electric charges and fields

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 A & B are rubbed, A loses its electrons and B gains the electrons which results in A being positively charged and B being negatively charged

(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, one becomes -vely charged and another becomes +vely charged

• 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)