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| Electrostatic Potential |
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When a force  acts
on a particle that moves from a point 'a' to a point 'b' the work done by
the force is given by the integral  |
where
 is infinitesimal
displacement along the particles path and q is
the angle is the angle and
at each point along the
path. |
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| Since electrostatic forces are
conservative forces, work done by
can be in terms of potential at the intial and the final points. When a charge moves from a point where the potential energy is Va to a point where the potential energy is Vb, then the change in the potential energy is given by DV = Vb - Va |
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| And the work done by the force is given by |
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| Wa+b = Vb - Va = - (Vb - Va) - DV |
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| where Wa-b = is negative, Vb > Va the DV is positive and the potential energy increases. |
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Just as the electric field is described as force per unit charge, electric potential at a point can be described as electrical potential energy per unit charge. This concept is useful in calculations involving energies of charged particles. Not only that, as force is related to work, electric
potential is closely related to the electric field  |
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| Potential being a scalar quantity, it is easy to deal with, than electric field. Hence, when we need to determine an electric field, it is easy to find the potential first and find the field from it. |
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| The electrostatic potential at any point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that point against the electric force of the field. |
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| V = W/q |
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| where V is the potential |
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| W is work done |
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| q the charge |
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| S.I. unit is Joule/Coulomb = 1 Volt |
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| Consider two points in an electric field. If a free test charge +qo was placed in the field, it would flow from b to a. So we say b is at higher potential. The electric field does negative work in moving the test charge qo from a to b. |
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| If a and b are two points in an electric field having a potential difference of 12V and if VB>VA, it means 12 Joules/C of work has to be done in moving a test charge from A to B. Conversely this 12 J/C of energy will be released if it moves from B to A. This process does not depend on the route traversed and like gravitational electrostatic forces are also conservative. |
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| If a is at infinity, then the potential due to a single point charge |
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| Electric potential at any point in vacuum due to group of point charges q1,q2,q3……….qn is equal to the algebraic sum of the potentials due to q1,q2,q3……….qn at point P. Algebraic sum is one in which, sign of the physical quantity is taken into account. |
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| Problem: |
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| An infinite number of charges each equal to 'q' are placed along with X-axis at x = 1m, x = 2m, x = 4m, x = 8m, …………… and so on. Find the electric potential at the point x = 0 due to this set of charges. |
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| Suggested answer: |
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| Step 1: |
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| Potential is a scalar, hence no direction is involved. |
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| Step 2: |
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| The point where the potential is to be found is x = 0. |
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| Step 3: |
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| Which are the charges and where? All the charges are equal to 'q'. Position at 1m, x = 2m, x = 4m, x = 8m, …………… |
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| Step 4: |
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| The sign of all the charges is positive. |
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| Step 5: |
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| Step 6: |
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| Using the superposition principle, net potential at |
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| Step 7: |
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Need to simplify  represents
infinite geometric progression |
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