Electric Potential and Energy
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Potential Difference (p.d.) (V)
Derivation
Let us consider a positive charge +1 is being taken from point B at distance R from charge q in the electric field to A. Suppose at any instant it reaches at point P at distance x from charge q. Then force acting on the unit charge at point P is given by:
Also, suppose the unit charge is displaced by small distance dx toward charge q at point N. Then small amount of work done is given by:
Negative sign shows that unit positive charge moved in opposite direction of electrostatic force.
Now, the total amount of work done in bringing the unit test charge from B to A is given by:
WBA = ∫BA dW
= ∫Rr −q4πε0x2 dx
= −q4πε0 ∫Rr x−2 dx
= −q4πε0 [x−2+1−2+1]Rr
= q4πε0 [1x]Rr
WBA = q4πε0 [1r − 1R]
(iii)From definition of p.d.:
This is required expression for p.d. between two points in electrostatic field.
An Electron Volt (eV)
From definition of p.d.:
If an electron of charge 1.6 × 10−19 C is accelerated through a p.d. of 1 volt, then:
W = 1 V × 1.6 × 10−19 C
= 1.6 × 10−19 J
Unit of p.d.
If W be the amount of work done in moving the test charge +q0 from one point to the other point, then p.d. between two points is given by:
Thus, the SI unit of p.d. is J/C or volt.
Hence, the p.d. between two points is said to be 1 volt if 1 joule of work is to be done in bringing 1 coulomb of charge from one point to other against the electrostatic force.
Electric Potential
Derivation
Let us consider a positive +1 charge is being taken from infinity towards point A at distance r from charge +q in the electric field in air. Suppose at any instant it reaches at point P at distance x from charge q. Then force acting on the unit charge at point P is given by:
Suppose the unit charge is displaced by small distance dx towards point A. Then small amount of work done is given by:
Negative sign shows that unit positive charge moved opposite direction of electrostatic force.
Now, the total amount of work done in bringing the unit test charge from infinity to point A is given by:
W∞A = ∫∞A dW
= ∫∞r −q4πε0x2 dx
= −q4πε0 ∫∞r x−2 dx
= q4πε0 [1x]∞r
= q4πε0 [1r − 1∞]
⇒ W∞A = q4πε0r
(iii)From definition of electric potential:
This is required expression for electric potential.
Potential Gradient
Relation Between Electric Field Intensity & Potential Gradient
Let us consider two points A and B in the electric field of charge q as shown in figure. Let dx be the small distance between two points A and B. Suppose points A and B are so close to each other that the electric field intensity E between them is constant.
Now, work done in moving unit positive charge from B to A:
If dV is the p.d. between A and B, then work done in moving a unit charge from B to A:
Comparing equation (i) and equation (ii), we get:
This is required relation between electric field intensity and potential gradient, and this relation shows that electric field intensity at a point is equal to the negative of potential gradient at that point.
Equipotential Surface
If A and B are two points on an equipotential surface, then:
Thus, no work is done to move a unit positive charge on the surface of an equipotential surface.
Note: Charge Between Parallel Plates
At stationary (rest) or equilibrium condition:
Also:
Where:
V = p.d. between two plates
d = separation between two plates
Discussion
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