ELECTROSTATICS: Electric Field at the Center of an Equilateral Triangle

Q: Three point charges located at the corners of an imaginary equilateral triangle carry charges of +8 µC, +3 µC, and -5 µC, respectively. A distance of 0.5 m separates the charges from one another. What net electric field ( E-field ) will a positive test charge experience when placed at the triangle’s center?

A: The solution protocol involves the following steps:

( a ) Determining the direction/orientation of each charge.

( b ) Determining the distance separating each charge from the test charge.

( c ) If necessary, breaking E-fields into their x/y-components.

( d ) Adding E-field x/y components.

( e ) Using the Pythagorean Theorem to obtain a net E-field.

( f ) Using the tan ( θ ) function to obtain an angle for ( E_net ) relative to the x-axis.

Electric field values ( E ) represent how much force in Newtons ( N ) a charge ( or charges ) will experience at a given location. The orientation of each field above is determined by the charge that gives rise to it. A positive test charge is repelled by ( q₁ ) and ( q₂ ), and it is attracted to ( q₃ ). We need not concern ourselves with the signs of each charge when determining the magnitude of each E-field; however, the force ( F ) experienced by the test charge is a vector quantity, and thus, the direction of will indeed be dependent upon what charge is placed there:

E = F/q

F = Eq

We determine the magnitude of each E-field with the following equation:

E = kq/r² 

The triangle can be split into two right triangles. After this is done, the distance from the top of the triangle to the test charge can be labeled. Furthermore, the ( 0.5 m ) distance separating each point charge must be taken into account:

In a previous derivation, it was determined that the point charge ( x ) is located two-thirds of the distance ( d ) separating ( q₂ ) from the bottom of the triangle. The corner of each triangle is made of two sides separated by 60⁰ angles. As a consequence, the distance from ( q₂ ) to ( x ) is as follows:

cos θ = adj/hyp = ( d/0.5 m )

d = ( 0.5 m )( cos θ )

⅔ d = ( ⅔ )( 0.5 m )( cos θ )

⅔ d = ( ⅓ cos θ )

θ = ½ 60⁰ = 30⁰

⅔ d = ( ⅓ )( √3/2 )

⅔ d = √3/6

⅔ d = 0.289 m

The radial distance ( r = 0.289 m ) holds true for each point charge, and we use E = kq/r² to calculate the E-field created by each charge on ( x ):

E₁ = [ ( 9.0 x 10⁹ kg*m³s^-2C^-2 )( 8.0 x 10^-6 C ) ] / ( 0.289 m )² 

E₁ = ( 7.2 x 10⁴ kg*m³s^-2C^-1 / 8.35 x 10^-2 m² )

E₁ = ( 7.2 x 10⁴ kg*m³s^-2C^-1 / 8.35 x 10^-2 m² )

E₁ = 8.62 x 10⁵ N/C

The remaining E-field values are as follows:

E₂ = 3.24 x 10⁵ N/C

E₃ = 5.40 x 10⁵ N/C

Notice that fields E₁ and E₃ are both diagonal to the x-axis and y-axis. For this reason, we must break these fields into x/y-components. After doing so, we add these components together to obtain the E_net expressed in x/y-components. By symmetry, E₁ and E₃ are positioned above and below the x-axis by a magnitude of 30⁰:

It is immediately apparent that the x/y-components of E₁ are both oriented in the +x and +y-directions:

E_1x = ( E₁ )( cos 30⁰ )

E_1x = ( E₁ )( √3/2 )

E_1x = ( 8.62 x 10⁵ N/C )( √3/2 )

E_1x = 7.47 x 10⁵ N/C

We now use the ( sin θ ) function to find the value of the y-component of E₁:

E_1y = ( E₁ )( sin 30⁰ )

E_1y = ( 8.62 x 10⁵ N/C )( 0.5 )

E_1y = 4.31 x 10⁵ N/C

We use the same approach to find the x/y-components of E₃ . Notice, however, that the y-component of E₃ is oriented in the -y-direction. Similarly, the E₂ field is oriented in the downward direction. We must keep this in mind when doing our final summation of y-components:

E_3x = ( E₃ )( cos 30⁰ )

E_3x = ( E₃ )( √3/2 )

E_3x = ( 5.40 x 10⁵ N/C )( √3/2 )

E_3x = 4.68 x 10⁵ N/C

We now use the ( sin θ ) function to find the value of the y-component of E₁:

E_3y = ( E₃ )( sin 30⁰ )

E_3y = ( 5.40 x 10⁵ N/C )( 0.5 m )

E_3y = 2.70 x 10⁵ N/C

We are now ready to add the x-components of the E-fields to find the net E_{x net} value:

E_{x net} = ( E_1x + E_2x + E_3x )i

E_{x net} = [ ( 7.47 x 10⁵ N/C ) + ( 0.00 N/C ) + ( 4.68 x 10⁵ N/C ) ]i

E_{x net} = ( 1.22 x 10⁶ N/C )i

Adding the y-components of each E-field yields the following results:

E_{y net} = ( E_1y + E_2y + E_3y )j

E_{y net} = [ ( 4.31 x 10⁵ N/C ) – ( 3.24 x 10⁵ N/C ) – ( 2.70 x 10⁵ N/C ) ]j

E_{y net} = – ( 1.63 x 10⁵ N/C )j

We now find the net E-field by adding the x/y-components of E:

E_net = E_{x net} + E_{y net} 

E_net = ( 1.22 x 10⁶ N/C )i – ( 1.63 x 10⁵ N/C )j 

The ( + i ) and ( – j ) notation tell us that E_net points diagonally South of East. We use the Pythagorean Theorem to determine the magnitude of E_net:

 E_net = √[ ( 1.22 x 10⁶ N/C )² + ( 1.63 x 10⁵ N/C )² ]

E_net = √[ ( 1.49 x 10¹² N²/C² ) + ( 2.66 x 10¹⁰ N²/C² ) ]

E_net = √( 1.52 x 10¹² N²/C² )

E_net = 1.23 x 10⁶ N/C

We subsequently use the ( tan θ ) function to determine by what degree E_net falls below the x-axis:

tan θ = opp/adj

θ = tan^-1 ( opp/adj )

θ = tan^-1 ( E_{y net} / E_{x net} )

θ = tan^-1 ( – [ 1.63 x 10⁵ N/C ] / [ 1.22 x 10⁶ N/C ] )

θ = tan^-1 – ( 0.134 )

θ = -7.61^o