In: Physics
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 |
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.