ELECTROSTATIC

POTENTIAL








- BY

B . AMIRTHA ABHIRAMI
INTRODUCTION

•In this we understand how we move the charge in this field and what amount
of work is done to move the charge against this field and in what form the
work done is stored in the system and what is electrostatic potential.
• In this we first define electrostatic potential and then potential due to a
point charge, system of charges, electric dipole and potential energy of
charges and dipole in an external field. We got introduced to new concepts
like equipotential surface, electrostatics of conductors, dielectric and
polarization etc. and define electrostatic potential energy.
•In this we also get to learn about the capacitors and capacitance and the
effect of dielectric medium on it and their combination circuits and
electrostatic potential and capacitance all formulas.
 *point charge
located at • and when test
Work done point o and charge taken
against test charge q from infinity to
point B then work
electrostatic is taken from
done against potential
force of infinity to electrostatic
point A then energy at point
repulsion is force repulsion UB = WαB B is greater
UB >UA
called electro work done then potential
against energy is than at A
static
potential electrostatic
energy force of
repulsion is
UA=WαA
 ELECTROSTATIC POTENTIAL (OR)
ELECTRIC POTENTIAL
 UNIT = Volt
It is a scalar quantity
• Potential energy per unit charge

VA=wαA/q0=UA/q0
• VB=WαB/q0=UB/q0 = J/C
 1 volt = when 1 joule work done by the one coulomb of charge is
known as one volt
 ELECTRIC POTENTIAL
DIFFERENCE
 VB = WαB/qo

 VA = WαA/qo

 Potential difference = VA-VB ( or) VB-VA

= 1/qo [WαB – WαA]

VA – VB =WAB/qo

 Hence , amount of work done in bringing a positive charge from

the point of lower potential to point of higher potential is called potential difference
in electric field

 Unit – Volt(or)J/C
 LINE INTEGRAL OF ELECTRIC FIELD
Potential difference between two points in terms of line integral of electric field
F = -q0E
suppose test charge small , displacement dl towards point charge (towards
b )then we get small work done
• *d W = f . Dl
• *d w = -q0. E. dl
• *W = -q0 ꭍ E . dl
• *W/q0 = -ꭍE . dl
• * VB – VA = ʃ A to B . E . dl
•ʃ E . dl is the line integral . Hence , potential difference between two points is the
negative if line integral.
 IMPORTANT POINTS
1 . ʃ A to B is an integral that involves dot product of vector.

2 . Line integral of electric field represent work done per

unit charge .

3 . Potential difference is a property of two points and not of

charge .
ELECTROSTATIC FORCE IS A CONSERVATIVE
FORCE
•
ELECTRIC POTENTIAL DUE TO A SINGLE POINT
•
•
POTENTIAL DUE TO AN ELECTRIC
DIPOLE
 EQUIPOTENTIAL SURFACE

The equipotential surfaces are surface at which the potential at every point on
the surface is constant .

Also for any charged configuration , the electric field is always perpendicular or
equipotential surfaces otherwise it would have a component in the equipotential
surface , and hence the potential at the surface would not be constant .

Thus electric field should be perpendicular to the

equipotential surface.
CHARACTERISTICS OF EQUIPOTENTIAL
SURFACE
• No work is done in moving a test charge over an equipotential surface
• Electric field is always normal to the equipotential surface at every
1) point

• The spacing between equipotential surface identify the region of strong

and big field when equipotential surface are crowded then electric field
2) intensity is greater.

• Two equipotential surface never interact each other ,if they interact
then at the same point we get two values of electric potential which is
3) not possible.
ELECTRIC FIELD INTENSITY
AND POTENTIAL DIFFERENCE
•
 The electric potential
energy of a system of
point charges is the
work needed to bring
the charge from infinity
to their final position
ELECTRIC POTENTIAL ENERGY OF TWO
CHARGE SYSTEM
•
ELECTRIC FLUX
•
GAUSS LAW
•According to Gauss law, the total flux linked with a closed surface is 1/ε0

times the charge enclosed by the closed surface.

•∮��→.��→��=1∈0��
•For example, a point charge q is placed inside a cube of the edge ‘a’. Now,
as per Gauss law, the flux through each face of the cube is q/6ε .0

•The electric field is the basic concept of knowing about electricity. Generally,
the electric field of the surface is calculated by applying Coulomb’s law, but
to calculate the electric field distribution in a closed surface, we need to
understand the concept of Gauss law. It explains the electric charge
enclosed in a closed surface or the electric charge present in the enclosed
closed surface.
• Field due to an infinitely long straight
uniformly charged wire:
• Consider an infinitely long thin straight wire with uniform linear charge density λ. The direction of
electric field at every point must be radial (outward if λ > 0, inward if λ < 0).
• Consider a pair of line elements P1 and P2 of the wire. The electric fields produced by the two
elements of the pair when summed give a resultant electric field which is radial.
• Because the field is everywhere radial and the flux through the two ends of the cylindrical Gaussian
surface is zero.
• Direction of E is perpendicular to the wire and components of E normal to end faces of cylinder
makes no contribution to electric flux. Thus, its magnitude is constant.
• The surface area of the curved part is 2Пrl, where l is the length of the cylinder.
• Flux through the Gaussian surface = Flux through the curved cylindrical parts of the surface= E (2Пrl)
• Charge enclosed in cylinder is q = Linear charge density x length l of cylinder or q=λ.l
•
• Field due to a uniformly charged infinite plane sheet
• Consider a thin infinite plane uniformly charged sheet having a surface charge density.
• We take the x-axis normal to the given plane. By symmetry, the electric field will not depend
on y and z coordinates and its direction.
• We have to find the electric field intensity because of this sheet at any point which is at
distance r from the sheet.
• Let Gaussian surface be a rectangular parallelepiped of cross sectional area A. From the
figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel
to the other faces, and they, therefore, do not contribute to the total flux.
• No field lines crosses the side walls of rectangular parallelepiped i.e. component of E normal
to walls of parallelepiped is zero.
• The unit vector normal to surface 1 is in –x direction while the unit vector normal to
surface 2 is in the +x direction. Therefore, flux (E.ΔS) through both the surfaces is equal
and adds up. Therefore , the net flux through the Gaussian surface is 2 EA. The charge
enclosed by the closed surface is σ A.
•
• Field due to a uniformly charged thin spherical
shell
• Field outside the shell:
• Let σ be the uniform surface charge density of a thin spherical shell of radius R. Consider a point P
outside the shell with radius vector r. Consider Gaussian spherical surface radius r and with centre O,
passing through P. All points which correspond to sphere are equivalent relative to the given charged
configuration.
• The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is
along the radius vector at each point
• Thus, at every point on sphere E and ΔS are parallel and the flux through each element is E ΔS
• Therefore, the net flux will be summing over all ΔS.
• Hence, the flux through the Gaussian surface = E × 4 π r2.
• And the charge enclosed = σ × 4 π R2.
•
 Field inside the shell:

• In the given figure, point P is inside the shell

• . The Gaussian surface is again a sphere through P centered at O
• . When field is outside the shell, the flux through the Gaussian surface = E × 4 π r2
• . However, when field is inside the shell, the Gaussian surface encloses zero charge. Hence,
from Gauss’s law, flux is given as , E × 4 π r2 = 0 i.e., E = 0 (r < R).
• The field due to a uniformly charged thin shell is zero at all points inside the shell.
PRESENCE OF FREE CHARGE AND
CHARGE INSIDE A CONDUCTOR
 In normal state atoms have positive charge in the nuclease
and negative charge move around it .

 In a conductor electrons of outermost orbit are free charge

and positive one fixed in the latic are called bound charge .

 In an insulator before apply external electric field electrons

and positive ions are bound charge whenever after applying
external electric field negative charge is bound charge .
POTENTIAL ENERGY OF A DIPOLE IN AN
ELECTRIC FIELD
•
 DIELECTRICS AND POLARIZATION
•
LETS SUM UP
• While electric potential measures the ability to perform
work on a charge , capacitance measures the ability to store
charge .
• The unit of measurement for capacitance is coulomb per
voltage , which is the amount of charge present per voltage
applied .
• THANK
YOU