Q: Two subatomic particles have a charge ( q1 = q2 = 10-6 C ), and they are located on the x-axis at coordinates ( -1m, 0m ) and ( 1m, 0m ). Please calculate the following:
- The electric field due to the charges when a positive test charge ( P ) has x/y-coordinates of ( 0.0 m, 1.0 m ).
- The force experienced by a charge ( qo ) of -2.0 x 10-9 C placed at ( P ).
- The quantity of charge ( q3 ) that, when placed at the ( 0m, 0m ) origin, makes the net electric field value at ( P ) become zero.
Givens: k = 9.0 x 109 Nm2/C2
A: An electric field ( E ) measurement determines what force per unit charge ( q ) a positive test charge will experience within a region of space. If a charge that gives rise to an E-field is positive, the test charge will experience a repulsive force ( and vice versa ). This is represented above as E-fields that point away from the test charge located at ( P ); furthermore, the E-fields are collinear with the radial ( r ) distance vectors separating each point charge from ( P ).
The radii in question can be thought of as multiples of a “ base length “ referred to as a unit vector ( u̅ ). Unit vectors always point away from the charges that create an E-field towards the point ( P ) where we want to calculate the field. A unit vector has a value of 1, and it is used to assign a direction to the larger vector within which it resides:
It is important to note that unit vectors have magnitude ( scalar ) and direction ( vector ) components. In the diagram above, the unit vector ( u̅1 ) has a magnitude ( r ) that is equal to the vector ( r̅1 ):
u̅r1 = ( r̅1 / r )
The process of creating one vector from another is called normalization. As seen above, a unit vector is created via division by its magnitude.
The value ( r̅1 ) can be broken into x/y components. The standard notation of an x-component positioned on the +x-axis is ( +i ), and the standard notation of a y-component positioned on the +y-axis is ( +j ), and vice versa. The values of ( i ) and ( j ) are derived by subtracting the x/y coordinates of each point charge from the x/y coordinates of ( P ):
r̅1 = [ 0m – ( -1m )i + ( 1m – 0m )j ]
r̅1 = i + j
This tip-to-tail vector representation will allow us to simplify the problem at hand as follows:
u̅r1 = ( r̅1 / r )
u̅r1 = [ ( i + j ) / r ]
We may now use the Pythagorean Theorem to derive a numerical value for ( r ):
r = √( 12 + 12 )
r = √ 2
u̅r1 = [ ( i + j ) /√2 ]
u̅r2 = [ ( -i + j ) /√2 ]
Similarly, E-field vectors have both vector and scalar components, and they are normalized as follows:
u̅ = ( E̅ / E )
Thus, each E-field vector can be thought of as a multiple of a basic unit vector subcomponent:
E̅ = Eu̅
Carrying out the multiplication above will essentially divide ( E̅ ) into components of ( +/- i ) and ( +/- j ) that are positioned somewhere along the x/y axis. The magnitude of an E-field is summed up with the following equation:
E = kq/r2
Substitution of this value into the ( E̅ ) equation yields the following results:
E̅1 = ( kq1/r12 )( u̅r1 )
E̅1 = ( kq1/r12 )( [ ( i + j ) /√2 ] )
E̅1 = ( [ ( 9.0 x 109 Nm2/C2 )( 1.0 x 10-6 C ) ]/ √2m2 )( [ ( i + j ) /√2 ] )
E̅1 = [ ( 9.0 x 103 N )/( 2√2 C ) ]( i + j )
E̅2 = [ ( 9.0 x 103 N )/( 2√2 C ) ]( – i + j )
The net E-field at ( P ) is derived by adding the ( i ) and ( j ) components of E̅1 and E̅2:
E̅net = E̅1 + E̅2
E̅net = [ ( 9.0 x 103 N )/( 2√2 C ) ]( i + j ) + [ ( 9.0 x 103 N )/( 2√2 C ) ]( – i + j )
E̅net = [ ( 9.0 x 103 N )/( 2√2 C ) ]( j )( i – i )
E̅net = [ ( 9.0 x 103 N )/( 2√2 C ) ]( j )
Thus, the net E-field on a point charge located at ( P ) is a vector that is collinear with the ( +j ) axis.
B: The force experienced by a charge ( qo ) of -2.0 x 10-9 C placed at ( P ) is as follows:
F̅net = E̅netqo
F̅net = [ ( 9.0 x 103 N )/( 2√2 C ) ]( j )( -2.0 x 10-9 C )
F̅net = – 18 x 10-6 N/√2 j
Our force vector is situated on the +j-axis, but it is oriented downward from point ( P ):
We will use similar techniques to answer part ( C ).
RENAISSANCE PHYSICS 