Applied Physics Chapter 28 & 29 PDF

## Chapter 28: The Electrical Field ### Electric Fields The region or space around a charge in which it can exert the force of attraction or repulsion on other charged bodies is called electric field. ### Electric Field Intensity The electrostatic force on a positive charge at a specific field...

## Chapter 28: The Electrical Field ### Electric Fields The region or space around a charge in which it can exert the force of attraction or repulsion on other charged bodies is called electric field. ### Electric Field Intensity The electrostatic force on a positive charge at a specific field point is called the electric field intensity. In order to find out electric field intensity due a point charge 'q', a test charge 'q₀' is placed in its electric field at a field point. The electric field intensity E due to a point charge 'q' is expressed as, $E = \frac{F}{q_0}$ where F is the electrostatic force between the source charge 'q' and test charge 'q₀'. The test charge 'q₀' should be very small, so that it cannot disturb the field produced by source charge 'q'. Therefore the electric field intensity can be written as, $E = lim_{q_0 \to 0} \frac{F}{q_0}$ **S.I Unit:** The S.I unit of electric field N/C or Vm^-1 ### Problems **28-1:** An electron is accelerated eastward at 1.84 x 10¹⁰ m/s² by an electric field. Determine the magnitude and direction of the electric field? **Given Data:** * Acceleration = a = 1.84 x 10¹⁰ m/s² * Mass of electron = m = 9.1 x 10^-31 kg * Charge on electron = q = -1.6 x 10^-19 C **Magnitude and direction of electric field= E = ?** Since, electric force acting on electron is $F = qE$ (1) According to Newton's 2nd Law of motion $F = ma$ (2) Comparing equ (1) and (2), we have $qE = ma$ $E = \frac{ma}{q} = \frac{9.1 \times 10^{-31} \times 1.84 \times 10^{10}}{1.6 \times 10^{-19}}$ $E = 0.01 N/C$ **28-2:** Humid air breaks (its molecules become ionized) in an electric field of 3.0 x 10N/C. What is the magnitude of an electric force on (a) an electron and (b) an ion (with a single electron missing) in the field? **Given Data:** * Electric field intensity = E = 3.0 x 10⁶ N/C * Charge on electron = q = -1.6 x 10^-19 C **To Find**: Electric force = F = ? **Calculations?** **(a) For electron** $F = qE = 1.6 \times 10^{-19} \times 3.0 \times 10^{6}$ $F = 4.8 \times 10^{-13} N$ **(b) For ion** $F = qE = 1.6 \times 10^{-19} \times 3.0 \times 10^{6}$ $F = 4.8 \times 10^{-13} N $ **28-3:** An alpha particle, the nucleus of a helium atom has mass of 6.64 x 10^-27 kg and charge 2e. What is the magnitude of and direction of electric field that will balance its weight. **Given Data** * Mass of alpha particle = m = 6.64 x 10^-27 kg * Charge = q = 2e = 2 x 1.6 x 10^-19 C **To Find:** * Magnitude and direction of electric field = E = ? **Calculations** Since the electric force balances the weight of the particle. $F_e = F_g$ $qE = mg$ $E = \frac{mg}{q} = \frac{6.64 \times 10^{-27} \times 9.8}{2 \times 1.6 \times 10^{-19}}$ $E = 2.03 \times 10^{5} N/C$ **Direction:** The direction of electric field is vertically upward against the force of gravity. **28-4:** In a uniform electric field near the surface of the earth, a particle having a charge of -2.0 x 10^-9 C is acted upon by a downward force of 3.0 x 10^-6 N. (a) Find the magnitude of the electric field (b) What is the magnitude and direction of the electric force exerted on a proton placed in this field? (c) What is the gravitational force on the proton? (d) what is the ratio of the electric force to the gravitational force in this case? **Given Data:** * Charge on the particle = q = -2.0 x 10^-9 C * Force = F = 3.0 x 10^-6 N * Charge on proton = q_p = 1.6 x 10^-19 C * Mass of photon = m = 1.67 x 10^-27 kg **To Find:** * (a) Magnitude of electric field = E = ? * (b) Magnitude and directions of electric force on proton = F_e =? * (c) Ratio of electric to gravitational force = $\frac{F_e}{F_g}$ =? **Calculations** *(a) The magnitude of electric field is calculated as* $E = \frac{F}{q} = \frac{3.0 \times 10^{-6}}{2.0 \times 10^{-9}} = 1.5 \times 10^{3} N/C$ *(b) The magnitude of electric force acting on proton is calculated as* $F = qE = 1.6 \times 10^{-19} \times 1.5 \times 10^{3}$ $F = 2.4 \times 10^{-16} N$ **Direction:** Since proton is a positive charge particle, so the direction of electric force acting on proton is vertically upward. *(c) Ratio of electric to gravitational force for proton is* $\frac{F_e}{F_g} = \frac{2.4 \times 10^{-16}}{1.67 \times 10^{-27} \times 9.8}$ $\frac{F_e}{F_g} = 1.5 \times 10^{11}$ **28-3b:** (a) When acceleration would a proton experience if the electric field is 2.16 x 10⁴ N/C ? What speed the proton would attain if the field ads over a distance of 1.22cm **Given Data:** * Electric field Strength = E = 2.16 x 10⁴ N/C * Distance = d = 1.22cm = 1.22 x 10^-2 m * Charge on proton = q = 1.60 x 10^-19 C * mass of proton = m = 1.67 x 10^-27 kg **To Find:** * (a) Acceleration = a = ? * (b) Speed = v = ? **Calculations:** *(a)* $F_e = qE$ (1) comparing equ (1) and (2) we get $ma = qE$ $a = \frac{qE}{m} = \frac{1.6 \times 10^{-19} \times 2.16 \times 10^{4}}{1.67 \times 10^{-27}}$ $a = 2.07 \times 10^{12} m/s^{2}$ The speed of proton is calculated using third equ. of motion. Since v_i= 0 and v² = v_i² + 2as $v = \sqrt{2ad} = \sqrt{2 \times 2.07 \times 10^{12} \times 1.22 \times 10^{-2}}$ $v = 2.25 \times 10^{5} m/s$ **28-38:** A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of negatively charged plate and strike the surface of opposite plate, 195 cm away, 14.7 ns later. (1) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field? **Given Data:** * Distance b/w the plates = d= 1.95cm = 195 x 10^-6 m * Time taken = t = 14.7ns = 14.7 x 10^-9 s **To Find:** * (a) Speed of electron = v = ? * (b) Magnitude of electric field = E = ? **Calculations:** *(a)* The average speed b/w the plates is $v = \frac{d}{t} = \frac{195 \times 10^{-6}}{ 14.7 \times 10^{-9}} = 1.33 \times 10^{4} m/s$ Now the speed with which the electrons hits the plate is twice of the low $V = 2v_{au} = 2 \times (1.33 \times 10^{4}) m/s = 2.66 \times 10^{4} m/s$ *(b)* The electrons acceleration is $a = \frac{v}{t} = \frac{2.66 \times 10^{4}}{14.7 \times 10^{-9}} = 1.80 \times 10^{13} m/s^{2}$ Now, the magnitude of electric field is calculated by the relation $F = qE = ma$ $E = \frac{ma}{q} = \frac{9.1 \times 10^{-31} \times 1.20 \times 10^{13}}{1.6 \times 10^{-19}}$ $E = 1.03 \times 10^{6} N/C$ **28-3: Electric Field Intensity due to a point charge** Consider a test charge 'q₀' placed at point P in the electric field of a point charge 'q' at a distance 'r' apart. We want to find out electric field intensity at point 'p' due to a point charge 'q'. The electrostatic force 'f' between 'q₀' and 'q' can be find out by using expression, $F = \frac{1}{4\pi\epsilon_0} \frac{qq_0}{r^2}$ The electric field intensity 'E' due to a point charge 'q' can be obtained by putting the value of electrostatic force in equation (1), $E = \frac {1}{4\pi\epsilon_0} \frac {q}{r^2}$ This expression gives the magnitude of electric field intensity due to a point charge 'q'. In vector form, the electric field intensity 'E' will be: $E = \frac {1}{4\pi\epsilon_0} \frac {q}{r^2} \hat{r}$ Where $\hat{r}$ is the unit vector which gives the direction of electric field intensity. **Problem** **28-5:** What is the magnitude of a point charge chosen so that the electric field 75.0cm away has the magnitude 2.30 N/C? **Given Data:** * Distance = r= 75cm = 0.75 m * Magnitude of electric field = E = 2.3 N/C * K = $4\pi\epsilon_0 = 9 \times 10^{9} Nm^{2}/C^{2}$ **To Find:** Magnitude of point charge = q = ? **Calculations:** Since the magnitude of electric field is calculated as. $E = \frac{Kq}{r^2}$ $q = \frac{Er^2}{K} = \frac{2.30 \times (0.75)^{2}}{9 \times 10^{9}} = 1.44 \times 10^{-10} C$ **28-5: Electric Field due to a Dipole** **Electric Dipole:** Two equal and opposite charges separated by distance having same magnitude and direction is called electric dipole. **Electric dipole moment:** Two equal and opposite charges having same magnitude and same direction and separated by distance 'd' is called electric dipole moment. => Two equal and opposite charge consider two point charges +q and -q of equal magnitude lying distance 'd' apart as shown in the figure. We want to determine the electric field intensity due to a dipole at point 'p' the point 'p' is at distance 'x' along the perpendicular bisector of the line joining the charges. Let the electric field intensities at point P due to the charges +q and -q, be E₊ and E_- respectively. The total electric field intensity at point p due to the charges +q and -q is given by the expression- $E = E+ + E-$ For the present case, E₊ = E_- because the point P is equidistant for the charges +q and -q. Therefore. $E_+= E_- = \frac {1}{4\pi\epsilon_0} \frac {q}{r^2}$ , since r = $\sqrt {x^2 + (\frac{d}{2})^2}$ From the figure it is clear that the y-components of E₊ and E_- will cancel the effect of each other while the x-components of E₊ and E_- added up to give of resultant electric field intensity of electric dipole. Therefore $E = E+ COSA + E-COSO$ $E = 2E+COSO$ -(1) $E+ = E-$ $COSO = \frac{x}{r}$ $COSO = \frac{x}{\sqrt{x^2 + (\frac{d}{2})^2}}$ By putting the value of E₊ and COSO in equ. we get $E = 2E+ [\frac{x}{\sqrt{x^2 + (\frac{d}{2})^2}}]$ $E = 2 [\frac {1}{4\pi\epsilon_0} \frac {q}{x^2+(\frac{d}{2})^2}] [\frac{x}{\sqrt{x^2 + (\frac{d}{2})^2}}]$ $E = \frac {1}{4\pi\epsilon_0} \frac {qd}{(x^2 + (\frac{d}{2})^2)^{3/2}}$ The quantity P = qd is called dipole moment. Therefore, $E = \frac {1}{4\pi\epsilon_0} \frac{P}{(x^2 + (\frac{d}{2})^2)^{3/2}}$ $E = \frac {1}{4\pi\epsilon_0} \frac{P}{[x^2 + ( \frac{d}{2})^{2}]^{3/2}}$ $E = \frac {1}{4\pi\epsilon_0} \frac{P}{[x(1+ (\frac{d}{2x})^2]^{3/2}}$ By using binomial expansion, we get $E = \frac {1}{4\pi\epsilon_0} \frac{P}{[1+ (\frac{d}{2x})^2 + \frac{3}{2} (\frac{d}{2x})^4 + ........]}$ If we neglect the 2nd and higher terms, then $E = \frac {1}{4\pi\epsilon_0} \frac{P}{x^3} $ because d/2 is very small than x, so we get d/2 =0 $E = \frac {1}{4\pi\epsilon_0} \frac{P}{x^3} $ This is the expression of electric field due to a dipole. ### Problems **28-6:** Calculate the dipole moment of an electron and a proton 4.30nm apart. **Given Data:** * Distance b/w proton and electron = d = 4.3nm = 4.3 x 10^-9 m * Charge on proton as electron= q = 1.6 x 10^-19 C **To find:** * Dipole moment = P = ? **Calculation:** $P = qd = 1.6 \times 10^{-19} \times 4.3 \times 10^{-9}$ $P = 6.88 \times 10^{-28} Cm $ **28-9:** Calculate the magnitude of the electric field due to an electric dipole of dipole moment 3.56 x 10^-29Cm at a point 25.4 nm away along the bisector axis. **Dipole Moment** = P = 3.56 x 10^-29 Cm **Distance** = r = 25.4nm = 25.4 x 10^-9 m * K = $4\pi\epsilon_0 = 9 \times 10^{9} N/m^{2}/C^{2}$ **Magnitude of electric field = E = ?** Since the electric field due to adipole is calculated by $E = \frac {1}{4\pi\epsilon_0} \frac{P}{(x^2 + (\frac{d}{2})^2)^{3/2}} $ $E = \frac {KP}{(x^2 + (\frac{d}{2})^2)^{3/2}}$ $E = 9 \times 10^{9} \times 3.56 \times 10^{-29}$ $E = 19.5 \times 10^{3} N/C = 19.5kN/C$ **28-10:** If both charges are positive separated by a fixed distance d, show that electric field E at a point P a distance x along its perpendicular bisector of the line joining the charges, assuming x>> d is given by. **Solution:** If both the charges are positive then $E = 2E SIN\Theta$ $E = 2 [\frac {1}{4\pi\epsilon_0} \frac {q}{(x^2 + (\frac{d}{2})^2)^{3/2}}] [\frac{ \frac{d}{2}}{\sqrt{x^2 + (\frac{d}{2})^2}}]$ When x>>d then d»0, so $E = 2[\frac {1}{4\pi\epsilon_0} \frac {q}{x^3}] [\frac{\frac{d}{2}}{x}]$ $E = \frac {1}{4\pi\epsilon_0} \frac {2q\frac{d}{2}}{x^3}$ $E = \frac {1}{4\pi\epsilon_0} \frac {qd}{x^3}$ Which is the required result. **28-6: Electric Field Intensity due to an infinite line of charges** Consider an infinite line of positive charge We want to calculate the electric field intensity at Point 'p' at a perpendicular distance 'x' from the line of charge. As the charge is distributed uniformly over it, so it has constant linear charge density $\lambda = \frac{charge}{length} = \frac{q}{L}$ For an infinitesimal length element 'dy' having chargedy' $\lambda = \frac{dq}{dy} => dq = \lambda ds$ The electric field intensity due to this length element at point 'p' is given by; $dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{\lambda dy}{x^2 + y^2}$ , since r = $\sqrt {x^2 + y^2}$ The rectangular components of 'dE' are $dEx= dE COSO$ and $dEy$ = $dESINO$ If we consider identical charge elements symmetrically located on both sides of 'O', then the y-components of electrical field intensity will cancel out each other and its x-components are added up to give the total electric field intensity due to this continuous charge distribution. Therefore, the total electric field intensity will be $E = Ex = \int_{y=-\infty}^{y=\infty} dEx$ $E= \int_{y=-\infty}^{y=\infty} dE cos\theta = 2 \int_{y=0}^{y=\infty} dE COSO$ Pultting value of 'dE' from equ (1) we get $E = \frac{2}{4\pi\epsilon_0} \int_{y=0}^{y=\infty} \frac{\lambda dy}{x^2 +y^2} cos\theta $ $E = \frac{2\lambda}{4\pi\epsilon_0} \int_{y=0}^{y=\infty} \frac{dy}{x^2 + y^2} cos\theta $ $E = \frac{\lambda}{2\pi\epsilon_0} \int_{y=0}^{y=\infty} \frac{dy}{x^2+y^2} cos\theta $ Since y= x tan $\theta $ dy = x sec²$\theta$ d$\theta$ x = tan $\theta$ tan$\theta$ = sec$\theta$ Putting values in equ (2), we get $E = \frac{\lambda}{2\pi\epsilon_0} \int_{\theta=0}^{\theta=\frac{\pi}{2}} \frac{x sec^{2}\theta d\theta}{x^{2} + x^{2} tan^{2}\theta} cos\theta$ $E = \frac{\lambda}{2\pi\epsilon_0} \int_{\theta=0}^{\theta=\frac{\pi}{2}} \frac{x sec^{2}\theta d\theta}{x^{2}(1+ tan^{2}\theta)} cos\theta$ $E= \frac{\lambda}{2\pi\epsilon_0} \int_{\theta=0}^{\theta=\frac{\pi}{2}} \frac{x sec^{2}\theta d\theta}{x^{2} sec^{2}\theta} cos\theta$ $E= \frac{\lambda}{2\pi\epsilon_0 x} \int_{\theta=0}^{\theta=\frac{\pi}{2}} COS\theta d\theta = \frac{\lambda}{2\pi\epsilon_0 x}[SIN\theta ]_{0}^{1/2}$ Therefore, $E = \frac {1}{2 \pi \epsilon_0 x}[1 - 0] = \frac{\lambda}{2 \pi\epsilon_0 x}$ $E = \frac{\lambda}{2\pi\epsilon_0 x}$ $E = \frac{\lambda}{2\pi\epsilon_0 x} [SIN\theta ]_{0}^{1/2}$ This is the expression for electric field intensity at point 'p' due to an infinite line of charge. **28.7; Electric Field Intensity due to a Ring of charge.** Consider a ring of positive charge of radius 'R' as shown in the figure below. We want to find out the electric field intensity at point 'p' which is at the distance 'z' from the plan of sing. As the charge is distributed uniformly over it, so it has constant linear charge density $\lambda$. For an infinitesimal length element 'ds' of sing, $dq = \lambda ds$ The electric field intensity due to the charge $dq$ $dE = \frac {1}{4\pi\epsilon_0} \frac {dq}{r^2}= \frac{1}{4\pi\epsilon_0} \frac{\lambda ds}{r^2}$, since r = $\sqrt {Z^2 + R^2}$ The rectangular components of 'dE' are $dEz= dE COSO$ and $dEy$ = $dESINO$ We consides identical charge elements located on the opposite end of the diameter, then dey components will cancel out each other and dez components are added up to give the final value of electric field intensity at point P. Therefore, the total electric field will be $E = Ez = \int dE z = \int dE COSO$ Putting value of 'dE' from equ (1), we get $E = \int \frac {1}{4\pi\epsilon_0} \frac {\lambda ds}{\sqrt{2^2+R^2}} cos\theta$ $E = \frac{1}{4\pi\epsilon_0} \int \frac{\lambda ds}{2^2+R^2 } cos\theta $, From figure, r = $\sqrt{Z^2 + R^2}$ <br> $E = \frac {1}{4\pi\epsilon_0} \int \frac{\lambda ds}{(Z^2 + R^2)^{3/2}} cos\theta $ , $z^2 + R^2 = r^2$ $E = \frac {1}{4\pi\epsilon_0} \int \frac{ \lambda ds}{r^3} cos\theta$ , $r^2 = z^2 + R^2$ <br> $E = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{r^2} \int ds$ $E = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{r^2} (2\pi R) = \frac{QR}{4\pi\epsilon_0 r^{3}}$ $E = \frac{QR}{4\pi\epsilon_0 (Z^2+R^{2})^{3/2}} = \frac{QR}{2\pi\epsilon_0 (2^2 + R^2)^{3/2}}$, total charge q = $\lambda 2\pi R$, total length = 2$\pi R$ When the point P is far away from sing i.e Z>>R so that R can be neglected. $E = \frac{Q\sqrt{Z}}{4\pi\epsilon_0 (Z^2)^{3/2}} = \frac{Q\sqrt{Z}}{4\pi\epsilon_0 Z^{3}}$ <br> $E = \frac{Q}{4\pi\epsilon_0 Z^{2}}$ Which is the expression for the electric field intensity when the field point is far away from the sing. Thus the charged sing acts like a point charge when the field point is at the large distance. ### Problem **28.26:** At what distance along the axis of a charged ring of radius R is the axial electric field strength a maximum? Since the electric field at a point on the axis of the ring is given by the relation. $E_p = \frac {1}{4\pi\epsilon_0} \frac{Qz}{(Z^2 + R^2)^{3/2}}$ Since 'q' and $4\pi\epsilon_0$ are constants, so the electric field will be maximum when $(Z^2 + R^2)^{3/2}$ is maximum. Taking derivative of $\frac {1}{(2^2 + R^2)^{3/2}}$ with respect to Z $\frac{d}{dz}[\frac{1}{(Z^2 + R^2)^{3/2}} ]= \frac{1}{(Z^2 + R^2)^{3/2}} - \frac{3Z^2}{(Z^2 + R^2)^{5/2}}$ $\frac{-(2^2 + R^2) - 3Z^2}{(Z^2 + R^2)^{5/2}} = \frac{(Z^2 + R^2) + 3Z^2}{(Z^2 + R^2)^{5/2}} = \frac{2Z^2 + R^2}{(Z^2 + R^2)^{5/2}} $ This will vanish when the numerator vanishes. $2Z^{2}+R^2 = 0$ $R^2 - 2Z^2 = 0$ $R^2 = 2Z^2$ $Z^2 = \frac{R^2}{2}$ $Z = \frac{R}{\sqrt{2}}$ ## Chapter 29: Gauss's Law ### 29-1: Electric Flux **Statement:** The number of electric lines of force passing normally through a certain area is called the electric flux. It is measured by the product of area and the component of electric field intensity normal to the area. It is denoted by the symbol Φ_e Consider a surface 'S' placed in a uniform electric field of intensity 'E'. Let 'A' be the vectors area of the surface. The component of E perpendicular to the area A is ECOSO as shown in the figure below. The electric flux through the surface S is given by Φ_e = A (ECOSO) Φ_e = EACOSO Φ_e = EA Thus the electric flux is the scalar product of electric field intensity and the vectors area. The SI unit of the electric flux is Nm²/C. If the electric field is not uniform, then we divide the surface into a number of small patches each of area ΔA. Then equ (1) becomes, Φ_e = Σ E_i * ΔA_i When n→∞ or ΔA → 0, then the sigma is replaced by the surface integral i.e, Φ_e= ʃE.dA By convention, the outward flux is taken as positive and inward flux is taken as negative. ### Problems **29-1:** The square surface measures 3.2 mm on each side. It is immersed in a uniform electric field with E=1800 N/C. The field lines make an angle of 65° with the outward pointing normal. Calculate the flux through the surface. **Given Data:** * Length of each side = L= 3.2 mm = 3.2 x 10^-3 m * Electric field = E = 1800 N/C * Angle θ = 65° **To Find:** Electric flux = Φ_e =? **Calculations** Φ_e = EA = EACOSO Φ_e = 180 (3.2 x 10^-3)² cos(180°-65) Φ_e = 180 (3.2 x 10^-3)² cos(145°) Φ_e = -0.0078 Nm²/C = -7.8 x 10^-3 Nm²/C **29-2:** A cube with 1.4 m edges is oriented in a region of uniform electric field. Find the electric flux through the right face if the electric field, in N/C is given by (a) 6i (b) -2j (c) -3i + 4k (d) calculate the total flux through the cube for each of these fields. **Given Data:** * Length of each side of cube = L = 1.4 m * Electric field in N/C is given by (a) 6i (b) -2j (c) -3i +4k **To Find:** (a) Electric flux through right face in each case = Φ_e=? (b) Total flux through the cube = Φ_e= ? The right face has an area element given by A = (1.4m)². Now the flux through this face in each case is calculated as (a) Φ_e= EA = (6 N/C)i. (1.4m)² = 0 (b)Φ_e = EA = (-2 N/C)j. (1.4m)²j = 4m/C (c) Φ_e = EA = (-3i + 4k). (1.4m)²j = 0 (d) In each case the field is uniform so we can simply evaluate Φ_e = ẺA where Ả has been six parts, one for every face. The faces however, have the same size but are organized in pairs with opposite directions. These will cancel, so the total flux is zero in all three cases. ### 29.2: Grauss's Law. **Statement** The total electric flux through any closed surface is ε₀ times the total charge enclosed by the surface Φ_e = Total Charge enclosed **Explanation:** The Gauss's law gives the relation between total flux and total charge endosed by the surface. Consides a collection of positive and negative charges in a certain region of space. According to Gauss's law, Φ_e = Q/ε₀ ε₀Φ_e = Q where Q is the net charge enclosed by the surface But Φ_e= ʃE.dA ε₀ʃE.dA = Q ʃE.dA = Q/ε₀ **SI Unit:** The SI unit of Gauss's law is Nm²/C ### 29.3: Differential Form of Gauss's law If the charge is distributed into a volume having uniform volume charge density 'ρ', then according to Gauss's law. Φ_e = ε₀ʃ_Vρdv We know that by Divergence theorem, ʃ_SE.dA = ʃ_Vdiv E dv Equ (1) becomes ʃ_Vdiv E dv = ε₀ʃ_V ρdv (div E)dv = ε₀ʃ_V ρdv As dv ≠ 0 ʃ( div E - ε₀ρ)dv = 0 div E - ε₀ρ = 0 div E = ε₀ρ ### 29-4: Integral Form of Gauss's Law According to Gauss's law Φ_e = ʃE.dA = Q/ε₀ where 'Q' is the total charge enclosed. If the charge is uniformly distributed into a volume having charge density 'ρ' then. Φ_e = ʃE.dA = ε₀ʃ_Vρdv If the charge is uniformly distributed over a surface having a surface charge density ‘σ’, then Φ_e = ʃE.dA = ε₀ʃ_sσdA Eq (1) and (2) are the integral form of Gauss’s law. ### Problems **29-5:** A Point charge of 1.84 μC is at the center of a cubical Gaassian surface 55cm on edge. Find Φ_e through the surface. * Magnitude of charge = q = 1.84 μC = 1.84 x 10^-6 C * Length of each side of