Question

A positive charge of 5 nC is located 2 cm below the origin on y-axis, while a negative charge of -5 nC is located 2 cm above the origin on y-axis. What is the electric field of this electric dipole at the point (x=10 m, y=0) ?

         a. 1.8x10-3 N/c i                         b. - 1.8x10-3 N/c i

         c. 1.8x10-3 N/c j                         d. - 1.8x10-3 N/c j

The answer is C but how???

Answer 1

Use Coulomb's law

F = K q1 q2 / r²          F_T = F1+ F2

We assume that there is a positive charge of q₀ C at the point of (10, 0)

q1= 5 10^-9 C      r1= -2 10^-2 m j

. q2= -5 10^-9 C     r2= 2 10^-2 m j

. q3 = q₀ C             r3= 10 i + 0 j           Carga de prueba

K = 1/4πe₀ = 8.9875 10⁹ N m²/C²        

F13 = 8.98 10⁹   5 10^-9 q₀ /(10² + (-0.02)²) = 0.449/q_o N /C

F23 = 8.98 10⁹    (-5 10^-9) q₀ / (10² + 0.02²) = -0.449/q₀ N/C

The direction is

. tg angle = 0.02/10 = 0.01/5 angle =0.115°

The magnitude of force is the same for two load, so the X component is canceled and the component and adds that it is in the same direction

F_T = Fx i + Fy j =   0 i + 2 Fy

Fy= F13 Sin (0.115) = F23 Sin 0.115 =8.98 10^-4 /q₀ N

F_T = 1.76 10^-3 N