In the figure, four charges, given in multiples of 5.00×10-6 C form the corners

Question

In the figure, four charges, given in multiples of 5.00×10-6 C form the corners of a square and four more charges lie at the midpoints of the sides of the square. The distance between adjacent charges on the perimeter of the square is d = 5.80×10-2 m.

What are the magnitude and direction of the electric field at the center of the square?

The magnitude of E?

What is the X component of the electric field (Ex) at the center of the square?

What is the Y component of the electric field (Ey) at the center of the square?

Explanation / Answer

by symmetry, field due to +3q charges at either corners will cancel each other.

similarly electric field due to +q charges at the midpoints will cancel each other.

coordinates of remaining charges are:

+5q: (-d,-d)

-2*q: (d,0)

+q: (-d,0)

-5*q: (d,d)

field due to +5q at center:

as the charge is positive in nature, field direction will be away from the charge.

vector along the field=(0,0)-(-d,-d)=(d,d)

distance=sqrt(2)*d

unit vector=vector / distance

=(0.707,0.707)

field magnitude=k*charge/distance^2

where k=coloumb's constant

field magnitude=k*5*q/(2*d^2)

=2.5*k*q/d^2

so field due to +5q charge=E1=(2.5*k*q/d^2)*(0.707,0.707)

field due to -5q at center:

as the charge is negative in nature, field direction will be towards the charge.

vector along the field=(d,d)-(0,0)=(d,d)

distance=sqrt(2)*d

unit vector=vector / distance

=(0.707,0.707)

field magnitude=k*charge/distance^2

where k=coloumb's constant

field magnitude=k*5*q/(2*d^2)

=2.5*k*q/d^2

so field due to +5q charge=E2=(2.5*k*q/d^2)*(0.707,0.707)

field due to +q at center:

as the charge is positive in nature, field direction will be away from the charge.

vector along the field=(0,0)-(-d,0)=(d,0)

distance=d

unit vector=vector / distance

=(1,0)

field magnitude=k*charge/distance^2

where k=coloumb's constant

field magnitude=k*q/(d^2)

=k*q/d^2

so field due to +q charge=E3=(k*q/d^2)*(1,0)

field due to -2q at center:

as the charge is negative in nature, field direction will be towards the charge.

vector along the field=(d,0)-(0,0)=(d,0)

distance=d

unit vector=vector / distance

=(1,0)

field magnitude=k*charge/distance^2

where k=coloumb's constant

field magnitude=k*2*q/(d^2)

=2*k*q/d^2

so field due to -2q charge=E4=(2*k*q/d^2)*(1,0)

total field=E1+E2+E3+E4=(k*q/d^2)*(6.535,3.535) N/C

using the values of the symbols,

net field=(9*10^9*5*10^(-6)/0.058^2)*(6.535,3.535)

=1.3377*10^7*(6.535,3.535) N/C

part a:

magnitude=1.3377*10^7*(sqrt(6.535^2+3.535^2))=9.93889*10^7 N/C

angle with +ve x axis=arctan(3.535/6.535)=28.41 degrees

part b:

magnitude is 9.93889*10^7 N/C

part c:

x component is 8.74187*10^7 N/C

part d:

y component is 4.72877*10^7 N/C

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