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Chapter#01: Electric Field

Team Quanta gladly presents all possible short questions of Electricity & Magnetism – I’s Chapter#01: Electric Field.

Q.1 Can an electric field exist in a vacuum?

Answer: Yes, electric field can exist in vacuum.

Q.2 While defining the electric field, why is it necessary to specify that the Magnitude of the test charge be very small?

Answer: The magnitude of the test positive charge is kept very small so that it does not disturb the distribution of the charges.

Q.3 An insulated rod is said to carry an electric charge. How could we verify this and determine the sign of the charge?

Answer: If the rod is charged, it will attract small pieces of paper when it is kept near to them. To determine the sign of the electric charge on it, we an object of known electric charge, suppose positive. The sign of charge can be determined by the nature of force between rod and positively charged object.

Q.4 What are the limitations of Coulomb’s law?

Answer:

  1. Coulomb’s law only applies to point charges.
  2. It is not applicable where we have irregular shaped objects.

Q.5 We used a positive test charge to explore electric fields. Could we have used a Negative test charge?

Answer: Yes, we can use test positive charge to determine electric fields. But it may create problems while considering the direction of forces, because for test negative charge the direction of force is not same as the direction of electric field.

Q.6 For what orientations of an electric dipole in a uniform electric field is the potential energy of the dipole (a) the greatest and (b) least?

Answer: From the formula of potential energy of dipole P.E=-\vec{P}.\vec{E}=-PE\cos{\theta} , the potential energy is greatest when the angle between dipole and the field is  and is least when the angle is

Q.7 An electric dipole is placed in a non uniform electric field. Is there a net force on it?

Answer: Yes, a dipole kept in non-uniform electric field will experience both net force and torque.

Q.8 If the total charge inside a closed surface is known but the distribution of the charge is unspecified, can we apply Gauss’s law to find the electric field?

Answer: Yes, gausses law has only concern with the total charge enclosed. But in only case that the electric field remains constant over the surface.

Q.9 State and prove gausses law

Answer: Gausses law states that: “the flux passing through a closed surface is \frac{1}{\epsilon_0} times total charge enclosed by that the surface”

    \[\phi=\oint{E.dA}=\frac{q_{enc}}{\epsilon_0}\]

Q.10 How electric force is different from gravitational force?

Answer:

Gravitational forceElectric force
It is force caused by the inertial property of matter.It is a weak force.It is always attractiveIt is the force caused by the electric property of matter.It is a strong force.It could be attractive or repulsive.

Q.11 If net charge enclosed by a surface is zero, what will be the flux through it?

Answer: By Gausses law, if there is no charge enclosed by the surface, there will be no flux through that surface.

Q.12 Define one Coulomb

Answer: In system international coulomb is the unit of electric charge and it is equal to the charge of  electrons. It is defined as “one coulomb is the charge which passes through a conductor carrying 1 A electric current in 1 second”.

Q.13 Why electric field inside a hollow charged sphere is zero?

Answer: The flux through hollow charged sphere is zero because as Gauss’s law states that the flux is proportional to the total charged enclosed by the cavity. There is no charge inside the hollow sphere, therefore flux inside the sphere is zero, also, electric field.

Q.14 When the torque of the dipole is equal to zero.

Answer: From the following relation:

    \[\vec{\tau}=\vec{P}\times\vec{E}\]

The torque acting on the dipole is zero when the dipole moment and electric field are parallel or anti parallel.

Q.15 A surface encloses an electric dipole. What can you say about electric flux for this surface?

Answer: Through a surface enclosing dipole, the net flux is zero, because electric dipole consists of two equal and opposite charges. The net charge enclosed by the surface will be zero.

Q.16 What does it mean to say that a physical quantity is Quantized or conserved? Give some examples.

Answer: Quantization of a physical quantity means that its values exist in form of discrete packets such as charge, energy of photon and angular momentum of electron. Conservation of a physical quantity means that total amount of that quantity remains constant such as conservation of energy, conservation of linear momentum etc

Q.17 Is it precisely true that Gauss’s law states that the total number of lines of force crossing any closed surface in the outward direction is proportional to the net positive charge enclosed within the surface?

Answer: Yes the above statement is true. According to Gauss’s law,

    \[\phi_e=\frac{1}{\epsilon_\circ}\left(total\ charge\ enclosed\right)\]

    \[\Rightarrow\phi_e\propto total\ charge\ enclosed\]

It follows that total number of lines is proportional to the net positive charge enclosed.

Q.18 Is  necessarily zero inside a charged rubber balloon, which is spherical? Assume that charge is uniformly distributed over the surface.

Answer: Yes,  is necessarily zero inside a charged rubber balloon, if balloon is spherical. If Gaussian surface in imagined inside charged balloon, then it does not contain any charge i.e. . So by Gauss’s law,

    \[\phi_e=\frac{q}{\epsilon_\circ}=0\ \ \Rightarrow EA=0\]

    \[Since\ \ \ \ \ \ \ A\neq0,\ \ \ \ \ \ \ \ So\ \ \ \ \ \ \ \ E=0\]

Q.19 A surface encloses an electric dipole. What you can say about electric flux for this surface? Or, what is electric flux through a closed surface surrounding a dipole?

Answer: If a closed surface encloses an electric dipole, then total flux through this is zero. It is because an electric dipole consists of two equal and opposite charges separated by a small distance, hence algebraic sum of all charges within the surface is zero.

Q.20 A balloon is negatively charged by rubbing and then clings to a wall. Does this mean that the wall is positively charged?

Answer: No, balloon includes polarization of the molecules in the wall, so that a layer of positive charge exists near the balloon. The attraction between these charges and the negative charges on the balloon is larger than the repulsion between the negative charge on the balloon and the negative charge in the polarized molecules, so that there is a net attractive force toward the wall.

Q.21 What does it mean to say that a physical quantity is quantized or conserved? Give some examples.

Answer: Quantization of a physical quantity means that its values exist in form of discrete packets such as charge, energy of photon and angular momentum of electron. Conservation of a physical quantity mean that total amount of that quantity remains constant such as conservation of energy, conservation of linear momentum etc.

Q.22 When an object that was neutral becomes charged, does the total charge of universe increases?

Answer: Since charge is a conserved quantity, so charge of universe will remain constant. There is only transfer of charge from one object to the other.

Q.23 An atom is normally electrically neutral. Why then should an alpha particle be deflected by the atoms under any circumstances?

Answer: When an alpha particle is bombarded on an atom, it may be deflected due to nucleus because positive charge of atom is concentrated on nucleus of atom.

Q.24 What is Gaussian surface?

Answer: A hypothetical closed surface of any shape drawn in an electric field for the purpose of solving problem concerned like electric flux is called Gaussian surface. The shape of Gaussian surface is decided on the basis of symmetry of problem and values f electric flux be calculated.

Q.25 How many electrons will be produced by charge of 1C?

Answer:  No of electrons =\frac{1C}{1.6\times{10}^{-19}C}=6.25\times{10}^{18}  electrons.

Q.26 Gauss’s law is not very useful in calculating the electric field of charged disk. Explain.

Answer: We can find electric field of charged disk by using Gauss’s law but it is not more useful in this case, Gauss’s law can be used to find  electric field in those cases where problem possesses high symmetry so that on Gaussian surface, electric field is uniform or has no normal component.

Q.27 A point charge  of mass  is released from rest in a non-uniform field.

(a)- Will it necessarily follow the electric field lines that pass through the released point?

(b)- Under what circumstances, if any will a charged particle follow the electric field lines?

Answer: (a)- No, it will not necessarily follow electric field lines that passes through the released point.

(b)- If electric field is uniform, then charge particle will follow the electric field lines.

Q.28 Give a comparison of Coulomb’s law of electrostatics and Newton’s law of gravitation.

Answer:  In both laws, force is inversely proportional to square of distance between the centers of objects in case of Newton’s law and point charges in case of Coulomb’s law.

In case of Newton’s law, gravitational force is always attractive where as electrostatic force in Coulomb’s law may be attractive or repulsive.

Q.29 Compare the way  varies with  for a point charge, dipole and quadrupole.

Answer:  For a point charge (mono pole),

    \[E\propto1/r\]

For a dipole,

    \[E\propto1/r^2\]

For a quadrupole,

    \[\phi_e=\frac{q}{\epsilon_\circ}=\vec{E}\cdot\vec{A}\]

Q.30 Small bits of paper are attracted to an electrically charged comb, but as soon as the touch, often jump violently away from it. Explain.

Answer: The small bits of paper are attracted to the comb by polarization effect. When one of the bits of paper comes into contact with the comb, it acquires charge from the comb. Now the piece of paper and the comb have charge of the same sign. Hence due to repulsive force between them, bits of paper jump violently away from the comb.

Q.31 in Gauss’s law . Is necessarily the electric field attributable to the charge ?

Answer: By Gauss’s law,

    \[\phi_e=\frac{q}{\epsilon_\circ}=\vec{E}\cdot\vec{A}\]

Electric field intensity due to a point charge is given by,

    \[E=\frac{1}{4\pi\epsilon_\circ}\frac{q}{r^2}\\]

This equation shows that  is necessarily the electric field attributable to the charge .

Q.32 A uniform electric field of 1N/C is set up by a uniform distribution of charge in the -plane. What is the electric field intensity inside a metal ball placed 0.5m above the -plane?

Answer: In metallic ball, free charges within the ball will rearrange themselves to produce electrostatic equilibrium at all points within the ball. Due to electrostatic equilibrium, the electric field intensity is zero at all points within the ball.

Q.33 What is value of electric field in uniformly charged hollow cylinder?

Answer: The electric field intensity inside a uniformly charged hollow cylinder is zero.

Q.34 Is the electric force that one charge exerts on another charge changed, if other charges are brought nearby?

Answer: Coulomb’s force between two charges is a two body interaction and is indecent of other charges. Hence the electric force that one charge exerts on another charge remains unchanged if other charges are brought nearby.

Q.35 What is unit of charge? Define it.

Answer: Unit of charge is coulomb. We know that,

    \[i=\frac{q}{t}\ \ \ \ \Rightarrow q=it\]

When a current of 1 ampere passes through a conductor for 1s, then charge stored in conductor is said to be one coulomb.

Q.36 What happens when a charged insulator is placed near an uncharged metallic object?

Answer:  When a charged insulator is placed near a metallic object, metallic object will be polarized. Due to this, attractive force between charged insulator and the opposite type of charge in the metal is greater than repulsive force between the insulator and the same type of charge in the metal. Hence charged insulator attracts the uncharged metallic object.

Q.37 How would one show that electrostatic force between two charged objects is independent of their masses?

Answer: It can be proved by using Coulomb’s law,

    \[F=\frac{1}{4\pi\epsilon_\circ}\frac{q_1q_2}{r^2}\]

This relation shows that electrostatic force between two charged object is independent of masses and depends on magnitude of point charges.

Q.38 A charge  is inside a closed Gaussian surface; a second charge  is just outside the surface. Does electric flux through the surface depend on both charges?

Answer: The electric flux through the Gaussian surface depends on  and not on both charges because charge  is outside the Gaussian surface.

Q.39 Vehicles carrying inflammable material usually have metallic ropes touching the ground during motion. Why?

Answer: When vehicles move through the air, they get charged due to the friction and charges developed on tyres also. If charges are sufficient, they will produce spark. The vapors which escape from inflammable material carried by vehicle may catch fire. To prevent this, metallic ropes touching the ground are suspended which send the charges to earth.

Q.40 Explain what is meant by statement that electrostatic force obey the principle of superposition?

Answer: Suppose there are three point charges q1 , q2 and q3 and we have to find force exerted on charge  due to other two charges, then we make use of superposition principle i.e. force  exerted on charge  due to other two charges is,

    \[\vec{F_1}=\vec{F_{12}}+\vec{F_{13}}\]

Where  is force on particle one from particle two and is force on particle one from particle three.

Q.41 Electric lines of force never cross. Why?

Answer: Electric lines of force never cross each other.This is because  could not have more than one direction which is physically incorrect.

Q.42 Ordinary rubber is an insulator. But the special rubber tyres of aircraft are made slightly conducting. Explain.

Answer: Tyres of aircraft get charged due to friction. If tyres are non-conducting, it causes spark. Since tyres are conducting, the charge flows to earth.

Q.43 The quantum of charge is 1.60 x 10-19 C. Is there a corresponding quantum of mass?

Answer: Quantum charge is 1.60 x 10-19 C  but there is no quantum of mass, however there is a quantum of energy called photon.

 Q.44 Force of attraction between two point charges placed at a distance  in a medium is . At what distance apart should these charges be kept in same medium so that force between them becomes ?

Answer: According to given condition,

    \[\frac{1}{4\pi\epsilon_\circ}\frac{q_1q_2}{x^2}=\frac{1}{3}\left(\frac{1}{4\pi\epsilon_\circ}\frac{q_1q_2}{d^2}\right)\ \Rightarrow x^2=3d^2\Rightarrow x=\sqrt{3d}\]

Q.45 A point charge is moving in the electric field at right angle to the field lines. Does any force act on it?

Answer: Electric force is given by relation,

    \[F=qE\]

This relation shows that electric force is independent of angle. Hence electric force acts on a particle moving in electric field at right angle to the field lines.

Q.46 what prevents gravity from pulling you through the ground to the center of earth?

Answer: The outer regions of atoms making up ground contain electrons and same is the case in our body. According to Coulomb’s law same charges repel each other, thus electrons on the surface of ground and our feet repel each other.

Q.47 If a metallic object receives positive charge by losing electrons, does its mass remain same? What happens to mass if object receives negative charge?

Answer: Mass of object decreases vet slightly when it is given a positive charge because it loses electrons. When an object is given negative charge, its mass increases very slightly because it gains electrons.

Q.48 Two point charges of unknown magnitude and sign are a distance apart. The electric field is zero at any point between them. What can you conclude about the charges?

Answer: As the electric field is zero at a point between the two charges, either both charges are positive or negative. But if one charge is positive and other charge is negative, then both electric fields will ac in the same direction and their resultant will never be zero.

Q. 49 For what orientation of electric dipole in a uniform electric field is the potential energy of dipole;

(a)- the greatest value.          (b)- the least value.    (c)- the zero value.

Answer: Potential energy of a dipole in a uniform electric field is given by

    \[U=-\vec{p}\cdot\vec{E}\]

(a)- Thus U is maximum when \vec{p} and \vec{E} are anti parallel.

(b)- U is minimum when  \vec{p} and \vec{E} are parallel.

(c)- U is zero when \vec{p} and \vec{E} are at right angle.

Q. 50 Positive and negative charges of same magnitude are on a straight line. What is direction of 50 on this line between charges?

Answer: Electric field due to both charges will be in same direction i.e. from positive to negative charge.

Q.51 Suppose that an electric field in some region is found to have a constant direction but to be decreasing in strength to that direction. What do you conclude about the charge in the region?

Answer: Electric field intensity due to a point charge is given by,

    \[E=\frac{1}{4\pi\epsilon_\circ}\frac{q}{r^2}\]

For constant r,\ \ \ \ E\propto q, thus direction of electric field remains same.

Q.52 Suppose a scientist has chosen to measure small energy in proton volts rather than electron volts. IS there any difference in measurement?

Answer: There would be no difference in measurement of energy (except for sign) because charge on proton is equal to charge of electron. Proton would be moving in opposite direction to electron and more slowly because mass of proton is larger than election.

Q.53 An electron of charge -e  circulated around a helium nucleus of charge 2e in a helium atom. Which particle exerts the larger force on the other?

Answer: In this case, nucleus exerts larger force on electron.

Q.54 A proton is released from rest in uniform electric field. Does electric potential energy increases?

Answer: No, when proton is moving in direction of electric field from higher to lower potential, its electrical potential energy decreases and appears as gain in kinetic energy.

Q.55 Identify the element X in reaction \mathbf{C}^{\mathbf{12}}+\mathbf{H}^\mathbf{1}\rightarrow\mathbf{X}.

Answer: The element X can be identified by using principle of conservation OF CHARGE. Atomic number of carbon is six and that of hydrogen is one, so atomic number of X should be seven which is nitrogen.

Q.56 Explain why charge is usually transferred by electrons than protons.

Answer: Electrons can be easily removed from atoms because they are loosely bound than protons inside the nucleus. That is why charge is easily transferred by electrons than protons.

Q.57 Why is it safe to stay inside an automobile during light storm?

Answer: The interior of car is safe because the charges on car’s metal shell reside on its outer surface. Thus the person in automobile touching the inner surface is safe.

Q.58 Two point charges of unknown magnitude and sign are placed a distance  apart. Is it possible to have  at any point not between the charges but on the line joining them? Where is the point located?

Answer: When two point charges of opposite sign  q & – q  are placed on same line separated by some distance , then electric field will be zero at a point not between charges but on side of negative charge

Q.59 Two point charges of unknown magnitude and sign are placed a distance   apart. Is it possible for any arrangement of two point charges to find two points at which E=0 , if so under what conditions?

Answer: It is possible in two cases:

  • When charges are of same sign, then electric field is zero at a point between charges.
  • When charges are of opposite sigh, then electric field will be zero at a point not between charges but on side of negative charge. In this case, mathematically we get a quadratic equation whose solution gives two points.

Q.60 An electric dipole has its dipole moment P aligned with a uniform electric field E.

  1. Is the equilibrium stable or unstable?
  2. Discuss the nature of equilibrium if  and  point in opposite direction.

Answer:

  1. Equilibrium will be stable when an electric dipole has its dipole moment \vec{p} aligned with a uniform electric field \vec{E}
  2. Equilibrium will be unstable when dipole moment \vec{p} and electric field \vec{E} are aligned in opposite direction. In this case dipole moment \vec{p} tends to rotate towards \vec{E} through 180o.

Q. 61 Write in your own words the purpose of Millikan oil drop experiment.

Answer: With the help of Millikan oil drop experiment, one can find charge on electron. In this experiment, charged oil drops subjected to an electric field and to gravity between the plates of a capacitor are accelerated by application of voltage. The charge on electron is determined from velocities in direction of gravity and in opposite direction.

Q.62 A point charge is placed at the center of a spherical Gaussian surface. Is electric flux changed if the sphere is replaced by a cube of one-tenth the volume and if a second charge is placed inside the Gaussian surface?

Answer: When sphere is replaced by a cube of one-tenth the volume there will be no change in flux.

When a second charge is placed inside the Gaussian surface, flux will change due to presence of second charge.

Q.63 Suppose the Gaussian surface encloses no net charge. What will be the value of flux?

Answer: By Gauss’s law,

    \[\phi_e=\frac{q}{\epsilon_\circ}=\vec{E}\cdot\vec{A}\]

When q=0,\ \ \ \ \phi=0. Thus in the absence of charge, no flux will pass through Gaussian surface.

Q.64 A person is placed in a large, hollow metallic sphere that is insulated from ground. If a large charge is placed on the sphere, will it be harmful for the person if he touches the inside of the sphere?

Answer: No, according to one of application of Gauss’s law the charge on the metallic sphere resides on its outer surface, so the person may touch the surface without any danger.

Q. 65 Electric and gravitational fields are common in some aspects. A room can be electrically shielded so that there are no electric fields in the room by surrounding it with a conductor. Can a room be gravitationally shielded? Explain.

Answer: A room can be electrically shielded because both types of charges positive and negative are present in the conductor. A room cannot be gravitationally shielded because mass is always positive or zero and can never be negative.

Q. 66 While defining the electric field, why is it necessary to specify that the magnitude of the test charge be very small?

Answer: The test should be very small so that electric field created by the test charge should not disturb the electric field we are measuring.

Q.67 An insulated rod is said to carry an electric charge. How could we verify this and determine the sign of the charge?

Answer: If it attracts a positively charged object, then rod is negatively charged and if it repel a positively charged object, then the rod is positively charged.

Q. 68 An object with a negative charge is placed in a region of space where the electric field is directed vertically upward. What is the direction of the electric force exerted on this charge?

Answer: An object with a negative is placed in a region of space where the electric field is directed vertically upward then the direction of the electric force exerted on this charge is vertically downward.

Q.69 Explain why the electric flux through a closed surface with a given enclosed charge is independent of the size or shape of the surface.

Answer: Electric flux is defined as,

    \[\phi_e=\vec{E}\cdot\vec{A}\]

This equation shows that electric flux is independent of size of surface. For example, in case of sphere, as radius of sphere increases, its area increases but field decreases at its surface so that product \vec{E}\cdot\vec{A} remains constant.

Q. 70 Can an electric field exist in a vacuum?

Answer: Electric field can exist in vacuum, just as light propagate through vacuum.

Q.71 Gauss’s law given information about the magnitude of charge inside Gaussian surface. Can it tell us where inside the surface, it is located?

Answer: No, according to Gauss’s law flux through a surface depends on the total charge enclosed by the surface, but is independent of the location of the enclosed charges.

Q.72 Two solid spheres, both of the same radii, R carry identical total charges Q. One sphere is a good conductor while the other is an insulator. If the charge on the insulating sphere is uniformly distributed throughout its interior volume, how do the electric fields outside these two spheres compare?

Answer: Electric fields inside the spheres is zero and outside the spheres are identical.

Q.73 A charged rod is brought near a suspended object, which is repelled by the rod. Can we conclude that suspended object is charged?

Answer: No, here this is due to polarization.

Q.74 If the total charge inside a closed surface in known but the distribution of the charge is unspecified, can we apply Gauss’s law to find the electric field?

Answer: No, we cannot apply Gauss’s law.

Q.75 Consider two equal point charges separated by some distance . At what point would a third test charge experience no net force?

Answer: Consider two equal point charges separated by some distance . The point at which third test charge experience no net force is exactly midway between the two charges.

Q.76 Equation E= σ/∊o gives the electric field at points near a charge conducting surface. Apply this equation to a conducting sphere of radius  carrying a charge  on its surface and show that the electric field outside the sphere is same as the electric field of  a point charge at the position of the center of the sphere.

Answer: Given equation is,

    \[E=\frac{\sigma}{\epsilon_\circ}=\frac{q}{\epsilon_\circ A}\]

Area of sphere is 4\pi r^2 , so

    \[E=\frac{q}{\epsilon_\circ4\pi r^2}=\frac{1}{4\pi\epsilon_\circ}\frac{q}{r^2}\]

This is the required result.

Q.77 A spherical Gaussian surface surrounds a point charge q. What happens to electric flux when

(a)- Radius of sphere is tripled.

(b)- Surface is changed to a cone?

Answer: The flux will remain same in both cases because according to Gauss’s law flux neither depends on radius of sphere nor on shape of the surface.

The solution to this question is available in the PDF file, will be attached here soon.

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