Question

Use Coulomb's law to determine the magnitude of the electric field at points A and B in (Figure 1) due to the two positive charges (Q = 7.0 μC ) shown. Suppose that a = 5.9 cm .

a.Determine the magnitude of the electric field at A.

b.Determine the angle between the direction of the electric field at point A and the positive x-direction.

c.Determine the magnitude of the electric field at point B.

d.Determine the angle between the direction of the electric field at point B and the positive x-direction.

Answer 1

(a)

At point A -

From the figure,

tan$\theta$ = 5.9 / (2*5.9) = 5.9 / 11.8

$\theta$ = 26.565 deg

Magnitude of electric field due to both the charges,

E = 2(kQ / d^2)*sin$\theta$

where, d = sqrt (5.9² + 11.8²) = 13.192 cm

d = 0.1319 m

EA = 2*9*10⁹*7*10^-6*sin (26.56) / (0.1319)^2

E_A = 3.237*10⁶ N/C

(b)

angle between the direction of the electric field at point A and the positive x-direction = 90 deg

(c)

At point B,

Angle at left charge, $\theta$ 1 = tan^-1 (5.9 / 5.9) = 45 deg

Angle at right charge, $\theta$ 2 = tan^-1 (5.9 / 3*5.9) = 18.43 deg

d1 = sqrt (0.059² + 0.059²) = 0.08343 m

d2 = sqrt (0.059 + 0.177) = 0.186 m

net electric field at point B in x direction,

Ex = kQcos$\theta$1 / d1^2 - kQcos$\theta$2 / d2^2

Ex = 9*10⁹*7*10^-6 [(cos45 / 0.08343²) - (cos 18.43 / 0.186²)]

Ex = 4.68*10⁶ N/C

net electric field at point B in y direction,

Ey = kQsin$\theta$1 / d1^2 + kQsin$\theta$2 / d2^2

Ey = 9*10⁹*7*10^-6 [(sin 45 / 0.08343²) + (sin 18.43 / 0.186²)]

Ey = 6.97*10⁶ N/C

Magnitude of electric field at point B,

E_B = sqrt (Ex^2 + Ey^2)

E_B = 8.398*10⁶ N/C

(d)

angle between the direction of the electric field at point B and the positive x-direction

tan$\theta$ =  6.97*10⁶ / 4.68*10⁶

$\theta$ = 56.13 deg