10⁻³)/(2.52 x 10⁻³)²

E = 5.287 X 10¹³ N/C

Thus, the electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.

Learn more about electric field here: brainly.com/question/14372859

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