In this post we are going to discuss some most important Electric Charges And Fields NEET Questions With Solutions. Electric charges and fields are fundamental part of electrostatics. You have to know the type of electric charges, electric field, its direction, electric flux and its rate etc. In this article all the questions related to these topics with solutions are given. These these Electric Charges And Fields Neet Questions and Answers are mostly asked in your NEET, AIIMS, JEE-Mains, IIT-JEE and various other examinations. So try to solve these TOP NEET Questions On Electric Charges And Fields with Answers that will help you most.
Electric Charges And Fields NEET Questions With Solutions
1. What is the unit of electric flux?
a) V/m
b) N/m
c) V*m
d) N/Coulomb
Answer: c
Explanation: Flux is equal to the product of field intensity and area of the surface. But field intensity multiplied by a length gives the unit of electric potential as E=-\(\frac {dV}{dx}\). Therefore, flux also means electric potential multiplied by length. This gives us the unit V*m. There is another unit of flux N*m2*C-1.
2. Electric flux will be maximum if the angle between the field lines and area vector is ______
a) 45 degree
b) 135 degree
c) 90 degree
d) 0 degree
Answer: d
Explanation: We know that the electric field and area both are vector quantity and the electric flux is expressed as \(\vec{E}.\vec{s}\) (taking dot product). But if the angle between the two vectors is θ then the formula becomes E.s.cosθ.Cosθ will be maximum if theta is zero, in all other cases, the value of Cosθ is less than 1. Therefore flux will be the maximum if the angle is 0 degrees.
3. If a charge is placed outside a closed surface, flux due to that charge inside the surface will be ________
a) Positive
b) May be positive or negative, depending on the nature of the charge
c) Negative
d) Zero
Answer: d
Explanation: According to Gauss’s principle, if there is no charge bound inside a surface, net electric flux coming out of the surface will always be 0. In this case, the charge is kept outside the surface, so it will generate no field lines and hence no flux inside the surface. The situation will be the same if an electric dipole is placed inside a surface as dipole has equal positive and negative charges; the total charge inside the surface becomes 0.
4. Two separate charges q and 10q are placed inside two different spheres. In which case, the electric flux will be greater?
a) Flux will be same in both the cases
b) 1st sphere
c) 2nd sphere
d) No flux in any of the spheres
Answer: c
Explanation: The electric field line is directly proportional to the charge bound inside a sphere. We know that, field lines are the measure of electric flux i.e. number of field lines crossing through a surface are know an electric flux. Therefore flux will be lesser in the case of q charge and will be 10 times in case of a 10q charge.
5. Which of the following law explains the relation between the charge inside a surface and electric flux?
a) Gauss’s Law
b) Coulomb’s Law
c) Faraday’s Law
d) Pascal’s Law
Answer: a
Explanation: Gauss’s Law gives the relation between electric flux and charges inside a surface. It states that electric flux coming out from a closed surface is equal to \(\frac {1}{\varepsilon}\) times the charge inside the surface. Coulomb’s Law explains the force between two charges and Pascal’s Law is related to fluids.
6. Flux coming out from a balloon of radius 10 cm is 1.0*103 N.m2.C-1-1. If the radius of the balloon is doubled, the flux coming out from the balloon will be _______
a) 0.5 times
b) 2 times
c) Same
d) 4 times
Answer: c
Explanation: If the radius of the balloon is 2 times the initial value, the surface area of the balloon will be 4 times the initial value because of surface area=4π(radius)2. But in this case, flux coming out from the balloon is dependent only upon the charge inside the balloon, not on the area of the surface. As charge enclosed in the balloon is the same in both the cases, the flux will also be the same.
7. Which among the following molecule is not a dipole?
a) NH3
b) H2O
c) HCl
d) CH4
Answer: d
Explanation: Though water and ammonia are covalent molecules, they have a net dipole moment due to their distorted structure than the ideal one. Hydrochloric acid has a linear molecular structure but its net dipole moment is towards the Hydrogen atom. But methane has a symmetrical tetrahedral structure and its net dipole moment is zero hence it doesn’t behave as a permanent dipole. Methane solvents are therefore non-polar solvents.
8. What is the dimension of the dipole moment?
a) [L T I]
b) [L T I2]
c) [M L T I]
d) [L T I-1]
Answer: a
Explanation: The dipole moment is defined as the product of a charge and distance. The dimension of charge (current*time) is [I T] and the dimension of distance is [L]. Therefore the dimension of dipole moment is [L T I]. Its unit in the CGS and the SI system are esu*cm and C*m respectively.
9. If electric field intensity at a certain distance from a dipole on its axis is E1 and at the same distance but on the perpendicular bisector of the dipole is E2, then __________
a) E1=E2
b) E1=2 E2
c) E1=4 E2
d) E1=3 E2
Answer: b
Explanation: If the dipole moment of a dipole is p, then electric field intensity at a distance r from the dipole on its axis will be \(\frac {2p}{r^3}\). But electric field intensity at a distance r from the dipole on its perpendicular bisector is \(\frac {p}{r^3}\). Both the formulas are derived by assuming r >> length of the dipole. Thus E1=\(\frac {2p}{r^3}\) and E2=\(\frac {p}{r^3}\) and hence E1=2 E2.
10. Dipole moment depends on _______
a) Charge only
b) Charge and length of a dipole
c) Charge, length of a dipole and dielectric constant of the medium
d) Charge and dielectric constant of the medium
Answer: b
Explanation: Dipole is defined as two equal but opposite charges, kept at a small distance and having a dipole moment. The dipole moment is simply the product of the electric charge and length of the dipole. The dielectric constant of the medium doesn’t affect the dipole moment i.e. dipole moment of an electric dipole will be the same in water as well as in the air.
11. An electric dipole is placed inside a cube. What will be the nature of electric flux from the cube surface?
a) Coming out of the surface
b) Coming in towards the surface
c) No flux at all
d) Coming out from one half and coming inwards in another half
Answer: c
Explanation: An electric dipole consists of two point charges. The amount of charges is the same but their polarities are different. Therefore the sum of total charges in a dipole is always 0. But flux from a closed surface is related to the total charge inside a surface. As the total charge inside the cube is zero, so there will be no flux coming out or going in towards the surface.
12. If an electric dipole is placed in a uniform electric field ______ will act on the dipole.
a) A force but no torque
b) Both force and torque
c) Torque but no force
d) No torque or force
Answer: c
Explanation: Dipole is the combination of two equal but opposite charges, kept at a certain distance. If it is placed in a uniform electric field, both the charges will suffer the same but opposite forces on them. As a result, the net force on the dipole becomes zero, but due to equal and opposite forces acting on two different points, there is a net torque acting on the dipole.
13. If an electric dipole is placed in a non-uniform electric field ______ will act on the dipole.
a) A force but no torque
b) Both force and torque
c) Torque but no force
d) No torque or force
Answer: b
Explanation: In the case of a non-uniform electric field, the force acting on both the charges of the dipole will be unequal. So, there will be a net force acting on the dipole in a certain direction. Also, there will be a torque due to two forces acting at two different points. But in case of a uniform electric field, the net force on the dipole will be zero but net torque will be non-zero.
14. An electric dipole will be in stable equilibrium if the angle between the axis of the dipole and the electric field is ________
a) 0 degree
b) 180 degree
c) 90 degree
d) 45 degree
Answer: a
Explanation: Torque acting on a dipole is p*E*sinθ where E is the electric field. Now θ is 0 degree, so sinθ becomes 0 and hence no torque acts on the dipole. So in this case, no force or torque acts on the dipole. Therefore it will be the condition of stable equilibrium. In the case of θ=180 degree, sinθ is also 0 but the condition is known as unstable equilibrium i.e. if we rotate the dipole a bit, it will not come back to its initial position.
15. 1uC and -1uC are placed at a distance of 5 cm forming a dipole. What is the amount of torque required to place the dipole perpendicularly to an electric field of 3*105 N/C?
a) 5*10-3 N. m
b) 15*10-3 N. m
c) 1*10-3 N. m
d) 10*10-3 N. m
Answer: b
Explanation: Dipole moment of the dipole = 1*10-6*5*10-2 C. m=5*10-8 C. m. Given the electric field-intensity is E=3*105 N/C. Therefore required torque will be p*E*sinθ where θ is the angle between the electric field and the dipole moment = 90 degrees. Therefore torque required in this case will be 5 *10-8*3*105*sin 90 N. m=15*10-3 N. m.
16. If a non-polar substance is placed in an electric field, what will happen?
a) A net dipole moment will be observed
b) The substance will oscillate
c) The substance will orient itself perpendicular to the electric field
d) It will conduct electricity
Answer: a
Explanation: A non-polar substance consists of a huge number of dipoles in it, but they are oriented randomly and hence the net dipole moment of the substance becomes 0 and it acts as a non-polar substance. But if it is placed in an electric field, the dipoles present in it will orient themselves in the direction of the field and hence a net dipole moment will be observed, known as induced dipole moment. No current flow or oscillation of the substance will be observed.
17. A circular annulus of inner radius r and outer radius R has a uniform charge density a. What will be the total charge on the annulus?
a) a*(R2-r2)
b) π*a*(R2-r2)
c) a*(R-r)
d) π*a* R2
Answer: b
Explanation: Total surface area of the annulus = π*(R2-r2) because it has outer radius R and inner radius r. We know surface charge density is the amount of charge stored on the unit surface area. In this case, surface charge density is a. Therefore total charge on the annulus = π*a*(R2-r2).
18. What is the dimension of linear charge density?
a) [I T L-1]
b) [I T-1 L]
c) [I T L]
d) [I-1 T-1 L]
Answer: a
Explanation: Linear charge density λ=\(\frac {Amount \, of\, charge}{Total \, length}\). The dimension of electric charge = [I T] and the dimension of length is [L]. Hence the dimension of linear-charge-density = [I T L-1]. In the case of surface charge density, the dimension is [I T L-2] because it means charge stored on the unit surface area.
19. A solid non conducting sphere of radius 1m carries a total charge of 10 C which is uniformly distributed throughout the sphere. Determine the charge density of the sphere.
a) 10 C/m3
b) 4.76 C/m3
c) 0.1 C/m3
d) 2.38 C/m3
Answer: d
Explanation: Volume of the sphere = \(\frac {4}{3}\)πr3 where r is the radius of the sphere. Therefore, the charge density = \(\frac {total \, charge}{\frac {4}{3}\pi r^3}\). Now substituting the values, charge density = \(\frac {10}{\frac {4}{3}\pi r^3}\) = 2.38 C/m3. But if the sphere is conducting, we have to consider the surface charge density.
20. A non-conducting sphere has uniform charge density in it. The electric field at a point inside the sphere will be _______
a) Zero
b) Only due to the charge inside that point
c) Only due to the charge outside that point
d) Due to the entire charge of the sphere
Answer: b
Explanation: In the case of a non-conducting sphere, an electric charge is uniformly distributed throughout the sphere. Inside the sphere, the electric field at a point will be only due to the electric charge inside that point. There will be no effect of the remaining charge outside that point. In the case of the conducting sphere, the electric field at every point inside the sphere is zero.
21. Electric flux coming out of a single Na+ ion is _________ Nm2C-1
a) 1.8*10-8
b) 1.8*10-10
c) 5.4*10-8
d) 3.6*10-8
Answer: a
Explanation: A single Na+ ion means a single sodium atom without an electron. The charge of the ion is the same as the charge of an electron but positive i.e. +1.602*10-19C. Therefore, applying Gauss’s Law, electric flux coming out of the ion = \(\frac {charge \, of \, the \, ion}{\varepsilon}\). Substituting the values, we get flux = 1.8*10-8 Nm2C-1.
22. Gauss’s Law cannot be applied in ________
a) Hollow sphere
b) Solid sphere
c) Cube
d) Unbounded surface
Answer: d
Explanation: Gauss’s Law is valid only in case of a closed surface. Solid or hollow sphere, cube are closed surface and so Gauss’s Law can be applied in those cases. But in case of unbounded surface, there is no meaning of charge bounded by the surface and so Gauss’s Law cannot be applied in those cases.
23. If a 10C charge is placed in a medium having relative permittivity 5.4, what will be the amount of flux coming out of that charge?
a) 1*1011 Nm2C-1
b) 2.1*1011 Nm2C-1
c) 2.1*1010 Nm2C-1
d) 7.1*1011 Nm2C-1
Answer: b
Explanation: Gauss’s Law states that electric flux coming out of a charge is=\(\frac {amount \, of \, charge}{permittivity \, of \, the \, medium}\). In this case, the charge is 1C and permittivity of the medium=5.4*8.85*10-12. Substituting the values, we get the total flux coming out of the charge is 2.1*1011 Nm2C-1.
24. A non-conducting cube of side length 1m has a spherical cavity of radius 0.1m inside it. The system has a charge density of 2.5 C/m3. What is the total flux coming out of the surface of the cube?
a) 2.8*1031 Nm2C-1
b) 2.8*1011 Nm2C-1
c) 5.7*1011 Nm2C-1
d) 0.2*1011 Nm2C-1
Answer: b
Explanation: The volume of material inside the cube=13–\(\frac {4\pi*0.1^3}{3}\)=0.996m3. Therefore, the total charge inside the system=volume*charge density=0.996*2.5=2.49C, Therefore the net flux coming out of the system=\(\frac {charge}{\varepsilon} = \frac {2.49}{8.85*10^{-12}}\)=2.8*1011 Nm2C-1.
25. Net inward and outward flux of a closed surface are ∅1 and ∅2. What is the charge enclosed inside the surface?
a) Zero
b) -(∅1-∅2)εo
c) (∅1-∅2)εo
d) \(\frac {(\phi_1 + \phi_2)}{\varepsilon_o}\)
Answer: b
Explanation: The net outward flux from the surface is (∅2-∅1) = -(∅1-∅2). We know, according to Gauss’s Law, net outward flux coming out of a closed surface is \(\frac {1}{\varepsilon_o}\) times the charge stored in it. Therefore charge stored in the surface = εo*flux coming out of the surface = (∅2-∅1)εo.
Electric Charges And Fields NEET Questions With Solutions
In class 12 we learned about charges and how they interact with electric field. Understanding how charges a field works is important as it provides us with insight into how we can manipulate them to achieve desired results.
The term electric field is used to describe the space surrounding an object in which there are an unequal number of positive and negative charges. The electric field exerts a force on any object that transports electricity – be it a moving electron or a charged particle of dust.
Electric fields are continuous quantities, whose direction can vary through electrodes with different signs (negative and positive). Thus, it is vital to know the net charge that exists between any two bodies in order to understand the nature of their interaction. The unit for measuring electric field strength is Coulomb/meter^2 .
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