question_answer 1)
Directions : (1-5) Coulomb's Low Coulomb's law states that the electrostatic force of attraction or repulsion acting between two stationary point charges is given by \[F=\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] where F denotes the force between two charges \[{{q}_{1}}\] and \[{{q}_{2}}\]separated by a distance r in free space, eq is a constant known as perimittivity of free space. Free space is vaccum and may be taken to be air practically. If free space is replaced by a medium, then \[{{\varepsilon }_{0}}\]is replaced by \[\left( {{\varepsilon }_{0}}k \right)\] or \[\left( {{\varepsilon }_{0}}{{\varepsilon }_{r}} \right)\]where k is known as dielectric constant or relative perimittivity.
In coulomb's law, \[F=\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]then on which of the following factors does the proportionality constant k depends ?
A)
Electrostatic force acting between the two point charges done
clear
B)
Nature of the medium between the point stationary charges done
clear
C)
Magnitude of the two point charges done
clear
D)
Distance between the two point charges. done
clear
View Solution play_arrow
question_answer 2)
Dimensional formula for the permittivity constant go of free space of absolute permittivity is
A)
\[\left[ M{{L}^{-3}}\,\,{{T}^{2}}\,{{A}^{2}} \right]\] done
clear
B)
\[\left[ {{M}^{-1}}\,{{L}^{3}}\,{{T}^{4}}\,{{A}^{2}} \right]\] done
clear
C)
\[\left[ {{M}^{-1}}\,{{L}^{-3}}\,{{T}^{4}}\,{{A}^{2}} \right]\] done
clear
D)
\[\left[ ML{{\,}^{-4}}\,{{T}^{4}}\,{{A}^{-2}} \right]\] done
clear
View Solution play_arrow
question_answer 3)
The force of repulsion between two charges of 1 C each, kept 1 m apart in vaccum or free space is
A)
\[\frac{1}{9\times {{10}^{9}}}N\] done
clear
B)
\[9\times {{10}^{9}}N\] done
clear
C)
\[9\times {{10}^{7}}\,N\] done
clear
D)
\[\frac{1}{9\times {{10}^{12}}}N\] done
clear
View Solution play_arrow
question_answer 4)
Two identical charges repel each other with a force equal to 10 mgwt when they are \[0\centerdot 6m\] apart in air, \[\left( g=10\text{ }m{{s}^{-2}} \right)\]. Then value of each charge is
A)
\[2\,mC\] done
clear
B)
\[2\times {{10}^{-7}}mC\] done
clear
C)
\[2\,nC\] done
clear
D)
\[2\mu C\] done
clear
View Solution play_arrow
question_answer 5)
Coulomb's law for the force between electric charges most closely resembles with
A)
law of conservation of energy done
clear
B)
Newton's law of gravitation done
clear
C)
Newton's 2nd law of motion done
clear
D)
law of conservation of charge. done
clear
View Solution play_arrow
question_answer 6)
Directions : (6-10)
Quantization of Electric Charge
Smallest charge that can exist in nature is (i.e., quanta of charge) the charge of an electron. During friction or rubbing it is only the transfer of electrons which makes the body charged. Hence net charge on any body is an integral multiple of charge of an electron \[[1\centerdot 6\times {{10}^{-19}}C]\] i.e.,
\[q=\pm \,ne\]
where n = 1, 2, 3, 4,...
Hence no body can have a charge represent as \[2\centerdot 1e,\]
\[2\,.\,5e,\frac{3}{7}\]e etc.
Recently, it has been discovered that elementary particles such as protons or neutrons are composed of more elemental units called quarks.
Which of the following properties is not statisfied by an electric charge?
A)
Total charge conservation done
clear
B)
Quantization of charge done
clear
C)
Two types of charge done
clear
D)
Circular line of force done
clear
View Solution play_arrow
question_answer 7)
Which one of the following charges is possible?
A)
\[5\centerdot 6\times {{10}^{-18}}C\] done
clear
B)
\[3\centerdot 2\times {{10}^{-18}}C\] done
clear
C)
\[4\centerdot 5\times {{10}^{-19}}C\] done
clear
D)
\[8\centerdot 6\times {{10}^{-19}}C.\] done
clear
View Solution play_arrow
question_answer 8)
If a charge on a body is 1 nC, then how many electrons are present on the body?
A)
\[6\centerdot 25\times {{10}^{27}}\] done
clear
B)
\[6\centerdot 25\times {{10}^{19}}\] done
clear
C)
\[6\centerdot 25\times {{10}^{28}}\] done
clear
D)
\[6\centerdot 25\times {{10}^{9}}\] done
clear
View Solution play_arrow
question_answer 9)
If a body gives out \[{{10}^{9}}\] electrons every second, how much time is required to get a total charge of 1 C from it ?
A)
190.19 years done
clear
B)
150.12 years done
clear
C)
198.19 years done
clear
D)
188.21 years done
clear
View Solution play_arrow
question_answer 10)
A polythene piece rubbed with wool is found to have a negative charge of \[3\centerdot 2\times {{10}^{-7}}C\]. Calculate the number of electrons transferred.
A)
\[2\text{ }\times \text{ }{{10}^{12}}\] done
clear
B)
\[~3\text{ }\times \text{ }{{10}^{12}}\] done
clear
C)
\[2\text{ }\times \text{ }{{10}^{14}}\] done
clear
D)
\[3\text{ }\times \text{ }{{10}^{14}}\] done
clear
View Solution play_arrow
question_answer 11)
Directions : (11-15) Electric Flux Through a Cube Net electric flux through a cube is the sum of fluxes through its six faces. Consider a cube as shown in figure, having sides of length L = 10.0 cm. The electric field is uniform, has a magnitude \[E=4.00\times {{10}^{3}}N{{C}^{-1}}\]and is parallel to the xy plane at an angle of \[37{}^\circ \]measured from the 4-x-axis towards the +y-axis.
Electric flux passing through surface \[{{S}_{6}}\] is
A)
\[-24\,N\,{{m}^{2}}{{C}^{-1}}\] done
clear
B)
\[24\,N\,{{m}^{2}}\,{{C}^{-1}}\] done
clear
C)
\[32\,N\,{{m}^{2}}\,\,{{C}^{-1}}\] done
clear
D)
\[-32\,N{{m}^{2}}\,{{C}^{-1}}\] done
clear
View Solution play_arrow
question_answer 12)
Electric flux passing through surface Si is
A)
\[-24\,N\,{{m}^{2}}\,\,{{C}^{-1}}\] done
clear
B)
\[24\,N\,{{m}^{2}}\,{{C}^{-1}}\] done
clear
C)
\[32\,N\,{{m}^{2}}\,\,{{C}^{-1}}\] done
clear
D)
\[-32\,N\,{{m}^{2}}{{C}^{-1}}\] done
clear
View Solution play_arrow
question_answer 13)
The surfaces that have zero flux are
A)
\[{{S}_{1}}\] and \[{{S}_{3}}\] done
clear
B)
\[{{S}_{5}}\]and \[{{S}_{6}}\] done
clear
C)
\[{{S}_{2}}\] and \[{{S}_{4}}\] done
clear
D)
\[{{S}_{1}}\]and \[{{S}_{2}}\] done
clear
View Solution play_arrow
question_answer 14)
The total net electric flux through all faces of the cube is
A)
\[8\,N\,{{m}^{2}}\,{{C}^{-1}}\] done
clear
B)
\[-8\,N\,{{m}^{2}}{{C}^{-1}}\] done
clear
C)
\[24\,N\,{{m}^{2}}{{C}^{-1}}\] done
clear
D)
zero done
clear
View Solution play_arrow
question_answer 15)
The dimensional formula of surface integral \[\oint{\overrightarrow{E}}\,.\,d\overrightarrow{S}\] of an electric field is
A)
\[\left[ M{{L}^{2}}\,{{T}^{-2}}\,{{A}^{-1}} \right]\] done
clear
B)
\[\left[ M{{L}^{3}}\,{{T}^{-3}}\,{{A}^{-1}} \right]\] done
clear
C)
\[\left[ {{M}^{-1}}\,{{L}^{3}}\,{{T}^{-3}}\,A \right]\] done
clear
D)
\[\left[ M{{L}^{-3}}\,{{T}^{-3}}\,{{A}^{-1}} \right]\] done
clear
View Solution play_arrow
question_answer 16)
Directions: (16-20) Motion of Charged particle in Uniform Electric Field When a charged particle is placed in an electric field, it experiences an electrical force. If this is the only force on the particle, it must be the net force. The net force will cause the particle to accelerate according to Newton's second law. So \[{{\overrightarrow{F}}_{e}}=q\overrightarrow{E}=m\overrightarrow{a}\] If \[\overrightarrow{E}\]is uniform, then \[\overrightarrow{a}\] is constant and\[\overrightarrow{a}=q\,\overrightarrow{E}/m\]. If the particle has a positive charge, its acceleration is in the direction of the field. If the particle has a negative charge, its acceleration is in the direction opposite to the electric field. Since the acceleration is constant, the kinematic equations can be used.
An electron of mass m, charge e falls through a distance h metre in a uniform electric field E. Then time of fall.
A)
\[t=\sqrt{\frac{2hm}{eE}}\] done
clear
B)
\[t=\frac{2hm}{eE}\] done
clear
C)
\[t=\sqrt{\frac{2eE}{hm}}\] done
clear
D)
\[t=\frac{2eE}{hm}\] done
clear
View Solution play_arrow
question_answer 17)
An electron moving with a constant velocity v along X-axis enters a uniform electric field applied along Y-axis. Then the electron moves
A)
with uniform acceleration along Y-axis done
clear
B)
without any acceleration along Y-axis done
clear
C)
In a trajectory represented as \[y=a{{x}^{2}}\] done
clear
D)
in a trajectory represented as y = ax. done
clear
View Solution play_arrow
question_answer 18)
Two equal and opposite charges of masses \[{{m}_{1}}\]and \[{{m}_{2}}\]are accelerated in an uniform electric field through the same distance. What is the ratio of their accelerations if their ratio of masses is\[\frac{{{m}_{1}}}{{{m}_{2}}}=0.5\]?
A)
\[\frac{{{a}_{1}}}{{{a}_{2}}}=2\] done
clear
B)
\[\frac{{{a}_{1}}}{{{a}_{2}}}=0.5\] done
clear
C)
\[\frac{{{a}_{1}}}{{{a}_{2}}}=3\] done
clear
D)
\[\frac{{{a}_{1}}}{{{a}_{2}}}=1\] done
clear
View Solution play_arrow
question_answer 19)
A particle of mass m carrying charge q is kept at rest in a uniform electric field E and then released. Then kinetic energy gained by the particle, when it moves through a distance y is
A)
\[\frac{1}{2}q\,E{{y}^{2}}\] done
clear
B)
\[qEy\] done
clear
C)
\[qE{{y}^{2}}\] done
clear
D)
\[q{{E}^{2}}y\] done
clear
View Solution play_arrow
question_answer 20)
A charged particle is free to move in an electric field. It will travel
A)
always along a line of force done
clear
B)
along a line of force, if its initial velocity is zero done
clear
C)
along a line of force, if it has some initial velocity in the direction of an acute angle with the line of force done
clear
D)
none of these. done
clear
View Solution play_arrow
question_answer 21)
Directions: (21-25)
Millikan's Oil-Drop Experiment
In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled Mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of \[1.6\times {{10}^{-19}}\text{ }C\] the charge of the electron. For this, he won the Nobel prize.
If a drop of mass \[1\centerdot 08\times {{10}^{-14}}kg\] remains stationary in an electric field of \[1\centerdot 68\times {{10}^{5}}N{{C}^{-1}}\], then the charge of this drop is
A)
\[6\centerdot 40\times {{10}^{-19}}C\] done
clear
B)
\[3\centerdot 2\times {{10}^{-19}}C\] done
clear
C)
\[1\centerdot 6\times {{10}^{-19}}C\] done
clear
D)
\[4\centerdot 8\times {{10}^{-19}}C.\] done
clear
View Solution play_arrow
question_answer 22)
Extra electrons on this particular oil drop (given the presently known charge of the electron) are
A)
4 done
clear
B)
3 done
clear
C)
5 done
clear
D)
8 done
clear
View Solution play_arrow
question_answer 23)
A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field \[V\,{{m}^{-1}}\]. If the mass of the drop is \[1\centerdot 6\times {{10}^{-3}}g\], the number of electrons carried by the drop is \[\left( g=10\text{ }m{{s}^{-2}} \right)\]
A)
\[{{10}^{18}}\] done
clear
B)
\[{{10}^{15}}\] done
clear
C)
\[{{10}^{12}}\] done
clear
D)
\[{{10}^{9}}\] done
clear
View Solution play_arrow
question_answer 24)
The important conclusion given by Millikan's expeiment about the charge is
A)
charge is never quantized done
clear
B)
charge has no definite value done
clear
C)
charge is quantized done
clear
D)
charge on oil drop always increases. done
clear
View Solution play_arrow
question_answer 25)
If in Millikan's oil drop experiment, charges on drops are found to be \[8\mu C,\,12\mu C,\,20\mu C\], then quanta of charge is
A)
\[8\mu C\] done
clear
B)
\[20\mu C\] done
clear
C)
\[12\mu C\] done
clear
D)
\[4\mu C\] done
clear
View Solution play_arrow
question_answer 26)
Directions: (26-30)
Gauss' Low and Its Significance
Gauss's law and Coulomb's law, although expressed in different forms, are equivalent ways of describing the relation between charge and electric field in static conditions. Gauss's law is \[{{\varepsilon }_{0}}\phi ={{q}_{encl}}\] when \[{{q}_{encl}}\] is the net charge inside an imaginary closed surface called Gaussian surface, \[\phi =\oint\limits_{{}}{\overrightarrow {E}}\,.\,d\overrightarrow{A}\] gives the electric flux through the Gaussian surface. The two equations hold only when the net charge is in vaccum or air.
If there is only one type of charge in the universe, then (\[\overrightarrow{E}\to \]Electric field, \[d\,\overrightarrow{s}\to \]Area vector)
A)
\[\oint{\overrightarrow{E}}\,.\,d\,\overrightarrow{s}\ne \]on any surface done
clear
B)
\[\oint{\overrightarrow{E}}\,.\,d\,\overrightarrow{s}\]could not be defined done
clear
C)
\[\oint{\overrightarrow{E}}\,.\,d\,\overrightarrow{s}=\infty \]if charge is inside done
clear
D)
\[\oint{\overrightarrow{E}}\,.\,d\,\overrightarrow{s}=0\] if charge is outside, \[\oint{\overrightarrow{E}}\,.\,d\,\overrightarrow{s}=\frac{q}{{{\varepsilon }_{0}}}\]if charge is inside. done
clear
View Solution play_arrow
question_answer 27)
What is the nature of Gaussian surface involved in Gauss law of electrostatic ?
A)
Magnetic done
clear
B)
Scalar done
clear
C)
Vector done
clear
D)
Electrical done
clear
View Solution play_arrow
question_answer 28)
A charge \[10\,\mu C\] is placed at the centre of a hemisphere of radius R = 10 cm as shown. The electric flux through the hemisphere (in MKS units) is
A)
\[20\times {{10}^{5}}\] done
clear
B)
\[10\times {{10}^{5}}\] done
clear
C)
\[6\times {{10}^{5}}\] done
clear
D)
\[2\times {{10}^{5}}\] done
clear
View Solution play_arrow
question_answer 29)
The electric flux through a closed surface area S enclosing charge Q is \[\phi \]. If the surface area is doubled, then the flux is
A)
\[2\phi \] done
clear
B)
\[\phi /2\] done
clear
C)
\[\phi /4\] done
clear
D)
\[\phi \] done
clear
View Solution play_arrow
question_answer 30)
A Gaussian surface encloses a dipole. Then electric flux through this surface is
A)
\[\frac{q}{{{\varepsilon }_{0}}}\] done
clear
B)
\[\frac{2q}{{{\varepsilon }_{0}}}\] done
clear
C)
\[\frac{q}{2{{\varepsilon }_{0}}}\] done
clear
D)
zero done
clear
View Solution play_arrow
question_answer 31)
Directions: (31-35) Relation Between Strength of Electric Field and Density of Lines of Force Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in the given figure, the electric field at P is stronger that at Q.
Electric lines of force about a positive point charge are
A)
radially outwards done
clear
B)
circular clockwise done
clear
C)
radially inwards done
clear
D)
parallel straight lines. done
clear
View Solution play_arrow
question_answer 32)
Which of the following is false for electric lines of force ?
A)
They always start from positive charges and terminate on negative charges done
clear
B)
They are always perpendicular to the surface of a charged conductor done
clear
C)
They always form closed loops done
clear
D)
They are parallel and equally spaced in a region of uniform electric field. done
clear
View Solution play_arrow
question_answer 33)
Which one of the following pattern of electric lines of force is not possible in field due to stationary charges ?
A)
B)
C)
D)
View Solution play_arrow
question_answer 34)
Electric lines of force are curved
A)
in the field of a single positive or negative charge done
clear
B)
in the field of two equal and opposite charges done
clear
C)
in the field of two like charges done
clear
D)
both [b] and [c] done
clear
View Solution play_arrow
question_answer 35)
The figure below shows the electric field lines due to two positive charges. The magnitudes \[{{E}_{A}}\], \[{{E}_{B}}\] and \[{{E}_{C}}\] of the electric fields at points A, B and C respectively are related as
A)
\[{{E}_{A}}>{{E}_{B}}>{{E}_{C}}\] done
clear
B)
\[{{E}_{B}}>{{E}_{A}}>{{E}_{C}}\] done
clear
C)
\[{{E}_{A}}={{E}_{B}}>{{E}_{C}}\] done
clear
D)
\[{{E}_{A}}>{{E}_{B}}={{E}_{C}}\] done
clear
View Solution play_arrow
question_answer 36)
Directions: (36 - 40) Torque on Dipole in a Uniform Electric Field When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However some work is done in rotating the dipole against the torque acting on it.
The dipole moment of a dipole in a uniform external field \[\overrightarrow{E}\] is \[\overrightarrow{P}\]. Then the toruqe \[\overrightarrow{\tau }\] acting on the dipole is
A)
\[\overrightarrow{\tau }=\overrightarrow{P}\times \overrightarrow{E}\] done
clear
B)
\[\overrightarrow{\tau }=\overrightarrow{P}\,\,.\,\,\overrightarrow{E}\] done
clear
C)
\[\overrightarrow{\tau }=2\left( \overrightarrow{P}+\overrightarrow{E} \right)\] done
clear
D)
\[\overrightarrow{\tau }=\left( \overrightarrow{P}+\overrightarrow{E} \right)\] done
clear
View Solution play_arrow
question_answer 37)
An electric dipole consists of two opposite charges, each of magnitude \[1.0\,\mu C\] separated by a distance of 2.0 cm. The dipole is placed in an external field of \[{{10}^{5}}\,N\,{{C}^{-1}}\]. The maximum torque on the dipole is
A)
\[0\centerdot 2\times {{10}^{-3}}Nm\] done
clear
B)
\[1\times {{10}^{-3}}\,Nm\] done
clear
C)
\[2\times \,{{10}^{-3}}\,Nm\] done
clear
D)
\[4\times {{10}^{-3}}\,Nm\] done
clear
View Solution play_arrow
question_answer 38)
Torque on a dipole in uniform electric field is minimum when \[\theta \] is equal to
A)
\[0{}^\circ \] done
clear
B)
\[90{}^\circ \] done
clear
C)
\[180{}^\circ \] done
clear
D)
Both [a] and [c] done
clear
View Solution play_arrow
question_answer 39)
When an electric dipole is held at an angle in a uniform electric field, the net force F and torque \[\tau \] on the dipole are
A)
\[F=0,\,\tau =0\] done
clear
B)
\[F\ne 0,\,\tau \ne 0\] done
clear
C)
\[F=0,\,\tau \ne 0\] done
clear
D)
\[F\ne 0,\,\tau =0\] done
clear
View Solution play_arrow
question_answer 40)
An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle \[\theta \] with the direction of the field. Assuming that the potential energy of the dipole to be zero when \[\theta =90{}^\circ \], the torque and the potential energy of the dipole will respectively be
A)
\[pE\,\sin \theta ,\,-pE\,\cos \theta \] done
clear
B)
\[pE\,\sin \theta ,\,-2pE\,\cos \theta \] done
clear
C)
\[pE\,\sin \theta ,\,2pE\,\cos \theta \] done
clear
D)
\[pE\,\cos \theta ,\,-pE\,\sin \theta \] done
clear
View Solution play_arrow
question_answer 41)
Directions : (41-45) Continuous Charge Distribution In practice, we deal with charges much greater in magnitude than the charge on an electron, so we can ignore the quantum nature of charges and imagine that the charge is spread in a region in a continuous manner. Such a charge distribution is known as continuous charge distribution. There are three types of continuous charge distribution : Line charge distribution Surface charge distribution Volume charge distribution as shown in fig.
Statement 1: Gauss's law can't be used to calculate electric field near an electric dipole. Statement 2: Electric dipole don't have symmetrical charge distribution.
A)
Statement 1 and statement 2 are true done
clear
B)
Statement 1 is false but statement 2 is true done
clear
C)
Statement 1 is true but statement 2 is false done
clear
D)
Both statements are false. done
clear
View Solution play_arrow
question_answer 42)
An electric charge of \[8.85\times {{10}^{-13}}C\] is placed at the centre of a sphere of radius 1 m. The electric flux through the sphere is
A)
\[0\centerdot 2N\,{{C}^{-1}}{{m}^{2}}\] done
clear
B)
\[0\centerdot 1N\,{{C}^{-1}}{{m}^{2}}\] done
clear
C)
\[0\centerdot 3N\,{{C}^{-1}}{{m}^{2}}\] done
clear
D)
\[0\centerdot 01N\,{{C}^{-1}}{{m}^{2}}\] done
clear
View Solution play_arrow
question_answer 43)
The electric field within the nucleus is generally observed to be linearly dependent on r. So,
A)
\[a=0\] done
clear
B)
\[a=\frac{R}{2}\] done
clear
C)
\[a=R\] done
clear
D)
\[a=\frac{2R}{3}\] done
clear
View Solution play_arrow
question_answer 44)
What charge would be required to electrify a sphere of radius 25 cm so as to get a surface charge density of \[\frac{3}{\pi }C{{m}^{-2}}\]?
A)
\[0\centerdot 75\,C\] done
clear
B)
\[7\centerdot 5\,C\] done
clear
C)
\[75\,C\] done
clear
D)
zero done
clear
View Solution play_arrow
question_answer 45)
The SI unit of linear charge density is
A)
Cm done
clear
B)
\[C\,{{m}^{-1}}\] done
clear
C)
\[C\,{{m}^{-2}}\] done
clear
D)
\[C\,{{m}^{-3}}\] done
clear
View Solution play_arrow
question_answer 46)
Directions : (46 - 50)
Parallel Sheet of Charge
Surface charge density is defined as charge per unit surface area of surface charge distribution i.e.,
\[\sigma =\frac{dq}{dS}\]. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs having magnitude of \[17\centerdot 0\times {{10}^{-22}}\,C\,\,{{m}^{-2}}\] as shown. The intensity of electric field at a point is\[E=\frac{\sigma }{{{\varepsilon }_{0}}}\], where \[{{\varepsilon }_{0}}\]= permittivity of free space.
E in the outer region of the first plate is
A)
\[17\times {{10}^{-22}}N/C\] done
clear
B)
\[1\centerdot 5\times {{10}^{-25}}\,N/C\] done
clear
C)
\[1\centerdot 9\times {{10}^{-10}}\,N/C\] done
clear
D)
zero done
clear
View Solution play_arrow
question_answer 47)
E in the outer region of the second plate is
A)
\[17\times {{10}^{-22}}\,N/C\] done
clear
B)
\[1\centerdot 5\times {{10}^{-15}}\,N/C\] done
clear
C)
\[1\centerdot 9\times {{10}^{-10}}\,N/C\] done
clear
D)
zero done
clear
View Solution play_arrow
question_answer 48)
E between the plates is
A)
\[17\times {{10}^{-22}}\,N/C\] done
clear
B)
\[1\centerdot 5\times {{10}^{-15}}\,N/C\] done
clear
C)
\[1\centerdot 9\times {{10}^{-10}}\,N/C\] done
clear
D)
zero done
clear
View Solution play_arrow
question_answer 49)
The ratio of E from right side of B at distances 2 cm and 4 cm, respectively is
A)
1 : 2 done
clear
B)
2 : 1 done
clear
C)
1 : 1 done
clear
D)
\[1\,\,:\,\,\sqrt{2}\] done
clear
View Solution play_arrow
question_answer 50)
In order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is
A)
spherical done
clear
B)
cylindrical done
clear
C)
straight line done
clear
D)
none of these done
clear
View Solution play_arrow