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).

Ex =  N/C
Ey =  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,

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

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

Ex = - 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,

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

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

Ey = 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 = (Ex2 + Ey2)1/2

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

Direction of the electric field at this point is,

= tan-1 (Ey/Ex)

= 126.030 measured anticlockwise with respect to +x axis.


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