Question

In: Physics

A point particle with charge q = 4.2 ?C is placed on the x axis at...

A point particle with charge

q = 4.2 ?C

is placed on the x axis at

x = ?10 cm

and a second particle of charge

Q = 7.8 ?C

is placed on the x axis at

x = +25 cm.

(a) Determine the x and y components of the electric field due to this arrangement of charges at the point

(x, y) = (10, 10)

(the units here are centimeters).

E_x	= N/C
E_y	= N/C

(b) Determine the magnitude and direction of the electric field at this point.

magnitude	N/C
direction

Solutions

Expert Solution

q = 4.2 * 10^-6 C at x₁ = -10 cm = - 0.1 m

Q = 7.8 * 10^-6 C at x₁ = 25 cm = 0.25 m

k = 8.99 * 10⁹ N.m² / C²is the Coulomb's force constant.

(a) Determine the x and y components of the electric field due to this arrangement of charges at the point

(x, y) = (10, 10)

To determine the angle, we use the formula

= tan^-1 (y/x)

X component of the electric field due to this arrangement of charges at (x, y) = (10, 10) is,

E_x = kq cos 26.57/ [(10+10)² + (10)²] + kQ cos 146.31 / [(25-10)² + (10)²]

E_x = 8.99 * 10⁹ * 10^-6* [4.2 * 0.894 / 500 + 7.8 * -0.832 / 325 ]

E_x = - 1.12 * 10^-4 N/C = - 0.000112 N/C

Y component of the electric field due to this arrangement of charges at (x, y) = (10, 10) is,

E_y = kq sin 26.57/ [(10+10)² + (10)²] + kQ sin 146.31 / [(25-10)² + (10)²]

E_y = 8.99 * 10⁹ * 10^-6* [4.2 * 0.447 / 500 + 7.8 * 0.555 / 325 ]

E_y = 1.54 * 10^-4 N/C = 0.000154 N/C

(b) Determine the magnitude and direction of the electric field at this point.

Magnitude of the electric field at this point is,

E = (E_x² + E_y²)^1/2

E = 1.904 * 10^-4 N/C = 0.00019 N/C

Direction of the electric field at this point is,

= tan^-1 (E_y/E_x)

= 126.03⁰ measured anticlockwise with respect to +x axis.

genius_generous answered 8 months ago

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