Question

Use Coulomb's law to determine the magnitude of the electric field at points A and B in the figure due to the two positive charges (Q = 6.1 mu C, a = 47 cm) shown. Express your answers using two significant figures separated by a comma. Use Coulomb's law to determine the direction of the electric field at points A and B Express your answers using two significant figures separated by a comma. This question will be shown after you complete previous question(s).

Answer 1

At points A, we have

d² = (2a)² + a² = (0.094 m)² + (0.047 m)²

d² = 0.011045 m²

using a trigonometric identity, we have

$\theta$ = tan^-1 [(0.047 m) / (0.094 m)]

$\theta$ = 26.5 degree

using a formula, we have

E_A = k_e Q / d² = (9 x 10⁹ Nm²/C²) (6.1 x 10^-6 C) / (0.011045 m²)

E_A = 4970.5 x 10³ N/C

E_A = 4.97 x 10⁶ N/C

The x-component of an electric fied is given by,

E_A,x = (4.97 x 10⁶ N/C) cos 26.5⁰ = 4.44 x 10⁶ N/C

The y-component of an electric fied is given by,

E_A,y = (4.97 x 10⁶ N/C) sin 26.5⁰ = 2.21 x 10⁶ N/C

Since the magnitude left will eventually cancel out with the magnitude coming from the Charge on other side of A, we can ignore that number.

The Charge on left side will yield the same magnitude up. So at this point, we can just multiply by two to get the answer -

E = 2.21 x 10⁶ N/C

At points B, we have

For right side ; r² = (3a)² + a² = [(0.141 m)² + (0.047 m)²]

r² = 0.02209 m²

using a trigonometric identity, we have

$\theta$ = tan^-1 [(0.047 m) / (0.141 m)]

$\theta$ = 18.4 degree

We know that,   E_B = k_e Q / r² = (9 x 10⁹ Nm²/C²) (6.1 x 10^-6 C) / (0.02209 m²)

E_B = 2485.2 x 10³ N/C

E_B = 2.48 x 10⁶ N/C

The x-component of an electric fied is given by,

E_B,x = (2.48 x 10⁶ N/C) cos 18.4⁰ = 2.35 x 10⁶ N/C      (left)

The y-component of an electric fied is given by,

E_B,y = (2.48 x 10⁶ N/C) sin 18.4⁰ = 0.782 x 10⁶ N/C      (up)

For left side ;    r² = [(0.047 m)² + (0.047 m)²]

r² = 0.094 m²

using a trigonometric identity, we have

$\theta$ = tan^-1 [(0.047 m) / (0.047 m)]

$\theta$ = 45 degree

We know that,   E_B = k_e Q / r² = (9 x 10⁹ Nm²/C²) (6.1 x 10^-6 C) / (0.094 m²)

E_B = 584.04 x 10³ N/C

E_B = 5.84 x 10⁵ N/C

The x-component of an electric fied is given by,

E_B,x = (5.84 x 10⁵ N/C) cos 45⁰ = 4.12 x 10⁵ N/C       (right)

The y-component of an electric fied is given by,

E_B,y = (5.84 x 10⁵ N/C) sin 45⁰ = 4.12 x 10⁵ N/C      (up)

To sum the y-component of an electric field -

E_B,y = [(0.782 x 10⁶ N/C) + (4.12 x 10⁵ N/C) = 1194000 N/C

To sum the x-component of an electric field -

E_B,x = [(2.35 x 10⁶ N/C) + (-4.12 x 10⁵ N/C)]

E_B,x = 1938000 N/C

Magnitude :    E_B = _$\sqrt{}$(1938000 N/C)² + (1194000 N/C)²

E_B = 2.27 x 10⁶ N/C

Using a trigonometric identity, we have

$\theta$ = tan^-1 [(1194000 N/C) / (1938000 N/C)]

$\theta$ = 31.6 degree