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electric charge
it is the property due to which a particle experiences electrical & magnetic effects
unit → C
dimensions → [IT]
some common units of charge
mC = 10^-3 C
\muC = 10^-6 C
nC = 10^-9 C
properties of charge
scalar
conserved (total charge remains same in a system)
it is constant, Q = +-ne
like charges repel, unlike attract
always has mass
mass of -ve > mass of +ve
relativistically invariant (it doesnt change by its speed)
charging of a body by friction
when 2 neutral bodies are rubbed, they split their electrons and protons
when A and B are rubbed, A acquires the protons and B acquires the electrons, making them +vely and -vely charged respectively
(draw the diagram)
charging of a body by conduction
by connecting a charged conductor to an uncharged conductor, the charge will flow until the potential becomes equal
(draw the diagram)
charging of a body by induction
when a charged particle +Q is brought near a neutral conductor, the latter’s +ve and -ve charges get split, such that the -ve charges are pointing towards the +Q charge
if the result is connected to another conductor, they split their +ve and -ve charges
if the result is grounded, it acquires fully -ve charge
(draw the diagrams)
charging of a body by thermionic emission
when a metal is heated at high temperature, some electrons are ejected and it becomes +vely charged
(draw the diagram)
charging of a body by photoelectric effect
when a light of threshold frequency hits a metal, electrons gain energy and get ejected, making the metal +vely charged
(draw the diagram)
charging of a body by field emission
when a metal is near an electric field, the electrons move out, making it +vely charged
(draw the diagram)
coulomb’s law
the magnitude of electrostatic force between 2 points is directly proportional to the product of charges and inversely proportional to the square of distance between them
F = Kq1q2/ r² (K = 1/4\piEo)
in vector form,
F = Kq1q2 r> / |r|²
values of K & Eo
K = 9 × 10^9 Nm²/C²
Eo = 8.85 × 10^-12 C²/Nm²
where dimensions of Eo = [M^-1L^-3T^4A²]
electrostatic equilibrium
it is the point where resultant electrostatic force is zero
stable: if the particle has the tendency to return back to equilibrium when moved by dx
unstable: if it doesnt return back to equilibrium when moved by dx
neutral: if its still in equilibrium when moved by dx
electric field
it is the region around a charged body that exerts force on another test charge placed in it
E =F/qo (qo = test charge)
vector quantity
unit → N/C
absolute permittivity (dielectric constant)
K = Fvac/ Fmed = E/ Eo
derivation of electric field due to a point charge
E =Kq/ r²
derive this with diagram
electric field in rectangular components
r = xi + yj + zk
| r | = \sqrt{x^2+y^2+z^2}
hence,
E = Kq(xi + yj + zk)/ (x² + y² + z²)³/2
electric field lines
they give the direction of the electric field at a point
they start at positive charges and end at negative charges
properties of electric field lines
they start from +ve and end at -ve cchargges
the tangent to a point charge q gives the direction of field at that point
they never intersect, as if they did, it would give 2 directions of field at a point which is not possible
they dont form closed loops
they are continuous curves with no breaks (in a charge-free system)
they are perpendicular to the surface of a conductor
they contract lengthwise between unlike charges
they contract sideways between like charges
electric field lines between 2 like and 2 unlike charges
draw the diagrams in both cases
electric dipole
it is a pair of 2 equal and opposite charges +q and -q, separated by a distance 2l
an ideal dipole is one which has smallest length
electric field on the axial line due to dipole
E =2Kp/ x³
NOTE: in this, the direction of field is along p
dipole moment
the dipole moment of an electric dipole is the product of either charge and the distance between them
p = 2ql
unit → Cm
electric field on the equatorial line due to dipole
E =Kp/x³
derive this
NOTE: in this, the direction of field is opp. to p
electric field at any other point due to a short dipole
E =Kp(3cos²O + 1)^1/2 / x³
also,
angle between E and E1,
tan\alpha = ½ tanO
(derive this)
torque on an electric dipole
\tau = p x E = pEsin\theta
if O = 0, T = 0 (stable equilibrium)
if O = 90, T = pE (max.)
if O = 180, T = 0 (unstable equlibrium)
unit → Nm
(derive this)
work done on a dipole
work done by a dipole to attain stable equilibrium,
U = W = pE(cosO1 - cosO2) = -p.E
if O = 0, U = -p.E (min.)
if O = 180, U = p.E (max)
if O = 90, U = 0
area vector
it is the vector associated with every area element of a surface
it is taken in the direction of the normal
\Delta S = n^ (\DeltaS), where n^ is unit vector along the direction of normal
electric flux
the electric flux linked to any surface is the total no. of electric field lines that normally pass through that surface
electric flux d\phi through a small element dS due to to electric field E at an angle \theta with dS is
d\phi = E.dS = EdScos\theta
scalar quantity
unit → Nm²/C
dimensions → [ML³T^-3A^-1]
properties of electric flux
if the flux is negative, then the surface encloses a net negative charge
it is analogous to liquid flux
for O < 90, it is positive
for O = 90, it is zero
for 90 < O < 180, it is negative
difference between electric flux through dS and through a closed surface
through dS:
it is affected by charges present outside the surface
it can be changed if the position of charges inside are changed
through closed surface:
it depends only on the charges enclosed by the surface
it doesnt depend on the location of inside charges
gauss’ law
the surface integral of the electric field intensity over any closed surface (gaussian surface) in free space is equal to 1/Eo times the charge enclosed by it
\phi = \int E.dS = q/Eo
proof of gauss theorem for a spherically symmetric surface
prove this
features of gauss’ law
it is true for any type of closed surface
in a situation where the surface is chosen such that theres charges both inside and outside, the electric field is due to all the charges
it works better with a symmetric surface
deduction of coulomb’s law from gauss’ theorem
derive this
electric field due to an infinitely long thin straight wire
E = \lambda / 2\pi Eo r
the direction of E is radially outward from the positive line charge and vice versa
(derive this and draw the graph for relation between E and r)
electric field due to a thin infinite plane sheet of charge
E = \sigma/ 2Eo
(derive this)
electric field due to non-conducting charged solid sphere
at a distance r from sphere = r\rho/ 3E
at surface (R) = R\rho/3E
at centre = 0
(derive this)
electric field due to 2 thin infinite plane sheets
in region 1: E = -E1 - E2
in region 2: E = E1 - E2
in region 3: E = E1 + E2
(derive this with final conclusion)
electric field due to uniformly charged thin spherical shell
outside the shell: = Kq/r²
on the surface: = Kq/R²
inside the shell: = 0
(derive this and prove that it confirms 1/r² dependence in coulomb’s law along with graph)