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
=
tan-1 [(0.047 m) / (0.094 m)]
= 26.5
degree
using a formula, we have
EA = ke Q / d2 = (9 x 109 Nm2/C2) (6.1 x 10-6 C) / (0.011045 m2)
EA = 4970.5 x 103 N/C
EA = 4.97 x 106 N/C
The x-component of an electric fied is given by,
EA,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,
EA,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
=
tan-1 [(0.047 m) / (0.141 m)]
= 18.4
degree
We know that, EB = ke Q / r2 = (9 x 109 Nm2/C2) (6.1 x 10-6 C) / (0.02209 m2)
EB = 2485.2 x 103 N/C
EB = 2.48 x 106 N/C
The x-component of an electric fied is given by,
EB,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,
EB,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
=
tan-1 [(0.047 m) / (0.047 m)]
= 45 degree
We know that, EB = ke Q / r2 = (9 x 109 Nm2/C2) (6.1 x 10-6 C) / (0.094 m2)
EB = 584.04 x 103 N/C
EB = 5.84 x 105 N/C
The x-component of an electric fied is given by,
EB,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,
EB,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 -
EB,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 -
EB,x = [(2.35 x 106 N/C) + (-4.12 x 105 N/C)]
EB,x = 1938000 N/C
Magnitude : EB = (1938000
N/C)2 + (1194000 N/C)2
EB = 2.27 x 106 N/C
Using a trigonometric identity, we have
=
tan-1 [(1194000 N/C) / (1938000 N/C)]
= 31.6
degree
Use Coulomb's law to determine the magnitude of the electric field at points A and B...
Problem 21.42 19 of 20 Use Coulomb's law to determine lhe magnitude ot the electric held at points A and B in the t ure Hgure 1 shown C2-6 5 μ due to the two positive charges α-5 3 cm Constants | Penodic lable Express your answers using two significant figures separated by a comma. BA,B-4 106,10.26 10 N/C Submit Figure 1 of 1 xIncorrect; Try Agaln; 5 attempts remalning Part B Use Coulomb's law to determine the direction of...
Problem 16.35 Use Coulomb's law to determine the magnitude of the electric field at points A and B in due to the two positive charges Q 6.5 uC) shown. Suppose that a 4.8 cm 2a Part A Determine the magnitude of the electric field at A Express your answer to two significant figures and include the appropriate units. ANSWER Part B Determine the angle between the direction of the electric field at point A and the positive r-direction. Express your...
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...
Use Coulomb's law to determine the magnitude o f the electric
field at points A and B in (Figure 1) due to the two positive
charges (Q = 6.9 μC ) shown. Suppose that
a = 5.3 cm . Determine the magnitude of the electric field
at A. Determine the angle between the direction of the electric
field at point A and the positive x-direction. Determine
the magnitude of the electric field at point B. Determine the angle
between the...
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 = 4.3 μC ) shown. Suppose that a = 4.9 cm . Determine
the magnitude of the electric field at point A and point
B.
y B A x |-- a + +Q a a 2a
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 = 6.0 µC) shown.
Magnitude at point A___N/C
Magnitude at point B___N/C
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 = 6.3 μC ) shown. Suppose that a = 5.0 cm .
Problem 16.35
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 = 4.1 ?C ) shown. Suppose that
a = 4.9 cm .
Part A
Determine the magnitude of the electric field at A.
Part B
Determine the angle between the direction of the electric field
at point A and the positive x-direction.
Part C
Determine the magnitude of the electric field at...
12:00 PM Fri Jan 25 Assignments-20191-PHY207-OEK-University Physics Problem 21.42 1 of 5> Constants Periodic Table Part A Use Coulomb's law to determine the magnitude of the electric fiold at points A and B in the figure due to the two positive charges( Express your answers using two significant figures separated by a comma B A 2a N/C Part B Use Coulomb's law to determine the direction of the electric field at points A and B Express your answers using two...
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 = 5.3 μC ) shown. Suppose that a = 5.5 cm.Determine the magnitude of the electric field at point B.Determine the angle between the direction of the electric field
at point B and the positive x-direction.