Free NEET Physics multiple-choice questions on Electric Flux & Gauss's Law. Attempt each question and reveal the answer with a full explanation.
A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle with the horizontal side of the square as shown in figure. The electric flux linked to the surface is: Zero EL 2 EL 2 cos EL 2 sin A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:- Decreases Does not change Becomes zero Increases Two condensers, one of capacity C and other of capacity C/2 are connected to a V-volt battery, as shown. The work done in charging fully both the condensers is 1 4 CV 2 3 4 CV 2 1 2 CV 2 2CV 2 . Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be : 3C, 3V C 3 , V 3 3C, V 3 C 3 , 3V A , B and C are three points in a uniform electric field. The electric potential is Maximum at A Maximum at B Maximum at C Same at all the three points A , B and C A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are :- Zero and Q 4 0 R 2 Q 4 0 R and Zero Q 4 0 R and Q 4 0 R 2 Both are zero The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A , is Proportional to the square root of the distance between the plates Linearly proportional to the distance between the plates Independent of the distance between the plates Inversely proportional to the distance between the plates A parallel plate capacitor of capacitance 20 μF is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively Zero, zero Zero, 60 μA 60 μA, 60 μA 60 μA, zero In a certain region of space with volume 0.2 m 3 , the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is : 0.5 N/C 1 N/C 5 N/C zero A short electric dipole has a dipole moment of 16 10 -9 C m . The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60 with the dipole axis is : ( 1 4 0 =9 10 9 N m 2/C 2 ) 200 V 400 V zero 50 V The capacitance of a parallel plate capacitor with air as medium is 6 F. With the introduction of a dielectric medium, the capacitance becomes 30 F. The permittivity of the medium is : ( 0 = 8.85 10 -12 C 2 N -1 m -2 ) 1.77 10 -12 C 2 N -1 m -2 0.44 10 -10 C 2 N -1 m -2 5.00 C 2 N -1 m -2 0.44 10 -13 C 2 N -1 m -2 If a charge q is placed at the centre of an imaginary cube, the flux through any one face of the cube is: q/(6ε 0) q/ε 0 q/(3ε 0) q/(8ε 0) The equivalent capacitance of the combination shown in the figure is 2C C 2 3C 2 3C Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop. 1320 V 1520 V 1980 V 660 V The angle between the electric lines of force and the equipotential surface is 45° 90° 180° 0° Two hollow conducting spheres of radii R 1 and R 2 ( R 1 >> R 2 ) have equal charges. The potential would be More on smaller sphere Equal on both the spheres Dependent on the material property of the sphere More on bigger sphere The equivalent capacitance of the system shown in the following circuit is 2 μF 3 μF 6 μF 9 μF A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: (Take 1 4 0 =9 10 9 SI units) 3 10 5 1 10 5 0.5 10 5 Zero A capacitor of capacitance C is charged to a potential V . The flux of the electric field through a closed surface enclosing the capacitor is: zero CV 0 2CV 0 CV 2 0 A charge q is located at the center of a cube. The electric flux through any two adjacent faces is: q 3 0 q 6 0 q 0 q 2 0 If the electric flux entering and leaving an enclosed surface respectively are 1 and 2 , the electric charge inside the surface will be: ( 2 - 1) 0 ( 1 + 2) 0 ( 2 - 1)/ 0 ( 1 + 2)/ 0 An infinite line of charge produces a field of 9 10 4 N/C at a distance of 2 cm . The linear charge density is: 10 -7 C/m 10 -6 C/m 10 -8 C/m 10 -5 C/m A point charge q is placed at the center of the open face of a hemispherical surface as shown in the figure. The flux of the electric field through the surface is: q 2 0 q 0 q 4 0 zero An infinite line charge produces a field of 9 10 4 N/C at a distance of 2 cm . Calculate the linear charge density: 10 -7 C/m 2 10 -7 C/m 10 -8 C/m 4 10 -7 C/m An infinite line charge produces a field of 18 10 4 N/C at a distance of 4 cm. What is the linear charge density? 4 10 -7 C/m 2 10 -7 C/m 8 10 -7 C/m 1 10 -7 C/m A charge q is placed at the center of a cube. The flux through any three faces of the cube is: q 2 0 q 6 0 q 3 0 q 0 What is the flux through a cube of side a if a point charge of q is at one of its corners? q 8 ε 0 q ε 0 q 2 ε 0 2q ε 0 A hollow cylinder has a charge q coulomb within it. If Φ is the electric flux in units of voltmeter associated with the curved surface B , the flux linked with the plane surface A in units of voltmeter will be: 1 2 ( q ε 0 - Φ ) q 2 ε 0 Φ 3 q ε 0 - Φ A point charge Q is situated at a distance d/2 above the centre of a square of side d . The flux through the square is: Q 6 ε 0 Q ε 0 Q 4 ε 0 Q 8 ε 0 A point charge q is placed at the centre of a hemispherical surface of radius R . The electric flux through the curved surface is: q/(2ε 0) q/ε 0 q/(4ε 0) Zero Two charges q 1 and q 2 are placed 30 cm apart, as shown in the figure. A third charge q 3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is q 3 4 0 k , where k is – 8q 2 6q 2 8q 1 6q 1 A network of four capacitors of capacity equal to C 1=C , C 2=2C , C 3=3C and C 4=4C are conducted to a battery as shown in the figure. The ratio of the charges on C 2 and C 4 is – 7 4 22 3 3 22 4 7 The electric potential at a point (x, y, z) is given by V=-x 2y-xz 3+4 The electric field E at that point is : E = i (2xy-z 3)+ j xy 2+ k 3z 2x E = i (2xy+z 3)+ j x 2+ k 3xz 2 E = i 2xy+ j (x 2+y 2)+ k (3xz-y 2) E = i z+ j xyz+ k z 2 In a region, the electric field is given by E = E 0 i . The flux through a square of side L in the yz -plane is: E 0 L 2 E 0 L 2 / 2 E 0 L 2 / 2 zero Two thin dielectric slabs of dielectric constants K 1 and K 2 ( K 1<K 2 ) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field 'E' between the plates with distance 'd' as measured from plate P is correctly shown by :- In a region, the potential is represented by V(x, y, z)=6x-8xy-8y+6yz , where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is :- 6 5 N 30 N 24 N 4 35 N A capacitor of 2 F is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is: 0% 20% 75% 80% A parallel-plate capacitor is to be designed, using a dielectric of dielectric constant 5, so as to have a dielectric strength of 10 9 Vm -1 . If the voltage rating of the capacitor is 12kV, the minimum area of each plate required to have a capacitance of 80 pF is : 10.5 10 -6 m 2 21.7 10 -6 m 2 25.0 10 -5 m 2 12.5 10 -5 m 2 Two charged spherical conductors of radius R 1 and R 2 are connected by a wire. Then the ratio of surface charge densities of the spheres ( 1/ 2) is R 2 R 1 R 1 R 2 R 1 2 R 2 2 R 1 R 2 A parallel plate capacitor has a uniform electric field ' E ' in the space between the plates. If the distance between the plates is ' d ' and the area of each plate is ' A ', the energy stored in the capacitor is ( 0 = permittivity of free space ) 0EAd 1 2 0E 2Ad E 2Ad 0 1 2 0E 2 A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is 3.25 10 -6 J 2.25 10 -6 J 1.5 10 -6 J 4.5 10 -6 J An electric dipole is placed as shown in the figure. The electric potential (in 10 2 V) at point P due to the dipole is ( 0 = permittivity of free space and 1 4 , 0 =K ) ( 3 8 )qK ( 5 8 )qK ( 8 5 )qK ( 8 3 )qK In the following circuit, the equivalent capacitance between terminal A and terminal B is : 2 , F 1 , F 0.5 , F 4 , F Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector P of magnitude, 4 10 -6 C m , is 9 10 3 V . (Take 1 4 0 =9 10 9 SI units) Reason R: V = 2P 4 0 r 2 , where r is the distance of any axial point, situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below: Both A and R are true and R is the correct explanation of A. Both A and R are true and R is NOT the correct explanation of A. A is true but R is false. A is false but R is true. The electric flux through a closed surface depends on: Net charge enclosed only Shape of the surface Size of the surface Position of charge inside the surface If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. the charge stored in it, increases. B. the energy stored in it, decreases. C. its capacitance increases. D. the ratio of charge to its potential remains the same. E. the product of charge and voltage increases. Choose the most appropriate answer from the options given below: A, B and E only A, C and E only B, D and E only A, B and C only Five capacitors of capacitances C 1 = C 2 = C 3 = C 4 = 10 F and C 5 = 2.5 F are connected as shown, along with a battery of 50 V. The equivalent capacitance and the charges on each capacitor respectively are: 4 F , 250 C on C 1 to C 4 and 125 C on C 5 5 F , 250 C on all capacitors 5 F , 125 C on C 1 to C 4 and 25 C on C 5 5 F , 125 C on all capacitors Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is: 0.5 J 1.0 10 -6 J 0.5 10 -6 J 1.0 J The electric field components in a region are E x = α x 1/2 , E y = E z = 0 . The flux through a cube of side L with one face at x = L is proportional to: L 5/2 L 2 L 3/2 L 3 A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net electric field E at the center O is: q 2 2 0 r 2 j q 4 2 0 r 2 j q 2 0 r 2 j zero The electric field at the center of a uniformly charged semi-circular ring of radius R and total charge Q is: Q 2 2 0 R 2 Q 2 0 R 2 Q 4 2 0 R 2 zero Charges +q and -q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is qQ 2 0L qQ 6 0L - qQ 6 0L qQ 4 0L Three concentric spherical shells have radii a, b, and c (a < b < c) and have surface charge densities , - and respectively. If V A , V B and V C denote the potentials of the three shells, then, for c=a+b , we have : V C=V B=V A V C=V A V B V C=V B V A V C V B V A The plates of a parallel plate capacitor are separated by d . Two slabs of different dielectric constant K 1 and K 2 with thickness 3 8 d and d 2 , respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K 1 = 1.25 K 2 , the value of K 1 is: 1.33 2.66 2.33 1.60 A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A , B and C as shown in the figure, where the magnitudes of the electric fields are E A , E B , E C respectively. The points B and C are at the same distance from the center of the solid sphere. The correct option is: E A=0,E B=E C E A 0,E B=E C E A=0,E B>E C E A 0,E B<E C A point charge q is placed at the center of a cube of side L . What is the electric flux through any two opposite faces of the cube? q 3 0 q 0 q 6 0 q 2 0 The electric flux through a closed surface area S enclosing a charge Q is . If the surface area is doubled, then the flux becomes: 2 /2 4 If the electric flux entering and leaving an enclosed surface respectively is 1 and 2 , the electric charge inside the surface will be: ( 2 - 1) 0 ( 1 + 2) 0 ( 2 - 1)/ 0 ( 1 + 2)/ 0 As per this diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q from the point A coordinates (0,a) to another point B coordinates (a,0) along the straight path AB is ( -qQ 4 0 1 a 2 ) 2 a zero ( qQ 4 0 1 a 2 ) 1 2 ( qQ 4 0 1 a 2 ) 2 a