Chapter GAUSS LAW Recmmended Prblems: 1,4,5,6,7,9,11,13,15,18,19,1,7,9,31,35,37,39,41,43,45,47,49,51,55,57,61,6,69.
LCTRIC FLUX lectric flux is a measure f the number f electric filed lines penetrating sme surface in a directin perpendicular t that surface. A = A cs with is the angle between the and A. The directin f A, the area f a surface, is always nrmal t that surface and pints utward fr clsed surfaces, (by clsed surface we mean that surface which divides space int inside and utside regins). It is bvius frm the last euatin that is a scalar uantity with the SI unit f N.m /C.
If the electric field is nt cnstant ver the surface in uestin, surface da Nte that if is cnstant ver the surface it shuld be taken ut f the integratin and we recver the first euatin as expected. Cnsider nw the clsed surface in the figure, If is entering the surface, 90, and s the flux fr such an element is negative. If is utward and 90, hence the flux thrugh this element is psitive. da da Frm this argument, we can expect that if a field line entering and leaving the same clsed surface the net electric flux thrugh the clsed surface frm that line is zer.
xample 4.1 What is the e.flux thrugh a sphere that has a radius f R=1.0 m and carries a charge f =1.0 C at its center. Slutin: The e.field due t the pint charge at any pint n the surface f the sphere is k r This field is cnstant ver the surface s we write A = k R A cs The field pints radially utward and s & A are parallel, i.e., 0 cs 1 A cs0 4 R 110 6 4R cs0 1 8.8510 1.1310 5 N.m /C
xample 4. A cube f edge l is riented in a unifrm e.field, as shwn. Find the net e.flux thrugh the surface f the cube. Slutin: The net flux thrugh the cube is the sum f the fluxes thrugh the 6-faces f the cube: The 4-faces named in the figure and the ther tw unnumbered faces (the frth and the back faces), that is A 1 da 1 -A da 3 da 3 4 da 4 da back z da b y frth da 3 da 4 da The flux thrugh the faces 3, 4, the back, and the frth is zer because and da are perpendicular. Nw the angle between and da 1 is zer, while the angle between and da is 180 A A 0 f da 1 x
Test Yur Understanding (1) A charge Q is placed at the center f a spherical shell (the red ne). If the radius f the shell is increased (the black ne), what happens t the flux thrugh the shell and the magnitude f the electric field at the surface f the shell? Q a) The flux remains the same and the field decreases. b) The flux remains the same and the field increases. c) The flux decreases and the field remains the same. d) The flux and field bth decrease.
Gauss Law the electric flux thrugh any clsed surface is eual t the net charge inside that surface divided by, that is The integral is ver a clsed surface, called the Gaussian surface. The e. field due t the whle charge distributin. da in The surface f the Gaussian surface. The net charge inside the Gaussian surface.
Let us verify Gauss law by cnsidering a psitive pint charge surrunded by tw clsed surfaces: S 1 is spherical, whereas S is irregular. Culmb s law tells us that the magnitude f the electric field is cnstant everywhere n the spherical surface and given as S 1 S k r Since & A are parallel (=0), then the flux thrugh S 1 as da But k A 1 4 k r 4 r The figure shws that the number f field lines crssing S 1 is the same as that lines crssing S, that is, the flux thrugh the tw surfaces are eual and independent f their shapes.
If the charge exists utside a clsed surface, the electric field lines entering the surface must leave that surface. Hence, the electric flux thrugh that surface is zer. The practical utility f Gauss law lies largely in prviding a smart way t evaluate the electric filed fr a charge distributin. Fr this way t be as easy as pssible we must be able t chse a hypthetical clsed surface (Gaussian surface) such that the electric filed ver its surface is cnstant. This can be attained if the fllwing remarks are satisfied:
(i) The charge distributin must have a high degree f symmetry (spherical, cylindrical with infinite length, plane with infinite extends). (ii) The Gaussian surface shuld have the same symmetry as that f the charge distributin. (iii) The pint at which is t be evaluated shuld lie n the Gaussian surface. (iv) If is parallel t the surface r zer at every pint, then da 0 (v) If is perpendicular t the surface at every pint, and since is cnstant, then da A
Test Yur Understanding () Cnsider the charge distributin shwn in the figure. The charges cntributing t the electric flux,, thrugh S ' and t the electric filed,, at a pint n S ' is: a) All the fur charges cntributing t bth and. b) Only and 3 cntributing t bth and. c) Only and 3 cntributing t while the fur charges cntributing t. d) Nn f the fur charges cntributing t bth and.
Test Yur Understanding (3) Fur clsed surfaces, S 1 thrugh S 4, tgether with the charges Q, -Q, and -Q are sketched in the figure shwn. (The clred lines are the intersectins f the surfaces with the page.) The surface that has the largest electric flux is: a) S 1 b) S c) S 1 and S d) S 3 Referring t the same figure, the surface that has the smallest electric flux is: a) S 4 b) S c) S And S 4 d) S 1
xample 4.4 Find the e.f a distance r frm a pint charge. Slutin Since the charge distributin is spherical, we chse the Gaussian surface as a sphere f radius r. Nw da in Gaussian surface It is clear that and da are parallel, is cnstant ver the surface. is the ttal charge enclsed by Gaussian surface r da da r 4 4 r
xample 4.6 A thin spherical shell f radius a has a ttal charge Q unifrmly distributed ver its surface. Find the magnitude f the electric field at a pint a) utside the shell a distance ra b) inside the shell a distance ra r Slutin a) We chse an arbitrary pint utside the shell. Gaussian surface Q Since the charge distributin is spherical, we chse a spherical Gaussian surface cncentric with the shell. It is clear that is nrmal t the surface at every pint in the Gaussian surface, that is and da are parallel, s we write a da in A in Q
But A, the area f the Gaussian surface is A 4r Q 4 r Q b) Nw we chse a pint inside the shell. In this case the Gaussian surface is inside the shell. r a It is clear that there is n charge inside the Gaussian surface, that is in 0 0
xample 4.5 An insulating sphere f radius a has a ttal charge Q unifrmly distributed thrugh its vlume. Calculate the electric filed a) utside the sphere a distance ra b) inside the sphere a distance ra Gaussian surface Q a r Slutin a) Again, and because the spherical symmetry f the charge distributin, we select a spherical Gaussian surface f radius r, cncentric with the sphere. As fr the case in the previus example we write da in Substituting fr A by A A 4r Q Q 4 r
b) In this case we chse a spherical Gaussian surface f radius ra. T find the charge in within the Gaussian surface f vlume V in, we use the fact that Gaussian surface Q a r in V in where is the vlume charge density. Knwing that 3 3 V 3 4 in r and Q Qr 4 3 in a a 3 Nw, applying Gauss law we btain da in Qr 4r 3 a 3 3 Qr 4 a Nte that at the center f the sphere, r=0, =0. Is it reasnable? 3
xample 4.7 Find the electric field at a distance r frm an infinite line charge f unifrm density. Slutin As a Gaussian surface we select a circular cylinder f radius r with height h and caxial with the line charge. Since the cylinder has three surfaces, the integral in Gauss's law has t be split int three parts: h r da b da b da c the curved surface, and the tw bases. da da da b b c A in Frm the symmetry f the system, is parallel t bth bases. Furthermre, it has a cnstant magnitude and directed radially utward at every pint n the curved surface f the cylinder.
A rh in The ttal charge inside the Gaussian surface is nly the charge f the part inside the surface, i.e., h r in = h. Nw we write h rh r Nte that if the wire is nt t lng its ends will be clsed t any Gaussian surface. Since the electric field at, and clsed t the ends is nt unifrm it will be impssible t manage the integral f Gauss law.
xample 4.8 Find the electric field due t a nncnducting, infinite plane with unifrm surface charge density. da b Gaussian surface da b Slutin da c T slve this prblem we select as a Gaussian surface a small cylinder whse axis is perpendicular t the plane and whse ends each has an area A. As we d in the previus example we write Gauss law as A b da b da A c da in With in is given by in A
Furthermre, is directed nrmally utward and has a cnstant magnitude at each pint n the tw ends f the cylinder. This means that the third integral vanishes and the first tw integrals each reduce t A. A A Nte that this result agrees with the result f xample 1.9. It is left as an exercise t shw that the prblem can be slved using a Gaussian surface in the shape f parallelepiped.
Test Yur Understanding (4) A pint charge Q is lcated just abve the center f the flat face f a hemisphere f radius R as shwn. The electric flux thrugh the flat face f the hemisphere is: a) Q/ 4 b) Q/ c) -Q/ 4 d) -Q/
CONDUCTORS IN LCTROSTATIC QUILIBRIUM If a cnductr is charged, charges will mve a way frm each ther due t the repulsin frce between them. Fr the charges t be as far a way frm each ther as they can, they will mve t the uter surface f the cnductr. Cnductrs with n mtin f charges are said t be in electrstatic euilibrium. Such cnductrs have the fllwing prperty: (i) Any excess charge will reside entirely n the uter surface f an islated cnductr.
Bearing this prperty in mind, if a Gaussian surface is cnstructed inside such a cnductr, it will nt enclse any charge. Using Gauss s law, we cnclude that Gaussian surface (ii) The electric field must be zer inside any cnductr in electrstatic euilibrium. (iii) The electric field just utside a cnductr is always perpendicular t the surface f the cnductr and eual t If this is nt the case, the free charges will mve alng the surface and this vilate the cnditin f euilibrium.
Let us nw use Gauss s law t calculate the magnitude f the electric filed just utside a charged cnductr. T d s we draw a Gaussian surface in the shape f small cylinder as shwn. Gauss s law then gives b da b da c da in But the base inside the cnductr has n flux thrugh it since =0, and the flux thrugh the curved surface is zer since is nrmal t the area vectr f this surface. Hence, the last tw integrals vanish leaving us with A in A
xample 4.10 A cnducting sphere f radius a has a net charge Q. Cncentric with this sphere is a cnducting spherical shell f inner radius b and uter radius c and has a net charge f -Q. c Q a b -Q a) Find the electric field in the regins inside the sphere, between the sphere and the shell, inside the shell, and utside the shell. b) Determine the induced charge n the inner and uter surfaces f the shell. Slutin Since the sphere is cnducting we cnclude that the electric field in the first regin is zer, i.e., 1 =0.
In the secnd regin we select a spherical Gaussian surface with radius a r b. Since is cnstant in magnitude ver the Gaussian surface and nrmal t it, we find frm Gauss law r in 4 Q 4 r 4r Q a In the third regin the electric filed is again must be zer since this regin is inside the shell which is a cnductr, i.e., 3 =0. In the last regin utside the shell we cnstruct a Gaussian surface with radius r c. This surface enclse a ttal charge f (-QQ)=Q. Gauss law is then gives Q c b -Q
4 Q 4 r Induced charges zer Q Q -Q -Q b) The charge n the sphere induces a charge f -Q is n the inner sphere. Q The induced charge n the uter surface will be Q. Therefre, the net charge n the uter surface will be Q. If ne cnstruct a Gaussian surface in that regin with radius b < r < c the net charge inside that surface must be zer. S the electric field 3 is zer as expected.