On the other hand when going from C to B V BC 0 since the path is perpendicular

On the other hand when going from c to b v bc 0 since

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. On the other hand, when going from C to B , V BC 0 since the path is perpendicular to the direction of ? . Thus, the same result is obtained irrespective of the path taken, consistent with the fact that ? is conservative. Notice that for the path A C B , work is done by the field only along the segment AC which is parallel to the field lines. Points B and C are at the same electric potential, i.e., V B V C . Since U q V , this means that no work is required in moving a charge from B to C . In fact, all points along the straight line connecting B and C are on the same “equipotential line.” A more complete discussion of equipotential will be given in Section 2.5 . 2.3. Electric Potential due to Point Charges Next, let’s c ompute the potential difference between two points A and B due to a charge + Q . The electric field produced by Q is ? = ( ? 4?𝜀 0 ? 2 )?̂ , Where is a unit vector pointing toward the field point. Figure 2.3.1 Potential difference between two points due to a point charge Q . From Figure 2.3.1, we see that, ?̂ ∙ ?? = ???𝜃?? , which gives ∆? = ? ? − ? ? = − ∫ ? 4?𝜀 0 ? 2 ? ? ∙ ?? = − ∫ ? 4?𝜀 0 ? 2 ?? = ? 4?𝜀 0 ? ? ( 1 ? ? 1 ? ? ) (2.3.1)
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22 Once again, the potential difference V depends only on the endpoints, independent of the choice of path taken. As in the case of gravity, only the difference in electrical potential is physically meaningful, and one may choose a reference point and set the potential there to be zero. In practice, it is often convenient to choose the reference point to be at infinity, so that the electric potential at a point P becomes ? ? = − ∫ ? ? ∙ ?? (2.3.2) With this reference, the electric potential at a distance r away from a point charge Q becomes (2.3.3) When more than one-point charge is present, by applying the superposition principle, the total electric potential is simply the sum of potentials due to individual charges: ?(?) = 1 4?𝜀 0 ? ? ? ? = ? ? ? ? ? ? ? ? (2.3.4) 2.3.1 Potential Energy in a System of Charges If a system of charges is assembled by an external agent, then U W W ext . That is, the change in potential energy of the system is the work that must be put in by an external agent to assemble the configuration. A simple example is lifting a mass m through a height h . The work done by an external agent you, is mgh (The gravitational field does work mgh ). The charges are brought in from infinity without acceleration i.e. they are at rest at the end of the process. Let’s start with just two charges q 1 and q 2 . Let the potential due to q 1 at a point P be V 1 (Figure 2.3.2). Figure 2.3.2 Two point charges separated by a distance r 12 . The work W 2 done by an agent in bringing the second charge q 2 from infinity to P is then W 2 q 2 V 1 . (No work is required to set up the first charge and W 1 0). Since V 1 q 1 / 4  0 r 12 , where r 12 is the V ( r ) 1 Q 4  0 r
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23 distance measured from q 1 to P , we have ? 12 = ? 2 = 1 4?𝜀 0 ? 1 ? 2 ? 12 (3.3.5) If q 1 and q 2 have the same sign, positive work must be done to overcome the electrostatic repulsion and the potential energy of the system is positive, U 12 0 . On the other hand, if the signs are opposite, then U 12 0 due to the attractive force between the charges.
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