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Electric Charge

Charge
It is a physical quantity that appears when a body either losses or gains electron. Electrons are responsible for electrification of the body. Charges are of two types

  1. Positive charge 
  2. Negative charge

\(\rightarrow\) Charge is associated with mass. E.g. - When a glass rod rubbed with a silk it becomes +vely charged.
\(\rightarrow\) A particle or mass is said to be +vely electrified if it is repelled by a glass rod that has been freshly rubbed with the silk.
\(\rightarrow\) When rubber is rubbed with a fur it becomes -vely charged.
\(\rightarrow\) A particle of mass is said to be -vely electrified if it is repelled by rubber or amber that has been freshly rubbed with the fur.

image a

Electrostatics
Branch of physics in which study of phenomena due to stationary charges are studied is called electrostatics.
 me\(\rightarrow\) 9.1 \(\times\) 10-31 kg
mn \(\rightarrow\) 1.675 \(\times\) 10-27 kg
mp \(\rightarrow\) 1.673 \(\times\) 10-27 kg
|e| \(\rightarrow\) 1.6 \(\times\) 10-19 C
S.I. unit of charge \(\rightarrow\) coulomb
Dimensional formula \(\rightarrow\) M0L0A1T1

Properties of charge

  • Charge is transformable. If a charge body is put in contact with an uncharged body transfer of electron from one body to the other body may take place.
  • Charge is always associated with mass. The presence of charge itself is a convincing proof of existing of mass.
    Note
    Particles such as proton or neutron which have no rest mass can never have a charge.
  • Charge is quantized \(\rightarrow\) when a physical quantity can have only discrete values rather than any values that the quantity is said to be quantized.
  • Charge on body will always the integer multiple of charge on electron. Any body can never have charge which is \(\frac{2}{3}\)e or \(\frac{3}{5}\)e.
    so, charge on body will always be,
    q = ne, where n \(\in\) I
  • Charge is conserved. In an isolated system total charge does not change with time. In the system if a body loose e- then their will be another body to gain e-. Therefore, total charge remain's conserved.
  • Charge is invariant. This means that the charge is independent of frame of ref. & also independent of speed of body.
  • Accelerated charge radiates energy. A charged particle at rest will only produce electric field. If charged body is moving with constant speed it will produce both electric & magnetic field but doesn't radiate energy. If the motion of charged particle is accelerated, it will produce electric & magnetic field with energy in the space in the form of electromagnetic waves (Heat).

Coulombs Law
In free space / vaccum / Air
\(\dot{q_1}\) \(\leftarrow\) r \(\rightarrow\)\(\dot{q_2}\)
 F \(\propto\) q1q2
F \(\propto\) \(\frac{1}{r^2}\)
F = \(\frac{Kq_1q_2}{r^2}\)
Where, K = constant
k = \(\frac{1}{4\pi \varepsilon_{\circ}}\)
where, \(\varepsilon_{\circ}\) \(\rightarrow\) permittivity of free space.
k = 9 \(\times\) 109 \(\frac{Nm^2}{C^2}\)
\(\varepsilon_{\circ}\) = 8.85 \(\times\) 10-12 \(\frac{C^2}{Nm^2}\)
S.I. unit of K = \(\frac{Nm^2}{C^2}\)
Dimensional formula = \(\frac{M^1L^1T^-2L^2}{(A^1T^1)^2}\)
= \(M^1L^3T^{-4}A^{-2}\)
S.I unit of \(\varepsilon_\circ\)=\(\frac{C^2}{Nm^2}\)
Dimensional formula = M-1L3T4A2

Permittivity
K = \(\frac{1}{4 \pi \varepsilon_{\circ}}\)
For vaccum q1, for medium q2
Relative permittivity (k) \(\varepsilon_r\) = \(\frac{\varepsilon_{med}}{\varepsilon_{\circ}}\) = \(\frac{permittivity\;of\;medium}{permittivity\;of\;space}\)
Where,
For vacuum, \(\varepsilon_r\) = 1
Air, \(\varepsilon_r\) \(\simeq\) 1
For H2O, \(\varepsilon_r\) = 80
Conductor, \(\varepsilon_r\) = \(\infty\)

  • Permittivity of medium is the measurement of a fact have strongly a medium is influenced by external electric field.
  • On applying external electric field how much medium is polarizing. The measurement of this thing is permittivity.
  • On applying external electric field there are more polarizing effect than medium will have high permittivity.
    Note
    External electric field \(\rightarrow\) More
    Polarization of medium \(\rightarrow\) High Permittivity
    Note
    If two charges are placed in any medium other than vacuum force between two charges decreases due to polarization of medium therefore resultant force on charge get reduced by factor (k) or \(\varepsilon_r\). This (k) is called dielectric constant of medium or relative permittivity of the medium.

Dielectric constant / Relative permittivity
This is the ratio  of permittivity of the medium to the permittivity of free space.
\(\rightarrow\) Coulomb force of any medium 
Fmed = \(\frac{Kq_1q_2}{r^2}\)
K = \(\frac{1}{4\pi\varepsilon_{med}}\) = \(\frac{1}{4\pi\varepsilon_{\circ}\varepsilon_r}\)
Fmed = \(\frac{1}{4\pi\varepsilon_{\circ}\varepsilon_r}\) \(\frac{q_1q_2}{r^2}\) = \(\frac{1}{\varepsilon_r}\) \(\frac{1}{4\pi\varepsilon_{\circ}}\)\(\frac{q_1q_2}{r^2}\)
= \(\frac{F_{vacuum}}{k}\)
Where,
k = dielectric constant

Some points about Coulombs force

  • It is a central force.
  • Coulomb law is valid only for points charges.
  • The force between two point charges is independent of presence or absence of any other charges Hence, it will follow the principal of superposition.
  • This force will change by factor (R). If medium is changed. This is because of polarization of medium.
  • This force acts along line joining the 2 charges.
  • This force is conservative in nature.
  • Coulomb force between 2 charges is an action reaction pair.
    Note
    \(r_{12}\) = Dir(2 to 1).
    \(\vec{F_{12}}\) = Force on 1 due to 2, r12 = -r12

Vector form of coulomb's law 
\(\vec{F}\) = \(\frac{Kq_1q_2}{r^2}\)\(\hat{r}\)
\(\dot{q_1}\)                  \(\dot{q_2}\)
\(\vec{F_{12}}\) = \(\frac{Kq_1q_2}{r^2_{12}}\) \(\hat{r_{12}}\)
\(\vec{F_{12}}\) = \(\frac{Kq_1q_2}{r^2_{12}}\) \(\frac{\vec{r_{12}}}{|r_{12}|}\)
\(\vec{F_{12}}\) = \(\frac{Kq_1q_2}{r^3}\)\(\vec{r_{12}}\).
\(\vec{F_{12}}\) = \(\frac{Kq_1q_2}{r^3}\)\(\vec{r_{21}}\).

  • Case I
    If q1q2 > 0 or (+ve)
    It means q1 = +ve, q2 = +ve
    q1 = -ve, q2 = -ve
    q1 and q2 are of same nature 
    \(\vec{F_{12}}\) = \(\frac{Kq_1q_2}{r^3}\)\(\vec{r_{12}}\)
    \(\vec{F_{21}}\) = \(\frac{Kq_1q_2}{r^3}\)\(\vec{r_{21}}\)
    image b
  • Case II
    If q1q2 < 0 Or (-ve)
    It means q1 = +ve, q2 = +ve
    q1 = -ve, q2 = -ve
    q1 and q2 are of opposite nature 
    \(\vec{F_{12}}\) = \(\frac{Kq_1q_2}{r^3}\) \(\vec{r_{12}}\) = \(\frac{-Kq_1q_2}{r^3}\)\(\vec{r_{12}}\)
    = \(\frac{Kq_1q_2}{r^3}\) \(\vec{-r_{12}}\) = \(\frac{Kq_1q_2}{r^3}\) \(\vec{r_{21}}\)

Examples

  • The system is in rest in given situation the find out the value of q.
    Solution:
    r = 2lsin\(\theta\)
    Tcos\(\theta\) = mg -------  (1)
    Tsin\(\theta\) = Fe ------ (2)
    = \(\frac{Tsin\theta}{Tcos\theta}\) = \(\frac{Fe}{mg}\)
    Fe = mgtan\(\theta\).
    \(\frac{Kq_1q_2}{r^2}\) = mg tan\(\theta\).
    \(\frac{Kq^2}{(2\ell sin\theta)^2}\) = mg tan\(\theta\)
    q = \(\sqrt{\frac{4l^2sin^2\theta mgtan\theta}{\frac{1}{4}\pi \varepsilon_{\circ}}}\)
    image d
    image f
    Fe = \(\frac{Kq_1q_2}{r^2}\)
    Concept
    For stable equilibrium \(\vec{F_{net}}\) = 0
  • Two point charges q1 and q2 are placed at a distance 'r' and a medium of dielectric constant 'k' having thickness 'x' is placed between them (x < r). Find out the electric force between charges.
    image gh
    Solution:
    If q1  q2 are separated  by completely medium of Dielectric constant R & x.
    image fe
    \(\dot{q_1}\) \(\leftarrow x_{medium}\) \(\rightarrow\) \(\dot{q_2}\) 
    F = (\(\frac{1}{4\pi\varepsilon_\circ} \)\(\frac{q_1q_2}{x^2})\)x\(\frac{1}{\dot{R}}\)
    If xmed is equal to y length of air them
    image hi
    \(\dot{q}\) \(\leftarrow\) y \(\rightarrow\) \(\dot{q_2}\)
    F = \(\frac{1}{4\pi\varepsilon_{\circ}}\) \(\frac{q_1q_2}{y^2}\)
    For equivalent  length in air 
    F = \(\frac{1}{4\pi\varepsilon_{\circ}}\) \(\frac{q_1q_2}{x^2_{medium}R}\)
    = \(\frac{1}{4\pi\varepsilon_{\circ}}\) \(\frac{q_1q_2}{y^2_{air}}\)
    \(x^2_{med}\)k = \(y^2_{air}\)
    yair = \(\sqrt{k}\)xmed
    Distance between charges = \((r - x)_{air}\) + xmed
    = \((r - x)_{air}\) + \(\sqrt{R}\) xair
    = r – x + \(\sqrt{R}\)x
    Fnet = \(\frac{1}{4\pi\varepsilon_{\circ}}\) \(\frac{q_1q_2}{(r – x + \sqrt{R}x)^2}\)
    E.g.-
    image jk
    Fnet = \(\frac{1}{4\pi\varepsilon_{\circ}}\) \(\frac{q_1q_2}{(r - x_1 - x_2 - x_3 + x_1 + x_2 \sqrt{R} + x_3 \sqrt{R})^2}\)
    Concept
    If two objects of size are brought in contact and then separated charges will equally divided after contact.
  • Two point charges \(4e^-\) &\(e^-\) are placed at a distance 'a' and another charges of 2e is placed in between these charges if the net resultant force on this charges is zero than find out its position & discuss the equilibrium of this charges.
    Solution:
    \(\vec{F}\) = 0
    4e \(\leftarrow\) x \(\rightarrow\) 2e \(\leftarrow\) a – x \(\rightarrow\)e
    F2 \(\leftarrow\) \(\rightarrow\)  F1
    \(\leftarrow\) a \(\rightarrow\)
    image pq
    F1 = F2
    \(\frac{K(4e)(2e)}{x^2}\) = \(\frac{K(2e)(e)}{(a - x)^2}\)
    4(a - x)2 = x2
    Taking root both side
    2(a - x) = \(\pm\) x
    2a - 2x  =x
    x = \(\frac{2a}{3}\)
    2a - 2x =  -x
    x = 2a
    If 2e charges is +ve (x = \(\frac{2a}{3})\)
    image st
    4e \(\leftarrow\) \(\frac{2a}{3}\) \(\rightarrow\) 2e \(\leftarrow\) \(\frac{q}{3}\) \(\rightarrow\) e
    Fe \(\leftarrow\) \(\rightarrow\) F4e
    If displacement right 
    Fe \(\longleftrightarrow\) F4e less F4e < Fe
    It means it will move towards left stable eq.
    If 2e charges is (-ve)
    image st
    4e \(\leftarrow\) \(\frac{2a}{3}\) \(\rightarrow\) 2e \(\leftarrow\) \(\frac{a}{3}\) \(\rightarrow\) e
    F4e \(\longleftrightarrow\) Fe if it is displaced to right, it will move towards right unstable equal.
    F4e \(\longleftrightarrow\) Fe
    F4e < Fe

Electric field 
q test charge.
 \(\dot{Q}\)             \(\dot{E}\) = ?
\(\rightarrow\) Electric field is the force experienced by +ve test charge per unit charge is called electric field.
\(\rightarrow\) Vector quantity
\(\vec{E}\) = \(\frac{\vec{F}}{q}\)
\(\vec{E}\) = \(\lim\)\(\frac{\vec{F}}{q}\)
q \(\rightarrow\) 0
S.I. unit = \(\frac{N}{C}\)
Dimensional formula = \(\frac{N}{C}\) = \(\frac{M^1L^1T^{-2}}{\frac{C}{T}T}\) = \(\frac{M^1L^1T^{-2}}{A^1T^1}\)
D.F. = M1L1T-3A-1
\(\rightarrow\) Electric force on a charge can be understand in two steps .
i) Whether charge is placed in a speed in a space then it will develop electric field around it instant (With speed of light) .
ii
) When test charge is placed in this electric field than it will experience electric force due to electric field of first charge.
Electrostatic or coulombic force .
\(\rightarrow\) Non - contact force .
\(\rightarrow\) Action at distance.
Note

\(\vec{E}\) = \(\lim\frac{\vec{F}}{q}\)
With q \(\to\) 0 is taken because electric field we are calculating another charge will not be disturbed by electric field due to test charge.
F = Q1\(\vec{E_2}\)
\(\dot{Q_1}\)\(\vec{F}\) = Charge of its own \(\times\) Electric field due to another charge 
\(\dot{Q_2}\)
Uniform electric field \(\rightarrow\) If electric field does not change with the position this type of electric field is called U.E.F.
Constant electric field \(\rightarrow\) If electric field at a point does not change with time then this type of electric field is called C.E.F.
Uniform E.F. \(\rightarrow\) E \(\not = \) F(Position)
Non- uniform E.F. \(\rightarrow\)E = F(Position)
Constant E.F. \(\rightarrow\) E \(\not=\) F(Time)
Variable E.F. \(\rightarrow\) E = F(Time)

Electric field Lines (EFL)

  1. F.L. starts or diverge from '+ve' charge and ends or converge at '-ve' charge.
    image xy
  2. Tangent at any point on EFL show the direction of electric field .
  3. On'+ve' charge electric force will in the direction of electric field & on the '-ve' charge force will act opp. to direction of electric field.
    F= qE
  4. Electric field lines never from closed loop.
  5. Two EFL cannot intersect each other.
    Expalanation
    \(\rightarrow\) If two EFL intersect each other at any point then at that direction of electric field will be contradict.
  6. F.L do not pass through conductor because in conductor electric field is always zero.
    \(\rightarrow\) In electrostatic situation electric field on the surface of conductor will be in the \(\perp\) direction.
  7. F.L. have a tendency to contract longitudinally.
    \(\rightarrow\) They contract longitudinally like a stretched elastic string producing attraction b/w same charge & repel each other laterally resulting in repulsion.
    image po
    image pi
    image ou
  8. If EFL are closed to each other then their will be more electric field intensity.
    \(\rightarrow\) If they are at more separation than at that electric field unesity will be less.
    image wer

Different patterns of EFL 

image tyr
Direction constant
but magnitude diff.
image qop
Direction not constant
but magnitude constant.
image wi
Direction & magnitude
both not constant 
image pyt
Direction & magnitude 
both are constant.

Note
Neural point is the point at which electric field will become zero.
\(\rightarrow\) This point is Located near to small change or from the bigger charge.

If E = 0
q1 = +ve, q2 = -ve
r1 < r2
|q1| <|q2|

If E = 0
q1 = +ve, q2 = +ve
q1 = -ve, q2 = -ve 
\(\frac{ Rq_1}{r_1^2}\) = \(\frac{Rq_2}{r_2^2}\)
r1 < r2
|q1| < |q2|

      

If E= 0
q1 = +ve, q2 = +ve
q1 = -ve, q2 = -ve 
\(\frac{ Rq_1}{r_1^2}\) = \(\frac{Rq_2}{r_2^2}\)
r1 > r2
|q1| > |q2|

If E = 0
q1 = +ve, q2 = -ve
r1 > r2
|q1| > |q2|

image ere
q2 > q1

Examples

  • In following situation draw the electric field lines.
    Solution:
    image qqq
    image lll

Vector Form of Electric Field or Calculation of Electric Field for Point Charge
\(\vec{E}\) = \(\frac{Kq}{r^2}\)\(\hat{r}\)
\(\vec{E}\) = \(\frac{Kq}{r^3}\)\(\hat{r}\)
\(\hat{r}\) = \(\frac{\hat{r}}{|\hat{r}|}\)
Note
Electric field obey the principle of  superposition.
\(\vec{E_{net}}\) = \(\vec{E_1}\) + \(\vec{E_2}\) + ……

Examples

  • Two +ve charges Q & 2Q are placed at (1, 1, 1) & (4, 4, 4) respectively find out the net electric field at point (2, 2, 2).
    Solution:
    r1 = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
    r2 = 4\(\hat{i}\) + 4\(\hat{j}\) + 4\(\hat{k}\)
    r3 = 2\(\hat{i}\) + 2\(\hat{j}\) + 2\(\hat{k}\)
    image rrrr
    \(\vec{E_{net}}\) = \(\vec{E}\) \(\cdot\) field at 3 due to 1 + \(\vec{E}\)\(\cdot\) field at 3 due to 2.
    \(\vec(E_1)\) = \(\frac{KQ}{r^2}\)\(\hat{r}\) = \(\frac{KQ}{r_{31}}^3\)\(\hat{r_{31}}\) = \(\frac{KQ}{(\sqrt{3})^3}\)\(\cdot\)(- 2\(\hat{i}\) - 2\(\hat{j}\) - 2\(\hat{R})\)
    = \(\frac{-2KQ}{2(2\sqrt{3}})\)\((\vec{i}\)\(\vec{+j}\)\(\vec{+k})\)
    \(\vec{r_{31}}\)  = r3 – r1
    \(\vec{|r_{31}|}\) = \(\sqrt{3}\)
    \(\vec{r_{32}}\) = \(\vec{r_3}\) - \(\vec{r_2}\) = \(\hat{-2i}\)\(\hat{-2j}\)\(\hat{-2k}\)
    \(\vec{|r_{32}|}\) = 2\(\sqrt{3}\)
    \(\vec{E_3}\) = \(\vec{E_1}\) + \(\vec{E_2}\) = \(\frac{KQ}{2(3\sqrt{3})}\)\((\vec{i}\) + \(\vec{j}\) + \(\vec{R})\)
    Note
    \(\vec{E}\) on point A due to present = \(\frac{KQ}{(r_{AB})^3}\) \(\cdot\) \(\vec{r_{AB}}\) of charge at B.

Calculation for electric field for continuous charge Distribution
\(\lambda\) = \(\frac{charge}{Length}\), \(\sigma\) = \(\frac{charge}{Area}\), \(\rho\) = \(\frac{charge}{volume}\)  
Procedure 

  1. Consider a small charge element dQ.
  2. Write down the expression for de for small charge element dQ.
  3. Integrate de for appropriate limit.

Electric field at a point on the axis of a Ring 
We are calculating the electric field due to ring on its axis at distance x.

image zzzzzz

Radius = R
Charge = Q
Electric field due to dQ charge 
dE = \(\frac{KdQ}{r^2}\)
dEx = dEcos\(\theta\) = \(\frac{KdQ}{r^2}\) cos\(\theta\)
cos\(\theta\) = \(\frac{B}{H}\) = \(\frac{x}{r}\) = \(\frac{x}{\sqrt{x^2 + R^2}}\)
dEx = \(\frac{KdQ}{r^2}\)\(\frac{x}{r}\)
\(\int\)dEx = \(\int\)\(\frac{KdQ}{r^3}\).x
Ex = \(\frac{KQx}{r^3}\) = \(\frac{KQx}{r^3}\) = \(\frac{KQx}{(x^2 + R^2)^{\frac{3}{2}}}\)
All comp. of dEy will cancel out each other
Ey = 0, Enet = Ex = \(\frac{KQx}{r^3}\)
E = \(\frac{KQx}{(x^2 + R^2)^{\frac{3}{2}}}\)
Note

  1. For x = 0 means at center of Ring E = 0 due to symmetry.
  2. At large distance E will become zero.
    x = 0, x \(\rightarrow\)\(\infty\)
    therefore at any point between x = 0 & x \(\rightarrow\) \(\infty\). E will have maximum value k
    Fore maxima and minima 
    \(\frac{dE}{dx}\) = 0
    \(\frac{d}{dx}\) \(\frac{KQx}{(x^2 + R^2)^{\frac{3}{2}}}\)
    = KQ\(\bigg(\frac{(x^2 + R^2)^{\frac{3}{2}} – x \frac{3}{2}(x^2 + R^2)^{\frac{1}{2}}(2x)}{(x^2 + R^2)^3}\bigg)\)
    =KQ\(\bigg(\frac{(x^2 + R^2)^{\frac{3}{2}} – x \frac{3}{2}(x^2 + R^2)^{\frac{1}{2}}(2x)}{(x^2 + R^2)^3}\bigg)\)
    KQ\(\bigg(\frac{(x^2 + R^2)^{\frac{1}{2}}((x^2 + R^2) - 3x^2)}{(x^2 + R^2)^3}\bigg)\) = 0
    x2 + R2 - 3x2 = 0
    -2x2 = -R2
    x = \(\frac{R^2}{2}\)
    x = \(\frac{R}{\sqrt{2}}\)
    Electric field on the axis of Ring will be maximum at R = \(\pm\)\(\frac{R}{\sqrt{2}}\).
    on x = \(\pm\)\(\frac{R}{\sqrt{2}}\) E = Emax
    E = \(\frac{KQx}{(x^2 + R^2)^{\frac{3}{2}}}\).        
    image eee
    Emax = \(\frac{kQ\frac{R}{\sqrt{2}}}{((\frac{R}{\sqrt{2}})^2 + R^2)^{\frac{3}{2}}}\)
    = \(\frac{KQ\frac{R}{\sqrt{2}}}{(\frac{3R^2}{2})^{\frac{3}{2}}}\) = \(\frac{KQ\frac{R}{\sqrt{2}}}{(\frac{3R^2}{2})(\sqrt{\frac{3R^2}{2})}}\)
    Emax = \(\frac{KQ\frac{R}{\sqrt{2}}}{\frac{3\sqrt{3}}{2\sqrt{2}}R^2R}\) = \(\frac{KQR}{\sqrt{2}\frac{3\sqrt{3}{2\sqrt{2}}}R^3}\)
    Emax = \(\frac{2KQ}{3\sqrt{3}R^2}\)

Examples

  • A negative charge '-q' is placed at the axis of a ring having charge 'Q' initially charge is placed at the center of ring then prove that it will perform S.H.M.  and find the period of S.H.M. or oscillation. Radius of ring is 'R'.
    Solution:
    E = \(\frac{KQx}{(x^2 + R^2)^{\frac{3}{2}}}\)
    Force on -q charge 
    F = -qE
    F = \(\frac{KQqx}{(x^2 + R^2)^\frac{3}{2}}\)
    If x << R
    F = \(\frac{-Kqqx}{R^3}\)
    F = \(\propto{-x}\) it will execute S.H.M.
    F = ma = \(\frac{-KQqx}{R^3}\)
    a = \(\frac{-KQqx}{mR^3}\)
    a = -\(\omega^2\)x from SHM
    \(\omega^2\) = \(\frac{KQq}{mR^3}\)
    \(\omega\) = \(\sqrt{\frac{KQq}{mR^3}}\)
    T = \(\frac{2\pi}{\omega}\)

Electric Field on a Axis of Disc

image ttt

Radius = R
Charge = Q
Charge density \(\sigma\) = \(\frac{Q}{\pi R^2}\)
Consider a ring 
area of ring in which charge is distributed = 2\(\pi\) drarea of ring on ring =\(\sigma\)\(\times\)area of ring =2\(\pi\) rdr \(\sigma\) = dq
Electric field at point p due to ring 
dE = \(\frac{Kdqx}{(x^2 + R^2)^{\frac{3}{2}}}\)
dE = \(\frac{k(2\pi r\sigma d r)x}{(x^2+R^2)^{\frac{3}{2}}}\)
\(\int\) dE = \(\int _0^R\) \(\frac{K2\pi r\sigma x}{(x^2+r^2)^{\frac{3}{2}}}dr\)
Let  x2 + r2 = z2
Diff W.r.t. Z
 0 + 2r\(\frac{dr}{dz}\) = 2z
rdr = zdz  ------------------------ (2)
where r = 0, z2 = x2 + 02 = x2, z = x
where r = R, z2 = x2 + R2, z = \(\sqrt{x^2 + R^2}\)
From eqn(2)
 \(\int\) dE = \(\int\) \(\frac{K2\pi\sigma x}{z^2}\)(rdr)
E = K2\(\pi\)\(\sigma\) x \(\int _x^{\sqrt{x^2 + R^2}}\)\(\frac{zdz}{z^2}\)
E = K2\(\pi\)\(\sigma\)x \((\frac{z^{-2H}}{-2H})\)
E = 2K\(\pi\)\(\sigma\) x \((\frac{-1}{z})_x^{\sqrt{x^2+R^2}}\)
E = K2\(\pi\)\(\sigma \)x \([ \frac{-1}{\sqrt{x^2+R^2}}-\frac{-1}{x}]\)
E = \(\frac{1}{4\pi \varepsilon_{\circ}}\)2\(\pi\) \(\sigma\) \(\Big(\frac{-x}{\sqrt{x^2 + R^2}}\) + 1\(\Big)\)
E = \(\frac{\sigma }{2 \varepsilon_{\circ}}\)\(\Big(\)1 - \(\frac{x}{\sqrt{(x^2 + R^2)}}\Big)\)
From diagram 
cos\(\theta\) = \(\frac{B}{H}\) = \(\frac{x}{sqrt{x^2  +R^2}}\)
E = \(\frac{\sigma}{2\varepsilon_{\circ}}\)(1 – cos\(\theta\))

Some Important Result  

  • If disc is infinite long then r \(\rightarrow\)\(\infty\) and 0 \(\rightarrow\) 90\(^{\circ}\) then electric field will be \(\frac{\sigma}{2\varepsilon_{\circ}}\)\((1 - cos90)\) = \(\frac{\sigma}{ 2\varepsilon_{\circ}}\).
  • Electric field due to infinite long plane is equal to constant i.e. \(\frac{\sigma}{2\varepsilon_{\circ}}\).

Electric field due to a wire 
There is a wire which a uniform charge distribution \(\lambda\) we want to calculate electric field at point p & p makes angle \(\theta_1\) & \(\theta _2\) as shown in Fig from boundary point of wire.

image yyy
image qqqq

Consider a small charge element of length dl charge in element = \(\lambda\) dl = dq.
E.F. due to this small element dE = \(\frac{ Kdq}{r^2}\)
dE = \(\frac{K\lambda dl}{r^2}\) = \(\frac{ k\lambda dl}{(r_0 sec\theta)^2}\) = \(\frac{ k\lambda dl}{(r^2_0 sec^2\theta)^2}\)  ------------ (1)
tan\(\theta\) = \(\frac{l}{r_0}\)
l = r0tan\(\theta\)
diff w.r.t . l
1 = r0sec2\(\theta\) \(\frac{d\theta}{dl}\) ---------------- (2)
Putting eqn(2) in eqn(1)
dE = \(\frac{ k\lambda dl}{(r^2_0 sec^2\theta)^2}\)
dE = \(\frac{K\lambda dl}{r_0^2 sec^2\theta = \frac {K \lambda d\theta}{r_0}}\)
\(\int\) dEx = \(\int\) dEcos\(\theta\)
Ex = \(\int\limits_{\theta_1}^{\theta_2}\)\( \frac{K\lambda cos\theta d\theta}{r_\circ}\)= \(\big[\frac{K\lambda sin\theta}{r_\circ}\Big]^{\theta_1}_{\theta_2}\)
Ex = \(\frac{k\lambda}{r_\circ}(sin\theta_1+sin\theta_2)\)
\(\int\) dEy = \(\int\) dE sin\(\theta\)
Ey = \(\int\limits_{\theta_2}^{\theta_1}\)\(\frac{K\lambda sin\theta d\theta }{r_\circ}\) = \(\Big[\frac{k\lambda cos \theta}{r_\circ}\Big]_{\theta_1}^{\theta_2}\)
Ey = \(\frac{K\lambda}{r_\circ}\)(cos\(\theta_2\) – cos\(\theta_1\))
Result 
\(\rightarrow\) For infinite long wire
\(\theta_1\) = \(\theta_2\) = 90\(^{\circ}\)
Ex = \(\frac{k\lambda}{r_0}\) (sin\(\theta_1\) + sin\(\theta_2\)) = \(\frac{2k\lambda}{r_0}\)

image qww

Ey = \(\frac{k\lambda}{r_\circ}(cos\theta_2 - cos\theta_1)\) = 0
\(\rightarrow\) For semi infinite wire 
\(\theta_1\) = 90\(^{\circ}\), \(\theta_2\) = 0\(^{\circ}\)
 Ex = \(\frac{k\lambda}{r_0}\)(sin\(\theta_1\) + sin\(\theta_2)\) = \(\frac{k\lambda}{r_0}\)
Ey = \(\frac{k\lambda}{r_0}\)(cos\(\theta_2\) – cos\(\theta_1)\) = \(\frac{k\lambda}{r_0}\)
Enet = \(\sqrt{2}\frac{k\lambda}{r_0}\)

image opppp
image hhhhh

Ex = \(\frac{k\lambda}{r_0}\)(sin\(\theta\) + sin\(\theta\)) = \(\frac{2sin\theta K\lambda}{r_0}\)

Electric Field at the Center of a Circular Wire 

image bft
image wqqq

Consider a small element at \(\theta\) angle as shown in Fig 
dl = Rd\(\theta\)
dq = \(\lambda\)dl
E.F. due to this element dE = \(\frac{kdq}{R^2}\) = \(\frac{ kdl\lambda}{R^2}\) = \(\frac{k\lambda R d\theta}{R^2}\) = \( \frac{k\lambda d\theta}{R}\)
\(\int\)dEx = \(\int\) dE cos\(\theta\)
Ex = \(\int\limits_{\theta_2}^{\theta_1}\) \(\frac{k\lambda }{R}\) cos\(\theta\) d\(\theta\) = \(\frac{k\lambda}{R}\) (sin)\(_{\theta_2}^{\theta_1}\)
Ex = \(\frac{k\lambda }{R}\) (sin\(\theta_1\) + sin\(\theta_2)\)
\(\int\) dEy = \(\int\) dEsin\(\theta\)
Ey = \(\int\limits _{\theta_2}^{ d\theta_1}\) sin\(\theta\) d\(\theta\) = \(\frac{k\lambda}{R}\)\((-cos\theta)_{\theta_2}^{\theta_1}\)
Ey = \(\frac{k\lambda}{R}\)(cos\(\theta_2\) - cos\(\theta_1)\)
Result 
\(\rightarrow\) For  semicircular wire.

image were

\(\theta\) = 90\(^{\circ}\)
\(\theta_2\) = 90\(^{\circ}\)
 Ex = \(\frac{2k\lambda}{ R}\)
Ey = 0
For quarter circle 

image ytrt

\(\theta_1\) = 90\(^{\circ}\),
\(\theta_2\) = 0\(^{\circ}\),
Ex = \(\frac{\lambda k}{R}\),
Ey = \(\frac{k\lambda}{R}\)
Enet = \(\sqrt{E_x^2 + E_y^2}\) = \(\sqrt{2}\)\(\frac{K\lambda}{R}\)

Examples

  • Two straight wire are folded in given form find out electric field at point C.
    image ccccc
    Enet = 2\(\sqrt{2}\)\(\frac{k \lambda}{R}\)
  • Find electric field at point C.
    image tre


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#Things To Remember



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