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351.

There is a hemispherical shell having charge Q uniformly distributed on its surface. Radius of the shell is R. Find electric potential and field at the centre (of the sphere).

Answer» Correct Answer - `E=(q)/(8pi in_(0)R^(2));(q)/(4pi in_(0)R)`
352.

An infinitely long cylindrical surface density `sigma = sigma_(0) cos varphi`. Where `varphi` is the polar angle of the cylindrical coordinate system whose `z` axis coincides with the axis of the given surface. Find the magnitude and direction of the electric field strength vector on the `z` axis.

Answer» Correct Answer - `[E-1//2" "sigma_(0)//epsi_(0)`, with the direction of the vector E corresponding to the angle `phi=pi`]
353.

When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance(a) increases K times (b) increases K-1 times (c) decreases K times (d) remains constant

Answer»

Correct answer is (c) decreases K times

354.

Dielectric constant of metals is- (a) 1 (b) greater then 1 (c) zero (d) infinite

Answer»

(d) infinite

355.

Describe the working and uses of the Van de Graaff generator with a neat labelled diagram.

Answer»

Working : A high-voltage source, consisting of a high-voltage transformer rectifier circuit, is used to apply a potential difference of several thousand volts (about 50 kV to 100 kV) between spraycomb A and the ground. As a result of the high potential to spraycomb A, a continuous electric discharge takes place between comb A and the nonconducting belt that sprays electric charges onto the belt by corona or point discharge.

The moving belt carries this charge upward and transfers it to the hollow conductor. Collector comb B y inside the hollow conductor removes the charge from the belt by point discharge. Then, the charge flows to the outer surface of the hollow conductor where it accumulates as the process continues. The potential difference between the conductor and the Earth cumulatively increases until the energy density of the electric field builds up to such a high value that the insulating property of the surrounding gas breaks down and a corona discharge takes place through the surrounding gas between the conductor and the ground. If C is the capacitance of – the system and |Q| is the magnitude of the charge transferred from the ground to the conductor, the potential difference V between them is given by V = |Q|/C.

If the hollow conductor is well insulated from the Earth, the accumulated charge |Q| can be large enough and V can build up to several million volts. Since V is limited by the breakdown voltage of the surrounding medium, the entire apparatus is usually enclosed in a pressurized vessel containing a gas such as nitrogen or Freon. This raises the breakdown voltage considerably

Positively charged particles may be accelerated in the evacuated tube from a source at the same potential as the dome conductor toward a target at the ground potential.

Uses : Machines equipped with positive ion sources and arranged for positive ion acceleration are widely used for research in nuclear structure and nuclear reactions, and for production of radioisotopes. Machines equipped with electron sources are used for X-ray therapy, industrial radiography, food and drug sterilization as well as research.

[Note : in a Van de Graaff single-stage accelerator, an ion source is located inside the high-voltage terminal (the dome conductor). Since the electric field inside a charged conductor is zero, the source must be at the same potential as the terminal. The ions are accelerated to the target (at the ground potential) by repulsion. Early quan-titative data on nuclear properties and processes came from Van de Graaff positive ion accelerators, which had a positive ion (proton, deuteron, alpha particle) source and the high voltage terminal raised to a high positive potential. Van de Graaff electron accelerators had also been very successful as sources for X-rays for radiotherapy and industrial radiography. For an electrostatic electron accel-erator, the terminal is charged to a high negative potential and a thermionic cathode replaces the ion source. Accel-erated electrons impinging on watercooled gold targets provided for the first time high energy X-rays (~2 MeV) for cancer therapy. A number of such X-ray generators were developed by Prof. J. G. Trump and R.J. Van de Graaff which were installed in hospitals and industries. These instruments were about 3 ft in diameter and 6 ft long, and could be swung into any position to direct the X-ray beam.]

356.

The ratio of the force between two charges in air and that in a medium of dielectric constant K is (a) K : 1 (b) 1 : K (c) K : 1(d) 1 : K 

Answer»

Correct answer is  (a) K : 1

357.

When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.

Answer»

Charge produced in each object q = 50 nC 

q = 50 x 10-9

Charge of electron (e) = 1.6 x 10-9

Number of electron transferred, n = \(\frac {q}{e}\)\(\frac {50 \times 10^{-9}}{1.6 \times 10^{-19}}\) 

=31. 25 × 10-9 × 1019 

n = 31.25 x 1010 

electrons 

Ans. n = 31.25 x 1010 electrons

358.

What are non-polar molecules? Give examples.

Answer»

A non-polar molecule is one in which centers of positive and negative charges coincide. As a result, it has no permanent dipole moment. 

Examples of non-polar molecules are hydrogen (H2 ), oxygen (O2 ), and carbon dioxide (CO2 ) etc.

359.

What is Polarisation?

Answer»

Polarisation is defined as the total dipole moment per unit volume of the dielectric.

\(\vec P\)= Xe \(\vec P\)ext

360.

What is meant by dielectric?

Answer»

A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics.

361.

Write a short note on ‘electrostatic shielding’.

Answer»

Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be  protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

362.

What is meant by electrostatic energy density?

Answer»

The energy stored per unit volume of space is defined as energy density

uE = \(\frac{U}{Volume}\)

From equation u= 1/2 \(\frac {ε_0A}{d}\)(Ed)2 = 1/2 ε0 (Ad)E2  

or u= 1/2 ε0 E2

363.

What is meant by electrostatic shielding?

Answer»

During lightning accompanied by a thunderstorm, it is always safer to sit inside a bus than in open ground or under a tree. The metal body of the bus provides electrostatic shielding, since the electric field inside is zero. During lightning, the charges flow through the body of the conductor to the ground with no effect on the person inside that bus.

364.

Van de Graaff generator producers large electrostatic potential difference of the order of …………

Answer»

Van de Graaff generator producers large electrostatic potential difference of the order of 107 V

365.

Match the following :(i) Positive and negative(a) Michael faraday(ii) Electrostatic force(b) Gauss(iii) Concept of  field(c) Benjamin franklin(iv) q in arbitrary closed surface(d) Coulomb

Answer»

(i) → (c) 

(ii) → (d)

(iii) → (a) 

(iv) → (b)

366.

Which of the following statement(s) is/are correct?A. If the electric field due to a point charge variesas `r^(-2.5)` instead fo `r^(-2)`, then the Gauiss law will still be valid.B. The Gauss law can be used to calculate the field distributed around on electric dipole.C. If the electric field between two points charges is zero somewhere , then the sign of the two charges is not the same.D. Work done in moving a unit positive charge by the external force from point `A` at potential `V_(A)` to point `B` at potential `V_(B)` is `(V_(B) - V_(A))`

Answer» Correct Answer - C::D
(c ) Electric field is zero on line joining the two charges only, if both the charges are of same nature.
(d) `q_(0) = + 1C`
`V_(B) - V_(A) = (W_(AB))/(q_(0))`, So `V_(B) - V_(A) = W_(AB)`
367.

A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles:A. the stored energy of the capacitor increasesB. charge on capacitor increasesC. voltage of the capacitor increasessD. the capacitance increases

Answer» Correct Answer - A::C
As, `Q` reamains constant after disconnecting the battery , `C = (in_(0) A)/(d)` decreases, as `d` increases, As `Q` is constant `C_(i) V_(i) = C_(f) V_(f)`
`C_(f) lt C_(i)` So `V_(f) gt V_(i)`
so voltage increases
As `U = (1)/(2) QV, U prop V`, So `U_(f) gt U_(i)` as `V_(f) gt V_(i)`
368.

A parallel-plate capacitor is charged from a cell and then disconnected from the cell. The separation between the plates is now double.A. The potential difference between the plates will become doubleB. The field between the plates will not change.C. The energy of the capacitor doubles.D. Some work will have to be done by an external agent on the plates.

Answer» Correct Answer - A::B::C::D
369.

In a parallel-plate capacitor of plate area `A`, plate separation `d` and charge `Q` the force of attraction between the plates is `F`.A. `F propQ^(2)`B. `F prop 1/A`C. `F prop d`D. `F prop 1/d`

Answer» Correct Answer - A::B
370.

In the circuit shown each capacitor has a capcitance `C`. The emf of the celll is `E`. If the switch `S` is closed then A. some change will flow out of the positive terminal of the cellB. some charge will enter the positive terminal of the cellC. the amount of charge flowing through the cell will be `C varepsilon`.D. the amount of charge flowing through the cell will be `4/3 Cvarepsilon`.

Answer» Correct Answer - A::D
371.

In the circuit shown each capacitor has a capcitance `C`. The emf of the celll is `E`. If the switch `S` is closed then A. positive charge will flow out of the positive terminal of the cellB. positive charge will enter the positive terminal of the cellC. the amount of charge flowing through the cell will be CED. the amount of charge flowing through the cell will be `(4)/(3)CE`

Answer» Correct Answer - A::D
372.

The graph shows the potential difference ∆V between the plates of a capacitor versus the charge Q on its plates. Use the graph to find (1) the capacitance of the capacitor (2) the magnitude of the excess charge on its plates.

Answer»

C = = 0.05 nF = 50 pF

From the graph, for Q = 0.5 nC, ∆V = 10 V.

373.

Fig shows a parallel plate capacitor of plate area `A` and plate separation `d`. Its entire space is filled with three different dielectric slabs of same thickness. Find the equivalent capacitance of the arrangment.

Answer» Correct Answer - `C = (3 in_(0) A K_(1) K_(2) K_(3))/(d (K_(1) K_(2) + K_(2) K_(3) + K_(3) K_(1)))`
This arrangement in Fig, is equivalent to three capacitors connected in series
`C_(1) = (K_(1) in_(0) A)/(d//3) = (3 K_(1) in_(0) A)/(d) , C_(2) = (3 K_(2) A)/(d)`,
`C_(3) = (3 K_(3) in_(0) A)/(d)`
`(1)/(C ) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (d)/(3 in_(0) A) ((1)/(K_(1)) + (1)/(K_(2)) + (1)/(K_(3)))`
`C = (3 in_(0) A K_(1) K_(2) K_(3))/(d (K_(1) K_(2) + K_(2) K_(3) + K_(3) K_(1)))`
374.

An uncharged sphere of metal is placed in between two charged plates as shown. The lines of force look like

Answer» Correct Answer - B
Field terminate and originate in the surface pf metla perpendicular ot the surface so option `B` is is correct .
375.

An uncharged sphere of metal is placed in between two charged plates as shown. The lines of force look like A. AB. BC. CD. D

Answer» Correct Answer - C
376.

Two insulated identically sized charged copper spheres A and B have their centers separated by a distance of 50 cm. Charges on each sphere is `q=6.5xx10^(-7)C`. A third sphere of the same size but uncharged is brought in contact with the first, then in contact with the second and finally removed from both. What is the new force of repulsion between A and B?

Answer» Here, Charge on `A = 6.5xx10^(-7) C` , charge on `B = 6.5xx10^(-7) C`
Their sizes are equal . When third sphere C of same size is brought in contact with B carrying `6.5xx10^(-7)` coulomb charge. As their sizes are equal, therefore , charge on each of the spheres B and C becomes
`q_(2) = ((6.5+3.25) xx10^(-7))/(2) C = 4.875xx10^(-7) C`
As `F = (1)/(4pi in_(0)) (q_(1) q_(2))/(r^(2)) , r = 0.5m :. F = (9xx10^(9)xx3.25xx10^(-7)xx4.875xx10^(-7))/((0.5)^(2)) = 5.7xx10^(-3) N`
377.

Two positive point charges of 12 and 5 microcoulombs, are placed 10 cm apart in air. The work needed to bring them 4 cm closer isA. 2.4 JB. 3.6 JC. 1.6 JD. 6.0 J

Answer» Correct Answer - B
Potential energy of charges `q_(1) and q_(2)`, r distance aprat
`U=(1)/(4pi epsilon_(0))(q_(1)q_(2))/(r)`
For r = 0.1 m
`U_(1)=(1)/(4pi epsilon_(0))(12xx10^(-6)xx5xx10^(-6))/(0.1)`
`=(9xx10^(9)xx60xx10^(-12))/(0.1)=5.4J`
For r = 0.06 m,
`U_(2)=(9xx10^(9)xx60xx10^(-12))/(0.06)=9J`
`therefore" Work done"=(9-5.4)J=3.6J`
378.

Figure, shown above, shows three situations involving a charged particle and a uniformly charged spherical shell. The charges and radii of the shells are indicated in the figure. If `F_(1),F_(2)` and `F_(3)` are the magnitudes of the force on the particle due to the shell in situations (I),(II) and (III) then A. `F_(3) gt F_(2) gt F_(1)`B. `F_(2) gt F_(2)=F_(3)`C. `F_(3)=F_(2) gt F_(1)`D. `F_(1) gt F_(2) gt F_(3)`

Answer» Correct Answer - C
`F_(1)` is zero as charge lies inside the shell (Electric field inside the shell due to its own charge is zero). `F_(2)=F_(3)` as charge over the surface of the shell behaves like centerad at the centre.
379.

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch `S_1` is pressed first to fully charge the capacitor `C_1` and then released. The switch `S_2` is then pressed to charge the capacitor `C_2`. After some time, `S_2` is released and then `S_3` is pressed. After some time A. (a) The charge on the upper plate of `C_1` is `2CV_0`B. (b) The charge on the upper plate of `C_1` is `CV_0`C. (c) The charge on the upper plate of `C_2` is 0D. (d) The charge on the upper plate of `C_2` is `-CV_0`

Answer» Correct Answer - B::D
Step1: When `S_1` is pressed. The capacitor `C_1` gets charged such that its upper plate acquires a positive charge `+2CV_0` and lower plate `-2CV_0`.
Step 2: When `S_2` is pressed (`S_1` open). As `C_1=C_2` the charge gets distributed equal. The upper plates of `C_1` and `C_2` now take charge `+CV_0` each and lower plate `-CV_0` each.
(b) and (d) are correct option.
380.

Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities `rho_1` and `rho_2` respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio `rho_1/rho_2` can beA. `-4`B. `-32/25`C. `32/25`D. `4`

Answer» Correct Answer - B::D
381.

Two charged particle of equal mass are constrained to move along X and Y direction. The X – Y plane is horizontal and the tracks are smooth. The particles are released from rest when they were at positions shown in the figure. At the instant distance of q becomes 2r from the origin, find the location of charge Q.

Answer» Correct Answer - Q is at a distance of 4r from O.
382.

A polythene piece rubbed with wool is found to have negative charge of `3 xx 10^-7 C`. Estimate the number of electrons transferred.A. `1.8xx10^(15)`B. `1.8xx10^(12)`C. `1.2xx10^(11)`D. `1.2xx10^(10)`

Answer» Correct Answer - B
Here, total charge transferrred, `q=-3xx10^(-7)C`
Charge on an electron, `e=-1.6xx10^(-19)C`
From quantisation of charge, q = ne
Therefore, number of electrons transferred,
`n=(q)/(e)=(-3xx10^(7))/(-1.6xx10^(-19))`
Number of electrons, `n=1.8xx10^(12)`
383.

An electric dipole of diople moment `20xx10^(-6)C m` is enclosed by a closed surface. What is the net electric flux coming out of the surface ?

Answer» Net electric flux coming out of closed surface is zero, because net charge on electric dipole is zero.
384.

In the electric field shown in figure, the electric field lines on the left have twice the separation as that between those on the right. If the magnitudes of the fields at point `A` is `40NC^(-1)`, calculate the force experienced by a proton placed at point `A`. Also find the magnitude of electric field at point `B`

Answer» Correct Answer - (a) `6.4xx10^(-18)N (b) 20 N//C`
(a) `F = qE = 1.6xx10^(-19) xx40 = 6.4xx10^(-18) N`
(b) As separation of electric field lines at B is twice the separation at A, Fig
`:. E_(B) = (1)/(2) xx40 N//C = 20N//C`
385.

An electric dipole of length 4 cm, when placed with its axis making an angle of `60^(@)` with a uniform electric field, experiences a torque of `4 sqrt(3) N m`. Calculate the a. magnitude of the electric field, and b. potential energy of the dipole, if the dipole has charges of `+- 8 n C`.

Answer» Correct Answer - `-4 J`
Here, `2a = 4cm = 4xx10^(-2) m, theta = 60^(@)`.
`tau = 4 sqrt(3) Nm, q = +- 8nC = 8xx10^(-9)C`
Now, `p = q xx 2a = (8xx10^(-9)) xx (4xx10^(-2)) Cm`
`tau = pE sin theta or E = tau//p sin theta`
`PE = -pE cos theta = -p xx (tau)/(p sin theta) xx cos theta`
`= -tau cot theta = -4 sqrt(3)" " 60^(@)`
`= -4 sqrt(3) xx 1//sqrt(3) = -4 J`
386.

An electric dipole consists of charges `+ 2e and -2e` separated by 0.78mm. It is an electric field of strength `3.4xx10^(6) N//C`. Calculate the magnitude of the torque on the dipole when the dipole moment is (a) parallel to (b) perpendicular to, and (c) antiparallel to the electric field.

Answer» Correct Answer - (a) Zero, (b) `8.5xx10^(-22) N m` (c ) Zero
(a) `tau = pE sin theta = pE sin 0^(@) = 0`
(b) `tau = pE sin theta`
`= (2e) (0.78xx10^(-9)) 3.4xx10^(6) xx sin 90^(@)`
`= 2xx1.6xx10^(-19) xx0.78xx3.4xx10^(-3)`
`= 8.5xx10^(-22) Nm`
387.

An electric dipole is placed at an angle of `30^(@)` with an electric field intensity `2xx10^(5)N//C`. It experiences a torque equal to `4Nm`. The charge on the dipole, if the dipole is length is `2 cm`, isA. 8 mCB. 2 mCC. 5 mCD. `7 muC`

Answer» Correct Answer - B
388.

An electric dipole is placed at an angle of `30^(@)` with an electric field intensity `2xx10^(5)N//C`. It experiences a torque equal to `4Nm`. The charge on the dipole, if the dipole is length is `2 cm`, isA. 8mCB. 2mCC. 5mCD. `7muC`

Answer» Correct Answer - B
389.

A neutral water molecule `(H_(2)O)` in its vapour state has an electric dipole moment of magnitudes `6.4xx10^(-30)C-m`. How far apart are the molecules centres of positive and negative chargeA. `4 xx 10^(-10)`B. `4 xx 10^(-11)`C. `4 xx 10^(-12)`D. `4 xx 10^(-13)`

Answer» Correct Answer - B
390.

Three concentric conducting spherical shells carry charges as : `+4Q` on the inner shell, `-2Q` on the middle shell and `-5Q` on the outer shell. The charge on the inner surface of the outer shell is

Answer» Correct Answer - D
391.

Two thin conducting shells of radii `R` and `3R` are as shown in the fig. External shell carries a charge `+Q` and inner shell neutral. The inner shell is earthed with the help of a switch `S`. Then A. When the switch S is open, the potential of the inner sphere is equal to that of the outerB. When the switch S is closed, the potential of the inner sphere becomes zeroC. When the switch S is closed, the charge attained by the inner sphere is `-Q//3`D. By closing the switch the capacitance of the system increases.

Answer» Correct Answer - A::B::C::D
392.

Two concentric conducting thin shells of radius R and 2 R carry charges +Q and +3Q respectively. The magnitude of electric field at a distance x outside and inside from the surface of outer sphere is same. Then the value of x is

Answer» Correct Answer - [20]
393.

Fig shows three point charges `+2q, -q and + 3q`, What is the electric flux due to this configuration thorugh the surface S ?

Answer» `phi_(E) = (2q -q)/(in_(0)) = (q)/(in_(0))`
394.

Point charges `– q, 2q, – 3q, q, – q, 2q, – 3q, q, – q, 2q, – 3q`, and q have been placed at marks `1, 2, 3, 4, 5 .....12` respectively on the circular dial of a clock. Find the electric field intensity at the centre of the dial if distance of each charge from the centre is r.

Answer» Correct Answer - zero.
395.

Six charge of magnitude `+q` and `-q` are fixed at the corners of a regular hexagon of edge length a as shown in the figure .The electrostatic interaction energy of the charged particle is : A. `q^(2)/(pi in_(0) a) [sqrt(3)/8-15/4]`B. `q^(2)/(pi in_(0)a ) [sqrt(3)/2-9/4]`C. `q^(2)/(pi in_(0) a) [sqrt(3)/4-15/2]`D. `q^(2)/(pi in_(0) a) [sqrt(3)/2-15/8]`

Answer» Correct Answer - D
396.

(a) Six equal charges have been placed at the vertices of a regular hexagon. Charge at vertex A is moved to the centre of the hexagon and there it experiences a net electrostatic force of magnitude F. Charge at E is also moved to the centre so as to double the magnitude of the charge at the centre. Calculate the magnitude of the electrostatic force that this central charge experiences now.(b) Three charges of equal magnitude lie on the vertices of an equilateral triangle ABC. All of them are released simultaneously. The charge at A experiences initial acceleration along AD where D is the midpoint of the side BC. Find the direction of initial acceleration of the charge at B.

Answer» Correct Answer - (a). 2F towards point F
(b). In a direction making an angle of `60^(@)` with BA (NOT along BC).
397.

A uniform semicircular ring of radius R is in yz plane with its centre at the origin. The half ring carries a uniform linear charge density of `lamda`. (a) Find the x, y and z component of Electric field at a point P(x, 0, 0) on the axis of the ring. (b) Prove that the field at P is directed along a line joining the centre of mass of the half ring to the point P.

Answer» Correct Answer - (a) `E_(x)=(KlamdaRpi*x)/((R^(2)+x^(2))^(3//2))`
(b). `E_y=0`
(c). `E_(z)=-(2KlamdaR^(2))/((R^(2)+x^(2))^(3//2))`
398.

An infinitely long line charge is bent in U shape as shown in figure. The semicircular part has radius R and linear charge density is `lamda C//m`. Using the results obtained in last two problems, calculate the electric field intensity at centre of the circle (point O)

Answer» Correct Answer - zero.
399.

Two fixed, equal, positive charges, each of magnitude `5xx10^-5` coul are located at points A and B separated by a distance of 6m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C.

Answer» Correct Answer - B
From energy conservation
`U_C+K_C=U_D+K_D`
`or 2[(9xx10^9xx(5xx10^-5)(-5xx10^-5))/5]+4`
`=2 [((9xx10^9x(5xx10^-5)(-5x10^-5))/(AD)]+0`
solving we get `AD =9m`
`:.` maximum distance
`OD=sqrt((9)^2-(3)^2`
`=sqrt72m`
400.

Six point charges are arrange at the vertices of a regular hexagon of side length a (shown in figure). Find the magnitude of electric field at the centre of regular hexagon.

Answer» Correct Answer - Zero (By symmetry)