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What are the strength and direction of the electric field at the position indicated by the dot in FIGURE P23.35? Give your answer (a) in component form and (b) as a magnitude and angle measured cw or ccw (specify which) from the positive x-axis.

Short Answer

Expert verified

a. The electric field at the dot is1.0×105N/Ci^-3.6×104N/Cj^.

b. The angle made by the electric filed is 19.4°counter clockwise direction from the positive X-axis.

Step by step solution

01

Formula for electric field

The electric field is,

E=14πε0qr2

permittivity of free space is ε0.

02

Diagram for electric field position

In the diagram below, two positive charges and one negative charge are located in the three corners of a rectangle.

The electric field generated by the positive charge at the dot is directed away from the positive charge. The electric field generated by the negative charge at the dot is directed towards the negative charge.

03

Explanation (part a)

a.

Calculate the component form of the electric field strength at the place given by the dot in the diagram.

The angle formed by the vertical and the diagonal is,

θ=tan-12.0m4.0m

=26.56°

The component form of the resultant electric field at the dot is as follows:

Er=Exi^+Eyj^

From figure,

Ex=E1-E3cosθ

Ey=E3sinθ-E2

04

Explanation (part a) solution

a.

x-component of the net electric field at the dot.

The magnitude of the electric field due to the charge one is

E1=14πε0q1r12

The magnitude of the electric field due to the charge two is,

E2=14πε0q2r22

The magnitude of the electric field due to the charge three is,

role="math" localid="1651413236226" E3=14πε0q3r32

05

Explanation (part a)

The value,

r3=(2.0cm)2+(4.0cm)2

=4.47cm

Ex=14πε0q1r12-14πε0q3r32cosθ

So,

Ex=9×109C2/N·m2(5.0nC)(2.0cm)2-9×109C2/N·m2(5.0nC)(4.47cm)2cos26.56°

=9×109C2/N·m25.0nC10-9C1nC2.0cm1.0m100cm2-5.0nC10-9C1nC4.47cm1.0m100cm2cos26.56°

=1.0×105N/C

06

Explanation (part a)

For ycomponent,

Ey=14πε0q3r32sinθ-14πε0q2r22

So,

Ey=9×109C2/N·m2(5.0nC)(4.47cm)2(sin26.56)-(10.0nC)(4.0cm)2

=9×109C2/N·m25.0nC10-9C1nC4.47cm1.0m100cm2(sin26.56)-10.0nC10-9C1nC4.0cm1.0m100cm2

=-3.6×104N/C

So electric dot is,

Er=1.0×105N/Ci^+-3.6×104N/Cj^

07

Explanation (part b)

b.

Magnitude of the electric field at the dot is given by the expression,

E=Ex2+Ey2

Substitute all values,

E=1.0×105N/C2+-3.6×104N/C2

=1.1×105N/C

08

Explanation part b

angle of electric field is,

θ'=tan-1EyEx

θ'=tan-1-3.6×104N/C1.0×105N/C

=-19.4°

Here, negative sign indicates counter clockwise direction from the positivex-axis.

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