FORCES BETWEEN MULTIPLE CHARGES
The mutual electric force between two charges is given by Coulomb-s law. Consider a system of n stationary charges
`q_1, q_2, q_3, ..., q_n` in vacuum. Coulomb-s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin? Experimentally it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.
To better understand the concept, consider a system of three charges `q_1, q_2` and `q_3`, as shown in Fig.(a). The force on one charge, say `q_1,` due to two other charges `q_2, q_3` can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by `F_(12), F_(12) ` is given by Eq as following even though other charges are present.
`F_(12) = 1/(4pi epsilon) (q_1q_2)/r_(12)^2 hat(r_(12))`
In the same way, the force on q1 due to q3, denoted by `F_(13),` is given by
`F_(13) = 1/(4pi epsilon) (q_1q_3)/r_(13)^2 hat(r_(13))`
which again is the Coulomb force on `q_1` due to `q_3,` even though other charge `q_2` is present. Thus the total force `F_1` on `q_1` due to the two charges `q_2` and `q_3` is given as
`F_1 = F_(12) + F_(13) +......+F_(1n)`
`= 1/(4pi epsilon_0)[(q_1q_2)/r_(12)^2 hat(r_(12)) + (q_1q_3)/r_(13)^2 hat(r_(13)) +........+(q_1q_n)/r_(1n)^2 hat(r_(1n))]`
`= q_1/(4pi epsilon_0) sum_(i=2)^n q_i/r_(1i)^2 hat(r_(1i))`
The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb-s law and the superposition principle.
`q_1, q_2, q_3, ..., q_n` in vacuum. Coulomb-s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin? Experimentally it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.
To better understand the concept, consider a system of three charges `q_1, q_2` and `q_3`, as shown in Fig.(a). The force on one charge, say `q_1,` due to two other charges `q_2, q_3` can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by `F_(12), F_(12) ` is given by Eq as following even though other charges are present.
`F_(12) = 1/(4pi epsilon) (q_1q_2)/r_(12)^2 hat(r_(12))`
In the same way, the force on q1 due to q3, denoted by `F_(13),` is given by
`F_(13) = 1/(4pi epsilon) (q_1q_3)/r_(13)^2 hat(r_(13))`
which again is the Coulomb force on `q_1` due to `q_3,` even though other charge `q_2` is present. Thus the total force `F_1` on `q_1` due to the two charges `q_2` and `q_3` is given as
`F_1 = F_(12) + F_(13) +......+F_(1n)`
`= 1/(4pi epsilon_0)[(q_1q_2)/r_(12)^2 hat(r_(12)) + (q_1q_3)/r_(13)^2 hat(r_(13)) +........+(q_1q_n)/r_(1n)^2 hat(r_(1n))]`
`= q_1/(4pi epsilon_0) sum_(i=2)^n q_i/r_(1i)^2 hat(r_(1i))`
The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb-s law and the superposition principle.