Concepts and reason
The concepts used to solve the question is a diagram and Coulomb's law in electrostatics. At first, draw the diagram of the two-point charges and then determine the magnitude and direction of two charges at the required point using Coulomb's law in electrostatics.
Fundamentals
"Coulomb's law states that the force of attraction or repulsion between two point charges is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between the two charges" Mathematically, it is defined as, F=r2kq1q2
Here, k is the Coulomb constant q1 and q2 are the two-point charges and r is the distance between two charges.
(A)
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The diagram for the three-point charges is drawn below.
The electric force at point A due to the charge at point B is,
FAB=(AB)2kqAqB
Here, qΛ,qB are the two point charges and AB is the distance between charges.
Substitute 9×109 N⋅m2/C2 for k,1 m for AB,1nC for qA and 1nC for qB
FAB=(AB)2kqAqB=(1 m)2(9×109 N⋅m2/C2)(1nC(1nC10−9C))(1nC(1nC10−9C))=9×10−9 N
The electric force at point A due to the charge at point C is,
FΛC=(AC)2kqAqC
Here, qΛ,qc are the two point charges and AC is the distance between charges.
Substitute 9×109 N⋅m2/C2 for k,1 m for AB,1nC for qA and 4nC for qc
FΛC=(AC)2kqAqC=(2 m)2(9×109 N⋅m2/C2)(1nC(1nC10−9C))(4nC(1nC10−9C))=9×10−9 N
The net force acting on the charge at point A is,
F=FAB−FAC
Substitute 9×10−9 N for FAB and 9×10−9 N for FAC
F=FAB−FAC=9×10−9 N−9×10−9 N=0 N
According to Coulomb's law.
F∝q1q2
And,
F∝r21
Combining the above two relations.
F∝r2q1q2
To remove the proportionality sign, a constant of proportionality is introduced in the above equation.
F=kr2q1q2
Here, k is called Coulomb's constant and it has the value 9×109 N⋅m2/C2
(B)
The net electric force acting at the charge A is 0 N Therefore, the force acting at the charge A is zero.
There will be no direction of the net force acting on charge A as the net force acting on charge A is zero.
Part A
The magnitude of the net electric force acting on the charge at point A is 0 N.
Part B
The direction of the electric force on the charge A is zero as the force is zero.