ELECTROSTATIC POTENTIAL Flashcards
CAPACITANCE
AND SI UNIT
MATHEMATICALLY
RATIO OF ELECTRIC CHARGE TO THE POTENTIAL PRODUCED DUE TO THAT CHARGE
COULOMB/VOLT = FARAD
c= q/v
1 FARAD
THE CAPACITANCE OF A CONDUCTOR IS SAID TO BE 1 FARAD IF 1 C of charge raise its potential 1 volt
capacitance of paralled plate capacitor
1.consider a parallel plate capacitor of plate separation d and area of cross section A
2.let the capacitore be charged by CELL
- charge density
- the V potential between them
- since C= q/v
AE0/d
DIELECTRIC BREAKDOWN
WHEN THE INSULATOR IS KEPT IN VERY HIGH ELECTRIC FIELD
THE OUTER e- MAY GET DETACHED FROM THEIR ATOMS
THE DIELECTRIC BEHAVES LIKE CONDUCTORS
THIS PHENOMENON IS DIELECTRIC BREAKDOWN
WHY IS high potential difference not PREFFERED
IF there is high voltage, there will be HIGH E.F
THIS might ionize the surrounding air and accelerate the charges to oppositively charged capacitor plates
therefore TOTAL CHARGE OF CAPACITOR DECREASES
CAPACITANCE of Insulated Spherical Conductor
noneedofchargedensity
- consider an insulated spherical conductor of radius ‘r’
- let a charge +q be given to it. this charge spread uniformly over the entire surface
- suppose the whole charge +q be present at the cente of sphere
the potential V= kq/r - if C is the capacitance
** c= 4πεor**
energy stored in a CAPACITOR
1/2 CV^2
OR
1/2 q^2/C
energy density of CAPACITOR
E.D = Energy/ volume
1/2 εo E^2
energy =1/2 CV^2
volume - A d
C= AE0/d
V= E d
DIELECTRIC
Insulating material which transmit ELECTRIC EFFECT
2 TYPES of dielectric materials
- non polar dielectrics: H2, N2. O2. CO2
in such molecules centre of positive coincides with centre of negative charge - polar dielectrics: H2O,NH3, HCl
* in such molecules the centre of positive and negative don’t coincide
- each molecule have permentant dipole
- in absence of Eext, the individual dipole moment tend to orient Randomly due to THERMAL AGITATION
- so the net DIPOLE MOMENT IS ZERO
how DOES THESE DEPEND ON CAPACITANCE:
1. charge stored(q)
2. Voltage (V)
3. E.F
4. Area
5. distance
6. permittivity (E0)
- doesnt
- doesnt
- doesnt
- directly
- inversely
- directly
what happens to NON-polar DIELECTRIC when it is held in External ELectric field
e.g H2, N2, O2, CO2
urglasses😎
the centre of +ve charge of each molecule is pulled toward -ve plate
this dipole is induced in each molecule & hence the dielectric is POLARISED
what happens to polar DIELECTRIC when it is held in External ELectric field
H2O, NH3. HCl
😎😎
ALL molecules will reorient and tend to align themselves in the direction of External Electric Field
and hence get polarised
Capacitance of Capacitor when DIELECTRIC SLAB OF thickness t=d and dielectric constant k is INTRODUCED
dontaddd
🦎
- CONSIDER c- capacitance of capacitor
A- area
d - distance btwn plates
Eo - electric field
Vo - potential difference - when DIELECTRIC SLAB OF thickness t=d and dielectric constant k is INTRODUCED the slab gets polarised
- Electric field Ep is developed in dielectric region
- the net potential
V=Vp= Ept = Eo t /k
qt/AE0K - C= AEoK/d
- C= K Co
Co is the capacitance when air is present between plates
Capacitance of Capacitor when DIELECTRIC SLAB OF thickness t<d and dielectric constant k is INTRODUCED
- CONSIDER c- capacitance of capacitor
A- area
d - distance btwn plates
Eo - electric field
Vo - potential difference - when DIELECTRIC SLAB OF thickness t<d and dielectric constant k is INTRODUCED the slab gets polarised
- Electric field Ep is developed in dielectric region which is less than Eo
- therfore the net V between capacitor plate is
V =Vo+ Vp
Eo(d-t) + Ep t
5. C= EoA/(d-t) + t/k
Capacitance of Capacitor when CONDUCTING SLAB OF thickness t<d is introduced
- consider a capacitor of capacitance ‘C’ and area ‘A’. let ‘d’ be the separation between the plates
- let the capacitor be charged by using a cell
- the capacitance of capacitor is:
Co = AEo/d - suppose a conducting slab of AREA A AND thickness t<d is introduced btwn plates
then V =Eo(d-t) + o
5, C = A Eo/ (d-t)
= Co / (1-t/d)
if t=d then C= infinity
PARALLEL COMBINATION
ceff is
potential is
charge is
CEFF: C1+ C2
potential is same
charge is different
q1 = C1 V
q2 = C 2 V
derive for parallel combo
POTENTIAL SAME😭
- consider three capacitors of capacitance C1, C2, C3 all connected parallel
2.if q1,q2 and q3 are charges on each capacitor and V be the potential difference across PARALLEL combo
then q = q1 +q2+ q3
Ceq V = C1V +C2V +C3V
Ceq = C1+ C2 + C3
CAPACITANCE INCREASES
SERIES COMBINATION
ceff is
potential is
charge is
1/Ceq = 1/C1 +1/C2
different
same
derive for SERIES COMBO
- consider 3 capacitors of capacitance C1, C2, C3 cinnected in series
- if V1, V2, V3 are the potential difference across each capacitor then V be the total potential across the capacitors
- V= V1+V2+ V3
- q/Ceq = q/C1+q/C2+ q/C3
- 1/ Ceq =1/C1 + 1/C2 +1/C3
CAPACITANCE DECREASES
COMMON POTENTIAL
total charge before sharing
total charge after sharing
V = C1V1+ C2V2/ C1+ C2
* C1V1 +C2V2
* (C1+C2)V
WHEN CHARGES ARE SHARED THERE IS ALWAYS OF ENERGY
ΔU = Ui - Uf
- total energy before sharing= Ui= 1/2 C1V1^2 + 1/2 C1V2^2
- total energy after sharing= Uf= 1/2 (C1+C2)V^2
- ΔU =1/2 C1V1^2 + 1/2 C1V2^2 - 1/2 (C1+C2)V^2
- =** C1C2(V1 - V2)^2 /2 (C1+C2)**
positive quantity. there is always some energy loss
- inside a conductor Electrostatic field is
- at surface electrostaic field is
- potential difference inside a conductor
- potential difference on surface of conductor
be smarter😪
thus surface of conductor an equipotential SURFACE IN E FIELD
- 0
- normal to the surface
- 0
- 0
ELECRTOSTATIC SHIELDING 🛡
THE method / phenomena of protecting a certain region from ELECTRIC FIELD
REASON why it safe inside a car during⛈
ELECTROSTATIC POTENTIAL DIFFERENCE
unit and dimension
Is the amount of work done in bringing a +ve test charge from A TO B very slowly along any path between the 2 points without changing K.E
SI UNIT: Nm/C VOLT
D{: [ML2T-3A-1]
ELECTROSTATIC POTENTIAL DIFFERENCE
OF A UNIT CHARGE
Vb = Winfinty-b/ qo
V= k q/ r
when r= infinity V=0
ELECTROSTATIC POTENTIAL DIFFERENCE
OF A system of charges
V= k (q1/r1 + q2/r2+ q3/r3 …..+ qn/rn )
ELECTROSTATIC POTENTIAL DIFFERENCE
at a point due to ELECTRIC DIPOLE
V = K pcosθ/ r^2 - a^2 cos^2 θ
ELECTROSTATIC POTENTIAL DIFFERENCE
at a point due to ELECTRIC DIPOLE
where point is at axial line
cosθ
V= k p
