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Voltage and Potential
When the test charge q_{t} is moved in a direction toward the source charge, it is moved against the Coulomb force of repulsion and, energy is put into the system. On the other hand, if the test charge is moved in a direction
away from the source charge, the system gives up energy and the energy input to the system is negative. Hence, in moving the test charge qt through a differential distance ds the differential energy input to the system is expressed by
where θ is the angle between the direction of the force and the direction of the differential distance ds as shown in Fig. 24. The energy input to the system when the test charge is carried from a point P_{1} at a distance r_{1} from the source charge to a point P_{2} at a distance r_{2} from the source charge along any path whatsoever, as for example that illustrated in Fig. 25, is expressed by
From Fig, 24 it is evident that the scalar product i_{r}ds is the projection of the differential distance vector ds upon r and is dr because in Eq. 27
Hence
Equation 29 divided by q_{t} yields
Equation 210 expresses the voltage between the points P_{1} and P_{2}, the point P_{2} being at a potential higher than that of P_{1}. The voltage V_{21} is positive.
If the point P_{1} in Fig. 25 were removed to an infinite distance then the voltage between the point P_{2} and point P_{1} would be expressed by
However, from Eq. 29 it is evident that the energy required to move the test charge from an infinite distance
because r_{1} approaches infinity. The energy associated with the test charge at an infinite distance is zero. Hence, if Eq. 212 is divided by q_{t}, Eq. 211 results. Equation 211 expresses the potential of the point P_{2}. Similarly the potential of point P_{1} is expressed by
If Eq. 213 is subtracted from Eq. 211 the result is the difference of potential between point P_{2} and point P_{1} and Eq. 210 is obtained. From this it follows that
Thus the voltage between the points P_{1} and P_{2} is simply the difference of electrical potential, the point P_{2} being at a higher potential than P_{1} This means also that the voltage V_{21} is positive. Similarly the potential difference obtained by subtracting Eq. 211 from Eq. 213 results in
and V_{12} is negative. It is evident from Eq. 28 that the work done in carrying a test charge from one point to another in an electric field is independent of the path along which the charge is carried. By the same token if the test charge is carried from the second point back to the first along any path whatsoever, the same amount of energy is given up. Hence
This can also be expressed as a line integral as follows
from which
Equation 217 states that the summation of the electric field intensity around a closed path in an electrostatic field is zero.


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