Gauss Law Questions and Answers

3 The device shown in measure the pressure and volume flow rate when a person exhales There is a cylindrical pipe with inside radius R There is a slit of width b running down the length of the cylinder Inside the tube there is a light movable piston attached to an ideal spring of force constant K In equilibrium position the piston is at a position where the slit starts shown by line AB in the figure A person is made to exhale into the cylinder causing the piston to compress the spring Assume that slit width b is very small and the outflow area is much smaller than the cross section of the tube even at the pistons full extension A person exhales and the spring compresses by x Density of air pl a Calculate the gage pressure in the tube b Calculate the volume flow rate Q of the air A 1 R B X
Physics
Gauss Law
3 The device shown in measure the pressure and volume flow rate when a person exhales There is a cylindrical pipe with inside radius R There is a slit of width b running down the length of the cylinder Inside the tube there is a light movable piston attached to an ideal spring of force constant K In equilibrium position the piston is at a position where the slit starts shown by line AB in the figure A person is made to exhale into the cylinder causing the piston to compress the spring Assume that slit width b is very small and the outflow area is much smaller than the cross section of the tube even at the pistons full extension A person exhales and the spring compresses by x Density of air pl a Calculate the gage pressure in the tube b Calculate the volume flow rate Q of the air A 1 R B X
8 A long string with a charge of per unit length passes through an imaginary cube of edge a The maximum flux of electric field through cube will be 1 a Eo 6 a 2 LA 2 a Eo 32a
Physics
Gauss Law
8 A long string with a charge of per unit length passes through an imaginary cube of edge a The maximum flux of electric field through cube will be 1 a Eo 6 a 2 LA 2 a Eo 32a
A point charge q is placed in a cubical box of side a If all the dimensions of box are changed by a factor of 3 then by what factor the flux crossing the box will change O Decrease by 9 times O Increased by 9 times O Either 1 or 2
Physics
Gauss Law
A point charge q is placed in a cubical box of side a If all the dimensions of box are changed by a factor of 3 then by what factor the flux crossing the box will change O Decrease by 9 times O Increased by 9 times O Either 1 or 2
There is an infinitely long cylindrical region of radius a in which a uniform but time varying axial magnetic field is present whose magnitude increases at rate of k T s Figure shows one of the cross sections whose center is O In this cross section a concentric circular metallic loop ABCDA is fixed Resistance of semicircle part ACB of the loop is greater that resistance of semicircle part ADB Consider path 1 to be A CB and path 2 to be A D B E induced electric field and Enet is the net electric field in the loop Choose the incorrect inequalit y ies A End di Eind di path I OLE Path 2 101 D B path 1 Enet dl path 2 Enet di 15
Physics
Gauss Law
There is an infinitely long cylindrical region of radius a in which a uniform but time varying axial magnetic field is present whose magnitude increases at rate of k T s Figure shows one of the cross sections whose center is O In this cross section a concentric circular metallic loop ABCDA is fixed Resistance of semicircle part ACB of the loop is greater that resistance of semicircle part ADB Consider path 1 to be A CB and path 2 to be A D B E induced electric field and Enet is the net electric field in the loop Choose the incorrect inequalit y ies A End di Eind di path I OLE Path 2 101 D B path 1 Enet dl path 2 Enet di 15
a A conductor A with a cavity as shown in Fig 1 36 a is given a charge Q Show that the entire charge must appear on the outer surface of the conductor b Another conductor B with charge q is inserted into the cavity keeping B insulated from A Show that the total charge on the outside surface of A is q Fig 1 36 b c A sensitive instrument is to be shielded from the strong electrostatic fields in its environment Suggest a possible way a 9 A Bo b 9 9
Physics
Gauss Law
a A conductor A with a cavity as shown in Fig 1 36 a is given a charge Q Show that the entire charge must appear on the outer surface of the conductor b Another conductor B with charge q is inserted into the cavity keeping B insulated from A Show that the total charge on the outside surface of A is q Fig 1 36 b c A sensitive instrument is to be shielded from the strong electrostatic fields in its environment Suggest a possible way a 9 A Bo b 9 9
Point charge qo is placed at centre of a cylinder as shown in figure Find electric flux through curved surface DIR 1 Eo R 9 2R 2 q 2 o
Physics
Gauss Law
Point charge qo is placed at centre of a cylinder as shown in figure Find electric flux through curved surface DIR 1 Eo R 9 2R 2 q 2 o
E Xi x mt then flux through the a shaded area of a cube is 5 If E 1 E a 3 E a a 0 2 Zero 4 E a in the vol 1 43 2 3 43 2 Ans 2 Sol B 8 APV AV 12 Displace executin
Physics
Gauss Law
E Xi x mt then flux through the a shaded area of a cube is 5 If E 1 E a 3 E a a 0 2 Zero 4 E a in the vol 1 43 2 3 43 2 Ans 2 Sol B 8 APV AV 12 Displace executin
Questions 15 and 16 are case study based questions and are compuisory Attempt four sub parts from each question Each question carries 1 mark 15 Electric flux through an area is defined as E A Gauss law is used to assess amount of enclosed charge The concept of flux and Gauss law is very useful as in cases of non planar area calculation of flux is not easy Gauss law is also very important in calculating the electrostatic field Flux due to change is additive in nature Gauss law is not depend on the shape i Gauss law is dependent on internal charge on configuration a No it s not b Yes it depend on chanrge configuration d Both a and b c May be ii Is it necessary condition to have symmetric surface for Gauss law application a Yes b No c May be d Depends on situation iii Electric flux through a closed surface is due to a Total charged enclosed b Total charge present on the surface c Total charge induced charge d All of the above iv The Gaussian surface should not be passed through any discrete charge because a Electric field becomes zero b Electric field remains constant c Electric field due to a system of discrete charge is not well defined at location at any charge
Physics
Gauss Law
Questions 15 and 16 are case study based questions and are compuisory Attempt four sub parts from each question Each question carries 1 mark 15 Electric flux through an area is defined as E A Gauss law is used to assess amount of enclosed charge The concept of flux and Gauss law is very useful as in cases of non planar area calculation of flux is not easy Gauss law is also very important in calculating the electrostatic field Flux due to change is additive in nature Gauss law is not depend on the shape i Gauss law is dependent on internal charge on configuration a No it s not b Yes it depend on chanrge configuration d Both a and b c May be ii Is it necessary condition to have symmetric surface for Gauss law application a Yes b No c May be d Depends on situation iii Electric flux through a closed surface is due to a Total charged enclosed b Total charge present on the surface c Total charge induced charge d All of the above iv The Gaussian surface should not be passed through any discrete charge because a Electric field becomes zero b Electric field remains constant c Electric field due to a system of discrete charge is not well defined at location at any charge
If the magnitude of electric flux entering and leaving an enclosed surface respectively is 4 and T electric charge inside the closed surface will be wafa fa fa OPTIONS MARK FOR REVIEW CLEAR SELECTION OA 39 B 3 ED C 5 E
Physics
Gauss Law
If the magnitude of electric flux entering and leaving an enclosed surface respectively is 4 and T electric charge inside the closed surface will be wafa fa fa OPTIONS MARK FOR REVIEW CLEAR SELECTION OA 39 B 3 ED C 5 E
Three charges are placed fixed at the vertices of an isosceles right angle triangle ABC with point charge 0 20 Q are placed at points A B Crespectively The magnitude of electric field strength at point D mid point of AC is Find n qandang antas ABC is a c 1 M 0 20 q 42 1 114 3D AC OPTIONS MARK FOR REVIEW CLEAR SELECTION
Physics
Gauss Law
Three charges are placed fixed at the vertices of an isosceles right angle triangle ABC with point charge 0 20 Q are placed at points A B Crespectively The magnitude of electric field strength at point D mid point of AC is Find n qandang antas ABC is a c 1 M 0 20 q 42 1 114 3D AC OPTIONS MARK FOR REVIEW CLEAR SELECTION
Charge density on a triangle shown is 2 and moving towards a solid imaginary sphere is radius r so that the centre of Ring passes through the centre of sphere The maximum flux through the sphere in this process is E A 3 32r 80 C 0 3 B D 3hr o r 380
Physics
Gauss Law
Charge density on a triangle shown is 2 and moving towards a solid imaginary sphere is radius r so that the centre of Ring passes through the centre of sphere The maximum flux through the sphere in this process is E A 3 32r 80 C 0 3 B D 3hr o r 380
28 Assertion A A point charge is lying at the center of a cube of each side 1 The electric flux emanating from each surface of the cube is 1 6 th of total flux Reason R According to Gauss theorem total electric flux through a closed surface enclosing charge is equal to 1 EO times the magnitude of the charge enclosed
Physics
Gauss Law
28 Assertion A A point charge is lying at the center of a cube of each side 1 The electric flux emanating from each surface of the cube is 1 6 th of total flux Reason R According to Gauss theorem total electric flux through a closed surface enclosing charge is equal to 1 EO times the magnitude of the charge enclosed
field components 12 Electric E 100 x E E 0 Calculate net electric flux though the cube placed in electric field at shown position 3m D NCERT Pg 35 3m C x axis sheet is a C m 2 The kinetic ene charge after t second is NCERT 1 3 9 0 1 4m 9 0 1 8cm 2 4 9 0 1 m 9 0 1 43m 16 An electric dipole consists of two eq opposite charges 0 02 C separa The dipole is placed is 2 mm
Physics
Gauss Law
field components 12 Electric E 100 x E E 0 Calculate net electric flux though the cube placed in electric field at shown position 3m D NCERT Pg 35 3m C x axis sheet is a C m 2 The kinetic ene charge after t second is NCERT 1 3 9 0 1 4m 9 0 1 8cm 2 4 9 0 1 m 9 0 1 43m 16 An electric dipole consists of two eq opposite charges 0 02 C separa The dipole is placed is 2 mm
A uniform electric field of magnitude E 100 N C exists in the space in x direction Calculate the flux of this field through a plane square area of edge 10 cm placed in the y z plane Take the normal along the positive x axis to be positive A 1 0Nm C 1 B 10000Nm C 1 C 1 0Nm 2C 1 D DON 2
Physics
Gauss Law
A uniform electric field of magnitude E 100 N C exists in the space in x direction Calculate the flux of this field through a plane square area of edge 10 cm placed in the y z plane Take the normal along the positive x axis to be positive A 1 0Nm C 1 B 10000Nm C 1 C 1 0Nm 2C 1 D DON 2
Charge Outside of a solid sphere of radius R with uniformly distributed charge o the field is E r kQ F inside it is E r kQ T Compute using any coordinate system and plo
Physics
Gauss Law
Charge Outside of a solid sphere of radius R with uniformly distributed charge o the field is E r kQ F inside it is E r kQ T Compute using any coordinate system and plo
10 Two charges of magnitude 2Q and Q are located at points a 0 and 4a 0 respectively What is the electric flux due to these charges through a sphere of radius i 3a and ii Sa with its centre at the origin
Physics
Gauss Law
10 Two charges of magnitude 2Q and Q are located at points a 0 and 4a 0 respectively What is the electric flux due to these charges through a sphere of radius i 3a and ii Sa with its centre at the origin
The electric field calculated by Gauss s law is the field due to the charges which 1 Lie inside the Gaussian surface 2 Lie outside the Gaussian surface 3 Lie on the surface of the Gaussian surface 4 Lie either inside outside or on the Gaussian surface 161
Physics
Gauss Law
The electric field calculated by Gauss s law is the field due to the charges which 1 Lie inside the Gaussian surface 2 Lie outside the Gaussian surface 3 Lie on the surface of the Gaussian surface 4 Lie either inside outside or on the Gaussian surface 161
11 Consider a conducting shell as shown in figure Two point charges are inside the shell and two are outside the shell Then induced charge on the outer surface of conducting shell is 1 9 9 3 93 94 94 9 9 93 2 9 9 4 93 94
Physics
Gauss Law
11 Consider a conducting shell as shown in figure Two point charges are inside the shell and two are outside the shell Then induced charge on the outer surface of conducting shell is 1 9 9 3 93 94 94 9 9 93 2 9 9 4 93 94
92 Charge is distributed inside a long cylindrical volume of radius R such that charge density inside the volume varies as p ar where a is a constant and r the distance from the axis of the cylinder If the magnitude of electric field inside the cylinder is independent of r the value of n is a 1 b 2 c 2 93 In the circuit shown in figure find d 1
Physics
Gauss Law
92 Charge is distributed inside a long cylindrical volume of radius R such that charge density inside the volume varies as p ar where a is a constant and r the distance from the axis of the cylinder If the magnitude of electric field inside the cylinder is independent of r the value of n is a 1 b 2 c 2 93 In the circuit shown in figure find d 1
85 An infinite uniformly charged sheet with surface charge density o cuts through a spherical Gaussian surface of radius R at a distance x from its centre as shown in the figure The electric flux o through the Gaussian surface is a c TR 80 T R x 0 En R X O b 2 R x 80 d R x o En
Physics
Gauss Law
85 An infinite uniformly charged sheet with surface charge density o cuts through a spherical Gaussian surface of radius R at a distance x from its centre as shown in the figure The electric flux o through the Gaussian surface is a c TR 80 T R x 0 En R X O b 2 R x 80 d R x o En
Two identical parallel conducting plates each with area A having small separation between them are given positive charges Q and Q Q Q respectively The magnitude of electric field at point P between the two plates will be Q Q
Physics
Gauss Law
Two identical parallel conducting plates each with area A having small separation between them are given positive charges Q and Q Q Q respectively The magnitude of electric field at point P between the two plates will be Q Q
A charge q is placed at a distance r 2 from the centre of a hollow iron shell of radius ras shown What is electric flux through a closed imaginary spherical surface of radius 2r concentric with the shell 2r Iron shell Imaginary spherical surface Zero due to electrostatic shielding due to electrostatic shielding 0 due to induced charge appearing on 80 the surface of the ball
Physics
Gauss Law
A charge q is placed at a distance r 2 from the centre of a hollow iron shell of radius ras shown What is electric flux through a closed imaginary spherical surface of radius 2r concentric with the shell 2r Iron shell Imaginary spherical surface Zero due to electrostatic shielding due to electrostatic shielding 0 due to induced charge appearing on 80 the surface of the ball
A charge of 1 C is situated at the center of a hollow sphere of radius 2 m If the radius of the sphere reduces to 1 m without any shift in its center then the electric flux through the sphere would O Decrease by a factor of 2 O Increase by a factor of 2 O Remain unchanged O Reduce by a factor of 4
Physics
Gauss Law
A charge of 1 C is situated at the center of a hollow sphere of radius 2 m If the radius of the sphere reduces to 1 m without any shift in its center then the electric flux through the sphere would O Decrease by a factor of 2 O Increase by a factor of 2 O Remain unchanged O Reduce by a factor of 4
A very long uniformly charged thread oriented along the axis of a circle of radius R Im rests on its centre with one of the ends The charge per unit length on the thread is 16e Find the flux of the vector E through the circle area in V m
Physics
Gauss Law
A very long uniformly charged thread oriented along the axis of a circle of radius R Im rests on its centre with one of the ends The charge per unit length on the thread is 16e Find the flux of the vector E through the circle area in V m
A system consists of uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density p a r where a is a positive constant and r is the distance from the centre of the sphere The charge of the sphere for which electric field intensity E outside the sphere is independent of r is a 2 c 2 R b 2 d aR
Physics
Gauss Law
A system consists of uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density p a r where a is a positive constant and r is the distance from the centre of the sphere The charge of the sphere for which electric field intensity E outside the sphere is independent of r is a 2 c 2 R b 2 d aR
An infinite plane in the xz plane carries a uniform surface charge density 01 8 85 nC m A second infinite plane carrying a uniform charge density 02 17 7 nC m intersects the xz plane at the z axis and makes an angle of 60 with the xz plane as shown in figure The electric field in the xy plane N 4 60 at x 6m y 2m is 500v3 N C at x 5m y 0 is 500v3 N C at x 2m y 6 m is 500 7 N C at x 1m y 1m is 500v7N C X X
Physics
Gauss Law
An infinite plane in the xz plane carries a uniform surface charge density 01 8 85 nC m A second infinite plane carrying a uniform charge density 02 17 7 nC m intersects the xz plane at the z axis and makes an angle of 60 with the xz plane as shown in figure The electric field in the xy plane N 4 60 at x 6m y 2m is 500v3 N C at x 5m y 0 is 500v3 N C at x 2m y 6 m is 500 7 N C at x 1m y 1m is 500v7N C X X
1 2 m diameter Ch24 kh1 1 E The total electric flux through a closed cylindrical length 0 20 m surface is equal to 5 0 N m2 C Determine the net charge within the cylinder Select one a 44 pC b 16 pC c 62 pC d 71 pC e 53 pC
Physics
Gauss Law
1 2 m diameter Ch24 kh1 1 E The total electric flux through a closed cylindrical length 0 20 m surface is equal to 5 0 N m2 C Determine the net charge within the cylinder Select one a 44 pC b 16 pC c 62 pC d 71 pC e 53 pC
A conducting solid sphere of radius R has a total charge Q on it The electric potential at a point at a distance r from the center varies as r R Oro O
Physics
Gauss Law
A conducting solid sphere of radius R has a total charge Q on it The electric potential at a point at a distance r from the center varies as r R Oro O
A B B Two concentric conducting thin spherical shells A and B having radii r and rg rg r are charged with charges QA and Q 1Qg IQ 1 The electric field along a line passing through the centre varies with distance x from the centre as B A 1 LU E 0 F 17 A X 2 LU E LU E i ad 4 533 8 X
Physics
Gauss Law
A B B Two concentric conducting thin spherical shells A and B having radii r and rg rg r are charged with charges QA and Q 1Qg IQ 1 The electric field along a line passing through the centre varies with distance x from the centre as B A 1 LU E 0 F 17 A X 2 LU E LU E i ad 4 533 8 X
Paragraph Force between a point charge and uniformly charged non conducting a surface normal to surface can be calculated by method of flux F E dq E ods Eds o 1 k EN is component of electric field normal to the surface 4 5 Find the force between point charge q at axis of non conducting uniformly charged Q square plate of side d and square plate A kQq d kQq B 4kQq d 27 kQq Q d 2
Physics
Gauss Law
Paragraph Force between a point charge and uniformly charged non conducting a surface normal to surface can be calculated by method of flux F E dq E ods Eds o 1 k EN is component of electric field normal to the surface 4 5 Find the force between point charge q at axis of non conducting uniformly charged Q square plate of side d and square plate A kQq d kQq B 4kQq d 27 kQq Q d 2
Two long wires have uniform charge density per unit length each The wires are no coplanar and mutually perpendicular Shortest distance between them is d The interacti Find the value of k force between them is 2 KE
Physics
Gauss Law
Two long wires have uniform charge density per unit length each The wires are no coplanar and mutually perpendicular Shortest distance between them is d The interacti Find the value of k force between them is 2 KE
3 A half cylinder of radius R and length L R is formed by cutting a cylindrical pipe made of an insulating material along a plane containing its axis The rectangular base of the half cylinder is closed by a dielectric plate of length of length L and width 2R A charge Q on the half cylinder and a charge q on the dielectric plate are uniformly sprinkled Electro static force between the plate and the half cylinder is closest to qQ 2 E RL a c qQ 28 RL qQ 48 RL b d qQ 88 RL
Physics
Gauss Law
3 A half cylinder of radius R and length L R is formed by cutting a cylindrical pipe made of an insulating material along a plane containing its axis The rectangular base of the half cylinder is closed by a dielectric plate of length of length L and width 2R A charge Q on the half cylinder and a charge q on the dielectric plate are uniformly sprinkled Electro static force between the plate and the half cylinder is closest to qQ 2 E RL a c qQ 28 RL qQ 48 RL b d qQ 88 RL
5 Shown below are four coaxial with an isolated long line charge with a uniform linear charge density 2 The four surfaces are I ODD III the net electric flux III I a closed cylinder of length L II a sphere of diameter L III a closed cubical box of side L IV a two dimensional square sheet of side L whose plane is perpendicular to the line charge Through which of the above surface s is AL 0 A I II and III only B I and III only C I only IV
Physics
Gauss Law
5 Shown below are four coaxial with an isolated long line charge with a uniform linear charge density 2 The four surfaces are I ODD III the net electric flux III I a closed cylinder of length L II a sphere of diameter L III a closed cubical box of side L IV a two dimensional square sheet of side L whose plane is perpendicular to the line charge Through which of the above surface s is AL 0 A I II and III only B I and III only C I only IV
c 4 5 x 10 N m C d 2 5 x 10 N m C 1 6 The electric field components in the given figure are E ax2 E E 0 in which a 800 NC m2 9 The charge within the cube is if net flux through the cube is 1 05 N m C assume a 0 1 m a 9 27 x 10 12 C c 6 97 x 10 12 C b 9 27 x 10 2 C d 6 97 x 10 2 C
Physics
Gauss Law
c 4 5 x 10 N m C d 2 5 x 10 N m C 1 6 The electric field components in the given figure are E ax2 E E 0 in which a 800 NC m2 9 The charge within the cube is if net flux through the cube is 1 05 N m C assume a 0 1 m a 9 27 x 10 12 C c 6 97 x 10 12 C b 9 27 x 10 2 C d 6 97 x 10 2 C
b half that due c double that due to a single charge d dependent on the position of dipole A point charge 4 C is at the centre of a cubic 91 Two pa Gaussian surface 10 cm on edge Net electric flux placed The ne two lin a zem through the surface is a 25 x 10 N m C c 4 5 x 10 N m C b 4 5 x 10 Nm C 1 d 2 5 x 10 N m C of the r a 8 c 6 the ven figure are
Physics
Gauss Law
b half that due c double that due to a single charge d dependent on the position of dipole A point charge 4 C is at the centre of a cubic 91 Two pa Gaussian surface 10 cm on edge Net electric flux placed The ne two lin a zem through the surface is a 25 x 10 N m C c 4 5 x 10 N m C b 4 5 x 10 Nm C 1 d 2 5 x 10 N m C of the r a 8 c 6 the ven figure are
Just after launch from the earth the space shuttle orbiter is in the 50 x 136 mi orbit shown At the apogee point A its speed is 17309 mi hr If nothing were done to modify the orbit what would its speed be at the perigee P Neglect aerodynamic drag Note that the normal practice is to add speed at A which raises the perigee altitude to a value that is well above the bulk of the atmosphere The radius of the earth is 3959 mi 50 mi Answer V i 136 mi 17309 mi hr mi hr
Physics
Gauss Law
Just after launch from the earth the space shuttle orbiter is in the 50 x 136 mi orbit shown At the apogee point A its speed is 17309 mi hr If nothing were done to modify the orbit what would its speed be at the perigee P Neglect aerodynamic drag Note that the normal practice is to add speed at A which raises the perigee altitude to a value that is well above the bulk of the atmosphere The radius of the earth is 3959 mi 50 mi Answer V i 136 mi 17309 mi hr mi hr
A solid sphere made of insulating material has a radius R and has a total charge Q distributed uniformly in its volume What is the magnitude of the electric field intensity E at a distance r 0 r R inside the sphere 1 2 1 Qr 4 R 3 4 0 Qr R
Physics
Gauss Law
A solid sphere made of insulating material has a radius R and has a total charge Q distributed uniformly in its volume What is the magnitude of the electric field intensity E at a distance r 0 r R inside the sphere 1 2 1 Qr 4 R 3 4 0 Qr R
Let p r r be the charge density distribution for a solid sphere of radius R and total charge Q For a point P inside the sphere at distance r from the centre of the sphere the magnitude of electric field is A O e Qr 4 R B D Q 4 Qr 3 EO R4
Physics
Gauss Law
Let p r r be the charge density distribution for a solid sphere of radius R and total charge Q For a point P inside the sphere at distance r from the centre of the sphere the magnitude of electric field is A O e Qr 4 R B D Q 4 Qr 3 EO R4
24 Figure shows four identical currents i and an amperian loops encircling them We shall calculate B di in the direction marked The correct value is 1 B di 2 i 2 B di 2 B di 0 4 B di 4uni
Physics
Gauss Law
24 Figure shows four identical currents i and an amperian loops encircling them We shall calculate B di in the direction marked The correct value is 1 B di 2 i 2 B di 2 B di 0 4 B di 4uni
36 A charge Q is uniformly dist length 21 Consider cube of edge with the centre of cube at one end of the rod The minimum possible electric flux through the surface of the cube is 1 EO Q 2 Q 480 2Q n S
Physics
Gauss Law
36 A charge Q is uniformly dist length 21 Consider cube of edge with the centre of cube at one end of the rod The minimum possible electric flux through the surface of the cube is 1 EO Q 2 Q 480 2Q n S
ix A non conducting spherical ball of radius R contains a spherically symmetric charge with volume charge density p kr where r is the distance from the centre of the ball and n is a constant What should be n such that the electric field inside the ball is directly proportional to square of distance from the centre 1
Physics
Gauss Law
ix A non conducting spherical ball of radius R contains a spherically symmetric charge with volume charge density p kr where r is the distance from the centre of the ball and n is a constant What should be n such that the electric field inside the ball is directly proportional to square of distance from the centre 1
A cube of side length 1 m has one corner at origin of co ordinate axes and extends along positive x axis y axis and z axis The electric field in region is given by relation 3 V m electric flux through cube will be in Sl units 1 Zero 3
Physics
Gauss Law
A cube of side length 1 m has one corner at origin of co ordinate axes and extends along positive x axis y axis and z axis The electric field in region is given by relation 3 V m electric flux through cube will be in Sl units 1 Zero 3
and outer radius 2R has a uniform charge distribution and total charge Q 2R R Choose the correct answer The charges inside the Gaussian sphere for the given three regions are respectively 1 0 a R ii R a 2R
Physics
Gauss Law
and outer radius 2R has a uniform charge distribution and total charge Q 2R R Choose the correct answer The charges inside the Gaussian sphere for the given three regions are respectively 1 0 a R ii R a 2R
The prism in the figure is in a non uniform electric field E 2x 3y N C Here x and y are in meters a Find the total electric flux through the prism b What is the net charge inside the prism 4 y m 3 z m 2 A x m
Physics
Gauss Law
The prism in the figure is in a non uniform electric field E 2x 3y N C Here x and y are in meters a Find the total electric flux through the prism b What is the net charge inside the prism 4 y m 3 z m 2 A x m
Figure here shows a charge q placed at the centre of a hemisphere A second charge Q is placed at one of the positions A B C D and E In which position s of this second charge the flux of the electric field through the hemispherical curved surface remains unchanged 8 4 D E
Physics
Gauss Law
Figure here shows a charge q placed at the centre of a hemisphere A second charge Q is placed at one of the positions A B C D and E In which position s of this second charge the flux of the electric field through the hemispherical curved surface remains unchanged 8 4 D E
L 0 5 mole of each H SO and CH4 are kept in a container A hole was made in the container After 3 hours decreasing order of partial pressures of gases in the container will be b PH Po PCH a P Pat PH C PCH P PH d POH PH Pos
Physics
Gauss Law
L 0 5 mole of each H SO and CH4 are kept in a container A hole was made in the container After 3 hours decreasing order of partial pressures of gases in the container will be b PH Po PCH a P Pat PH C PCH P PH d POH PH Pos
and r is the distance from the center of the disc Electric flux through a large spherical surface that encloses the charged disc completely is oo Electric flux through another spherical surface of radius R and concentric with the disc is Then the ratio 40 is 4 Ans 6 40 18 Sol ALLEN CAREER INSTITU CHERE tum Jdq 70 r R R 4 Sool Jdq 10 1 12mr dr R Eg 2 0 R R dq 00 R 4 R 00 R 3 2xr dr R 32 3x 64 G 2 r 0 r 2 11 R 2 r dr r JEE Advanced 2020 Paper 1 Held on Sunday 27 September 2020 LR 32 5 r R dr R 27rdr JEE Advanced 2020 Paper 1 NET
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Gauss Law
and r is the distance from the center of the disc Electric flux through a large spherical surface that encloses the charged disc completely is oo Electric flux through another spherical surface of radius R and concentric with the disc is Then the ratio 40 is 4 Ans 6 40 18 Sol ALLEN CAREER INSTITU CHERE tum Jdq 70 r R R 4 Sool Jdq 10 1 12mr dr R Eg 2 0 R R dq 00 R 4 R 00 R 3 2xr dr R 32 3x 64 G 2 r 0 r 2 11 R 2 r dr r JEE Advanced 2020 Paper 1 Held on Sunday 27 September 2020 LR 32 5 r R dr R 27rdr JEE Advanced 2020 Paper 1 NET
Figure shows a closed dotted surface which intersects a conducting uncharged sphere If a positive charges is placed at the point P the flux of the electric field through the closed surface o D P Will remain zero Will be negative Will be positive Will be infinite
Physics
Gauss Law
Figure shows a closed dotted surface which intersects a conducting uncharged sphere If a positive charges is placed at the point P the flux of the electric field through the closed surface o D P Will remain zero Will be negative Will be positive Will be infinite
What condition can the electric TiUXE be found through a closed surface If the magnitude of the electric field is known everywhere on the surface If the total charge outside the surface is specified Only if the location of each point charge inside the surface is specified If the total charge inside the surface is specified
Physics
Gauss Law
What condition can the electric TiUXE be found through a closed surface If the magnitude of the electric field is known everywhere on the surface If the total charge outside the surface is specified Only if the location of each point charge inside the surface is specified If the total charge inside the surface is specified
the electric field E Eoxi Eoyj Consider an imaginary cubical volume of edge a with its edges parallel to the axes of coordinates Now Option A B C D 4 3 X The total electric flux through the faces 1 and 3 is Eoa The charge inside the cubical volume is 280 Ea The total electric flux through the face 2 and 4 is 2E0a The charge inside the cubical volume is E0E0a Rate this qu
Physics
Gauss Law
the electric field E Eoxi Eoyj Consider an imaginary cubical volume of edge a with its edges parallel to the axes of coordinates Now Option A B C D 4 3 X The total electric flux through the faces 1 and 3 is Eoa The charge inside the cubical volume is 280 Ea The total electric flux through the face 2 and 4 is 2E0a The charge inside the cubical volume is E0E0a Rate this qu