1638 + Interview Questions in ELECTROSTATICS IN GENERAL AWARENESS Page 13 InterviewSolution

601.	Two metallic spheres of radii `1 cm` and `2 cm` are given charges `10^(-2) C` and `5 xx 10^(-2) C` respectively. If they are connected by a conducting wire, the final charge on the smaller sphere isA. `2xx10^(-2) C`B. `3xx10^(-2) C`C. `4xx10^(-2) C`D. `1xx10^(-2) C`
Answer» Correct Answer - B Here, `r_(1) = 1 cm, r_(2) = 3 cm`, `q_(1) = -10^(-2) C, q_(2) = 5xx10^(-2) C` When the two spheres are connected by a conducting wire, charge flows second sphere to first sphere till their potentials become equal.If bigger sphere loses charge `x xx 10^(-2)`coulomb, then as `V_(1) = V_(2)` `:. (k (x -1) 10^(-2))/(1 xx 10^(-2)) = (k (5-x) 10^(-2))/(3xx10^(-2))` `3x - 3 = 5-x, 4x =8, x = 2` `:.` Charge left on bigger sphere `= (5-2) 10^(-2) C` `= 3xx10^(-2) C`

Discussion

602.	A small electric dipole having dipole moment ` ` vec p` is placed along x-axis as shown in the figure . A semi-infinite uniformly charged di-electric thin rod is placed along `x` axis , with one end coinciding with origin . If linear charge density of rod is `+ lambda` and distance of dipole from rod is `a` then calculate the electric force acting on dipole .
Answer» Correct Answer - ` (- P lambda)/( 4 pi varepsilon_0 a^2)`. Field at `P` due to elemental length is `dE = (K lambdadx)/(x^2)` `E = int dE = K lambda int _a^(infty) = (K lambda)/a = 1/(4 pi in_0) (lqambda)/a` Potential energy of dipole `U =- Pe cos theta` `U = - ( p lambda) /( 4 pi in_0 a)` `F= - (dU)/(da) rArr F= ( lambda )/(4pi in_0 a^2)`.

Discussion

603.	If linear charge density of a wire as shown in the figure is `lambda` A. potential at the centre is `(lambda)/(2 epsi_(0))`B. electric field at the centre of the loop is `(lambda)/(2 pi epsi_(0)R)`C. electric field at the centre of the loop is `(lambda)/(2 pi epsi_(0)R) + (lambda)/(2 epsi_(0)R)`D. None of the above
Answer» Correct Answer - B

Discussion

604.	A drop of `10^(-6)` kg water carries `10^(-6) C` charge. What electric field should be applied to balance its weight (assume `g = 10 ms^(-2)`)A. 10 V/m upwardB. 10 V/m downwardC. 0.1 V/m downwardD. 0.1 V/m upward
Answer» Correct Answer - A

Discussion

605.	Two capacitances of capacity `C_(1)`and `C_(2)` are connected in series and potential difference `V` is applied across it. Then the potential difference across `C_(1)` will beA. `VC_(2)/C_(1)`B. `(V(C_(1)+C_(2)))/C_(1)`C. `(VC_(2))/((C_(1)+C_(2)))`D. `(VC_(1))/((C_(1)+C_(2)))`
Answer» Correct Answer - C `V_(1)=Q/C_(1)=(C_(s)V)/C_(1)=((C_(1)C_(2))/(C_(1)+C_(2)))V/C_(1)=(VC_(2))/(C_(1)+C_(2))`

Discussion

606.	Let us assume that charges on Earth and Sum are not neutralised and net charges are of equal magnitude and similar nature. What must be the charge on each so that coulomb force just cancels gravitational force ? This charge corresponds to how many free electrons ? Mass of sun `= 2xx10^(30) kg` Mass of earth `= 6xx10^(24) kg`
Answer» Correct Answer - `29.8xx10^(16)C ; 1.86xx10^(36)` Here, electrostatic force of repulsion must be equal to gravitational force. `F_(e) = F_(g)` `(9xx10^(9)xx Q^(2))/(r^(2)) = (GMm)/(r^(2))` `9xx10^(9)xx Q^(2) = 6.67xx10^(-11)xx2xx10^(30)xx6xx10^(24)` `Q = 29.8xx10^(16)C` As `Q =n e` `n = 1.86xx10^(36)`

Discussion

607.	Charges `q_(1) = 1.5 mC, q_(2) = 0.2 mC and q_(3) = -0.5 mC`, are placed at points A,B,C respectively as shown in Fig. If `r_(1) = 1.2m and r_(2) = 0.6m`, calculatae magnitude of resultant force on `q_(2)`.
Answer» Correct Answer - `3.1xx10^(3)N` Refer to Fig. `F_(1) = (q_(1) q_(2))/(4pi in_(0) r_(1)^(2))` `= (1.5xx10^(-3)xx9xx10^(9))/((1.2)^(2))` `= 1.875xx10^(3) N` …. along AB produced `F_(2) = (q_(2) q_(3))/(4pi in_(0) r_(2)^(2)) = (0.2xx10^(-3)xx9xx10^(9))/((0.6)^(2))` `= 2.5xx10^(3) N` along `BC _\|_ AB`. Resulant force on `q_(2) = sqrt(F_(1)^(2) + F_(2)^(2))` `= 3.1xx10^(3)N`

Discussion

608.	Equal charges each of `20 muC` are placed at `x = 0, 2,4,8,16 cm` on X-axis. Find the force experienced by the charge at `x = 2cm`.
Answer» Correct Answer - `1.2xx10^(3) N` Force on charge at `x = 2cm` due to charge at `x = 0 cm and x = 4cm` are equal and opposite. They cancel. Net force on charge at `x = 2cm` is resultant of repulsive forces due to two charges at `x = 8 cm and x = 16cm`. `:. F = (q xx q)/(4pi in_(0)) xx [(1)/((0.08 - 0.02)^(2)) + (1)/((0.16 - 0.02)^(2))]` `F = 9xx10^(9) (20xx10^(-6))^(2) [(1)/((0.06)^(2)) + (1)/((0.14)^(2))]` `F = 1.2xx10^(3) N`

Discussion

609.	A charges Q is placed at each of the two opposite corners of a square. A charge q is placed to each of the other two corners. If the resultant force on each charge q is zero, thenA. `Q = sqrt2 q`B. `Q = -sqrt2 q`C. `Q = 2 sqrt2q`D. `Q = - 2 sqrt2 q`
Answer» Correct Answer - D

Discussion

610.	Which of the following is correct regarding electric lines of force ? (i) Electric lines of force diverge from `+ve` and converge at a `-ve` charge (ii) Electric lines of force never cross each other , otherwise there will be two directions of electric field at a point of intersection which is not possible (iii) Closer are the electric lines of forces , stronger is the field , and further apart lines of force ,weaker is electric field (iv) The electric lines of forces are parallel for a uniform electric fieldA. `(i),(ii)`B. `(ii),(iii)`C. `(ii),(iv)`D. all
Answer» Correct Answer - 4

Discussion

611.	A charged ball with mass m and charge q is dropped from a height h over a non-conducting smooth horizontal plane. There exists a uniform electric field `E_0` in vertically downward direction and the coefficient of restitution between the ball and the plane is e. Find the maximum height attained by the ball after `n^(th)` collision.
Answer» Correct Answer - `he^(2n)`

Discussion

612.	In the last question, the electric field in vertical direction is switched off and a field of same strength`(E_0)` is switched on in horizontal direction. Find the horizontal velocity of the ball during the `n^(th)` collision. Also calculate the time interval between `n^(th)` and `(n + 1)^(th)` collision
Answer» Correct Answer - `V_(xn)=(qE)/(m)sqrt((2h)/(g))[1+2e((1-e^(n-1))/(1-e))}` and `T=2e^(n)sqrt((2h)/(g))`

Discussion

613.	An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will(A) be attracted to the +ve plate (B) be attracted to the -ve plate (C) remain stationary (D) will move parallel to the plates
Answer» Correct answer is (A) be attracted to the +ve plate

Discussion

614.	A charge of + 7 µC is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be(A) 1 : 4 (B) 1 : 2 (C) 1 : 1 (D) 1 : 16
Answer» Correct answer is (C) 1 : 1

Discussion

615.	A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.To create the spark, an electric field of magnitude 3 x 106 Vm-1 is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm
Answer» Separation gap between two electrodes, d = 0.6 mm d = 0.6 × 10^-3 m Magnetude of electric field Electric field = E = 3 × 10⁶ V m^-1 Electric field E =V/d (a) Applied potential difference, V = E . d = 3 × 10⁶ × 0.6 10^-13 = 1.8 × 10³ V = 1800 V (b) From equation, V = E . d If the gap (distance) between the electrodes increased, the potential difference also increases. (c) Gap between the electrodes, d = 1mm = 1 x 10^-3 m Potential difference, V = E.d = 3 × 10^-6 × 1 × 10^-3 = 3 × 103 V = 3000 V

615.

A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.To create the spark, an electric field of magnitude 3 x 106 Vm-1 is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm

Answer»

Separation gap between two electrodes,

d = 0.6 mm

d = 0.6 × 10^-3 m

Magnetude of electric field Electric field = E = 3 × 10⁶ V m^-1

Electric field E =V/d

(a) Applied potential difference, V = E . d

= 3 × 10⁶ × 0.6 10^-13 = 1.8 × 10³

V = 1800 V

(b) From equation, V = E . d

If the gap (distance) between the electrodes increased, the potential difference also increases.

(c) Gap between the electrodes, d = 1mm = 1 x 10^-3 m

Potential difference, V = E.d

= 3 × 10^-6 × 1 × 10^-3 = 3 × 103 V

= 3000 V

Discussion

616.	In fig. `C_(1) = 20 muF, C_(2) = 30 muF and C_(3) = 15 muF` and the insulated plate of `C_(1)` is at a potential of 90 V, one plate of `C_(3)` being earthed. What is the potential difference between th plates of `C_(2)` three capacitors being connected in series ?A. 10 e.s.u.B. 30 e.s.u.C. 40 e.s.u.D. 20 e.s.u.
Answer» Correct Answer - D `C_(1)=20esu, C_(2)=30esu,C_(3)=15esu,` `V=90esu, V_(2)=?` `1/C_(s)=1/C_(1)+1/C_(2)+1/C_(3)` `=1/20+1/30+1/15=(3+2+4)/60` `C_(s)=60/9esu` Now `V_(2)=Q/C_(2)=(C_(s)*V)/C_(2)=60/9xx90/30` `V_(2)=20esu`

Discussion

617.	What is T.N.E.I. through the surface A and B respectively ? A. (q, 2q)B. (-q, -2q)C. (0, q)D. (q, 0)
Answer» Correct Answer - C T.N.E.I. over the surface A = + q - q = 0 T.N.E.I. over the surface B = 2q - q = q

Discussion

618.	The figure shows some of the electric field lines corresponding to an electric field. The figure suggests A. `E_(A) gt E_(B) gt E_(C )`B. `E_(A) = E_(B) = E_(C) `C. `E_(A) = E_(C) gt E_(B)`D. `E_(A) = E_(C) lt E_(B)`
Answer» Correct Answer - 3

Discussion

619.	Two non-conduction hollow uniformly charged spheres of radil `R_(1)` and `R_(2)` with charge `Q_(1)` and `Q_(2)` respectively are places at a distance r. Find out total energy of the system.
Answer» `U_("total")=U_("self")+U_("Interaction")=Q_(1)^(2)/(8piepsi_(0) R_(1))+Q_(2)^(2)/(8pi epsi_(0) R_(2))+(Q_(1)Q_(2))/(4pi epsi_(0) r)`

Discussion

620.	A sperical shell of radius R with uniform charge q is expanded to a radius 2R. Find the work perfomed by the electric forces and external agent against electric forces in this process.
Answer» `W_("ext")=U_(f)-U_(i)=q^(2)/(16 pi epsi_(0)R)-q^(2)/(8 pi epsi_(0)R)=-q^(2)/(16 pi epsi_(0) R)` `W_("elec")=U_(i)-U_(i)=q^(2)/(8 pi epsi_(0)R)-q^(2)/(16 pi epsi_(0)R)=q^(2)/(16 pi epsi_(0) R)`

Discussion

621.	In the above problem , the value of `Q` for which charges will remain stationary isA. `q sqrt(3)`B. `(q)/(sqrt(3))`C. `-(q)/(sqrt(3))`D. `-q sqrt(3)`
Answer» Correct Answer - 3 If net force on each `q` is zero , for this `sqrt(3) vec(F) + vec(F)_(3) = 0` `sqrt(3) .(1)/(4 pi in_(0)) (q^(2))/(L^(2)) + (1)/(4 pi in_(0)) (Qq)/((L//sqrt(3))^(2)) = 0` `sqrt(3) q + 3 Q = 0` `Q = -(q)/(sqrt(3))`

Discussion

622.	Three equal charges q are placed at the corners of an equilateral triangle of side A. (i) Find out potential energy of charge system. (ii) Calculate work required to decrease the side of triangle to `a//2`. (iii) If the charges are released from the shown position and each of them has same mass m, then find the speed of each particle when they lie on triangle of side `2 a`. .
Answer» (i) Method I (Derivation) Assume all the charges are at infinity. Work done in putting charge q at corner A `implies W_(1)=q(v_(f)-v_(i))=q(0-0)` Since potential at A is zero in absence of charges, work done in putting q at corner B if presence of charge at A : `implies W_(2)=((Kq)/a-0) q=(Kq^(2))/a` similarly work done in putting charge q at corner C in presence of charge at A and B. `implies W_(3)=q(v_(f)-v_(i))=q [((Kq)/a+(Kq)/a)-0]=(2Kq^(2))/a` So net potential energy `PE=W_(1)+W_(2)+W_(3)=0+(Kq^(2))/a+(2Kq^(2))/a=(3Kq^(2))/a` Method II (using direct formula) : `U=U_(12)+U_(13)+U_(23)=(Kq^(2))/a+(Kq^(2))/a+(Kq^(2))/a=(3Kq^(2))/a` (ii) Work required to decrease the sides `W=U_(f)-U_(i)=(3Kq^(2))/(a//2)-(3Kq^(2))/a=(3Kq^(2))/a` Joules (iii) Work done by electrostatic forces - Change is kinetic energy of particles. `U_(i)-U_(f)=K_(f)-K_(i)" "implies (3Kq^(2))/a-(3Kq^(2))/(2a)=3 (1/2 mv^(2))-0 implies v= sqrt((Kq^(2))/(am))`

Discussion

623.	Six equal point charges q each are placed at six corners of a hexago of side a. Find out potential energy of charge system
Answer» `U_("Net")-(U_(1)+U_(2)+U_(3)+U_(4)+U_(5)+U_(6))/(2)` Due to symmetry `U_(1)=U_(2)=U_(3)=U_(4)=U_(5)=U_(6)" "`So `U_("net")=3U_(1)=(3Kq^(2))/a[2+2/sqrt(3)+1/2]`

Discussion

624.	A charge Q is place at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then `Q//q` equals:A. `-1`B. `1`C. `-1/sqrt(2)`D. `-2sqrt(2)`
Answer» Correct Answer - D

Discussion

625.	Four identical point charges q are placed at four corners of a square of side a. Find the potential energy of the charge system. .
Answer» Method 1 (using direct formula) : `U=U_(12)+U_(13)+U_(14)+U_(23)+U_(24)+U_(34)` `(Kq^(2))/a+(Kq^(2))/(asqrt(2))+(Kq^(2))/a+(Kq^(2))/a+(Kq^(2))/(asqrt(2))+(Kq^(2))/a=[(4Kq^(2))/a+(2Kq^(2))/(asqrt(2))]=(2Kq^(2))/a[2+1/sqrt(2)]` Method 2 `["Using, "U=1/2(U_(1)+U_(2)+....)]` : `U_(1)=` total P.E. of charge at corner 1 due to all other charges. `U_(2)=` total P.E. of charge at corner 2 due to all other charges. `U_(3)=`total P.E. of charge at corner 3 due to all other charges. `U_(4)=` total P.E. of charge at corner 4 due to all other charges. since, due to symmetery, `U_(1)=U_(2)=U_(3)=U_(4)` `U_("Net")=(U_(1)+U_(2)+U_(3)+U_(4))/(2)=2U_(1)=2[(Kq^(2))/a+(Kq^(2))/a+(Kq^(2))/(sqrt(2)a)]=(2Kq^(2))/a[2+1/sqrt(2)]`

Discussion

626.	In the previous question , if the conductor has a charge per unit length `lambda` , the particle has mass `m` and charge `q` then (choose the correct option)A. `v prop sqrt(q)`B. `v prop sqrt(lambda)`C. `v prop sqrt(m)`D. `v prop (1)/(sqrt(m))`
Answer» Correct Answer - 3 `v prop sqrt(q) , v prop sqrt(lambda) , v prop (1)/(sqrt(m))`

Discussion

627.	In the previous question , the equilibrium isA. stableB. unstableC. neutralD. none
Answer» Correct Answer - 3 Since in `F_(e ) = F_(g) ,r` cancels out , charges are in equilibrium at every separation ,hence equilibrium is neutral.

Discussion

628.	In the previous question , if `e = ` electronic charge , the minimum magnitude of `q` isA. `e`B. `2e`C. `4 e`D. none of these
Answer» Correct Answer - 3 `Q = -(q)/(4)` `\|q\| = 4e` The charge on any body is `+- "ne"` , where `n`: whole number.

Discussion

629.	Two identical charged spheres suspended from a common point by two mass-less strings of length `l` are initially at a distance d ( `d ltlt l`) apart because of their mutual repulsion . The charge begins to leak from both the spheres at a constant rate. As a result the charge approach each other with a velocity `v`. Then as a function of distance `x` between them .A. `x^(-1)`B. `x^(1//2)`C. `x^(-1//2)`D. `x`
Answer» Correct Answer - C

Discussion

630.	A certain charge Q is divided into two parts q and `Q-q`, wheich are then separated by a cetain distance. What must q be in terms of Q to maximum the electrostatic repulsion between the two charges?
Answer» Correct Answer - B `F=(kq(Q-q))/r^2` (where `k=1/(4piepsilon_0))` For F to be maximum `(dF)/(dq)=0` By putting `(dF)/(dq)=0` we get `q=Q/2`

Discussion

631.	A block of mass `m` is attached to a spring of force constant k .Charges on the block is `q`. A horizontal electric field E is acting in the directions as shown.Block is released with the spring in unstretched position (a) block will execute `SHM` (b) time period of osciallation is `2pisqrt(m)/(k)` (c) amplitude of oscillation is `(qE)/(k)` (d) Block will oscillate but not simple harmonically choose the correct answerA. Block will execute SHMB. Time period of oscillation is `2pi sqrt((m)/(k))`C. Amplitude of oscillation is `(qE)/(k)`D. Block will oscillate but not simple harmonically.
Answer» Correct Answer - A::B::C

Discussion

632.	An electric dipole of dipole moment `10^(-6)` C-m is released from rest in uniform electric field `10^(2)V//m` at angle `theta = 60^(@)`. Maximum rotational kinetic energy of the dipole is say K and maximum torque during the motion is `tau`, thenA. `K = 5.0 xx 10^(-5)J`B. `K=2.0 xx 10^(-4) J`C. `tau = 5.0 xx 10^(-4)` N-mD. `tau = 8.7 xx 10^(-5)` N-m
Answer» Correct Answer - A::D

Discussion

633.	Two concentric shells of radii `R` and `2R` have given charge `q` and `-2q` as shown in figure in a region `r lt R` (a) `E=0`,(b) `E!=0`,(c ) `V=0` (d) `V!=0`A. `E=0`B. `E ne 0`C. `V=0`D. `V ne 0`
Answer» Correct Answer - A::C

Discussion

634.	In the following diagram the work done in moving a point charge from point `P` to point `A, B` and `C` is respectively as `W_(A), W_(B) and W_(C)`, then A. `W_(A) lt W_(B) lt W_(C)`B. `W_(A) gt W_(B) gt W_(C)`C. `W_(A) = W_(B) = W_(C)`D. None of these
Answer» Correct Answer - 3 `V_(A) = V_(B) = V_(C )` `W_(A) = q(V_(A) - V_(P)) , W_(B) = q(V_(B) - V_(P)) , W_(C ) = q(V_(C ) - V_(P))` `W_(A) = W_(B) = W_(C )`

Discussion

635.	The work done in carrying a charge q once round a circle of radius r with a charge Q at the centre isA. `(qQ)/(4pi epsilon_(0)a)`B. `(qQ)/(4pi epsilon_(0)a^(2))`C. `(q)/(4pi epsilon_(0)a)`D. zero
Answer» Correct Answer - D The work done is zero due to equipotential surfaces.

Discussion

636.	A particle of mass `m` and carrying charge `-q_(1)` is moving around a charge `+q_(2)` along a circular path of radius `r` period of revolution of the charge `-q_(1)` about `+q_(2)` is
Answer» Here, force of attraction between charges = centripetal force `(1)/(4pi in_(0)) (q_(1) q_(2))/(r^(2)) = (mv^(2))/(r)` So `v = sqrt((1)/(4pi in_(0)) (q_(1) q_(2))/(mr)` Time period of revolution `T = (2pi r)/(v) = (2pi r) sqrt((4pi in_(0) mr)/(q_(1) q_(2)))` `T = sqrt((16pi^(3) in_(0) mr^(3))/(q_(1) q_(2)))`

Discussion

637.	An electric field vector E = 10 x i exists in a certain region of space. Then the potential difference V = V0 – VA , Where V0 is the potential at the origin and V is the potential at x = 2 m is- (a) 10 J (b) -20 J (c) + 20 J (d) – 10 J
Answer» Correct answer is (a) 10 J

Discussion

638.	A rectangular tank of mass `m_(0)` and charge Q is placed over a smooth horizontal floor. A horizontal electric field£ exist in the region. Rain drops are falling vertically in the tank at the constant rate of n drops per second. Mass of each drop is m. Find velocityoftank as function of time.
Answer» Two forces will act on the tank a. Electrostatic force b. Thrust force. let v bet the velocity at any instat. Then, `F_("net")=QE-mnv` `or `(m_0+mnt)(dv)/(dt)=QE=mnv` or `int_0^v (dv)/(QE-mnv)=int_0^t (dt)/(m_0+mnt)` or `"ln" ((QE)/(QE-mnv))="In" ((m_0+mnt)/m_0)` or `(QE)/(QE-mnv)=(m_0+mnt)/m_0` or` v=QE(t/(m_0+mnt))`

Discussion

639.	Two small balls having the same mass and charge and located on the same vertical at heights `h_(1)` and `h_(2)` are thrown in the same direction along the horizontal at the same velocity v. The first ball touches the ground at a horizontal distance R from the initial vertical position. At what height `h_(2)` will the second ball be at this instant? Neglect any frictional resistance of air and the effect of any induced charge on the ground.
Answer» Correct Answer - `[H_(2)=h_(1)+h_(2)-g((1)/(v))^(2)]`

Discussion

640.	Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order-(a) D < C < B < A (b) A < B = C < D (c) C < A = B < D (d)D > C > B > A
Answer» Correct answer is (a) D < C < B < A

Discussion

641.	Intensity of an electric field (E) depends on distance r due to a dipole, is related asA. `Eprop (1)/(r)`B. `E prop (1)/(r^(2))`C. `E prop (1)/(r^(3))`D. `E prop (1)/(r^(4))`
Answer» Correct Answer - C

Discussion

642.	A simple pendulum of length `l` has a bob of mass `m` , with a charge `q` on it. A vertical sheet of charge , with the vertical . Its time period of oscillation is `T` in this position (i) `tan theta = (sigma q)/(2 epsilon_(0) mg)` (ii) `tan theta = (sigma q)/(epsilon_(0) mg)` (iii) `T lt 2 pi sqrt((l)/(g))` (iv) `T gt 2pi ((l)/(g))`
Answer» Correct Answer - (a). `sqrt(4gL)` (b). `sqrt(4.5gL)`

Discussion

643.	A certain charge distribution produces electric potential that varies along the X axis as shown in figure. [There is no field in y or z direction] (a) At which point (amongst A, B, C, D and E) does a negative charge feel the greatest force in positive X direction? (b) Find the upper limit of the speed that a proton can have, as it passes through the origin, and still remain bound near the origin. Mass and charge of a proton are m and e. How will your answer change for an electron?
Answer» Correct Answer - (a). D (b). `sqrt((2eV_(0))/(m))` Electron cannot remain bound for any speed.

Discussion

644.	What is the electrostatic potential due to an electric dipole at an equatorial point?
Answer» Zero electrostatic potential due to an electric dipole at an equatorial point.

Discussion

645.	The same Gaussain surface is used to surround two charged objects. The net number of field lines penetracting the surface is same in both the cases, but the lines are oppositely directed. What can you say about the net charge on the two objects?
Answer» Since, lines are equal and oppositely directed, this shows charges are equal but of opposite signs. So, net charge is zero.

Discussion

646.	Charge `q_(1)` is inside the Gaussain surface , charge `q_(2)` just outside the surface. Does the electric flux through the surface.Does the electric flux through the surface depend on `q_(1)` ? Does it depend on `q_(2)` ? Explain.
Answer» Electric flux through the surface depends on `q_(1)` ( as `q_(1)` is inside the surface). But it does not depend on `q_(2)` , it being outside Gaussain surface. `q_(2)` does not contribute to electric flux.

Discussion

647.	Define electric flux. Write its SI unit. A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change ?
Answer» Electric flux over an area in an electric field represents the total number of electric lines of force crossing this area. SI unit of electric flux are `N m^(2) C^(-1)`. When R is reduced to half, electric flux through the surface remains the same. This is because electric flux through the surface .depends only on charge enclosed.

Discussion

648.	Define electric flux. Write its S.I.unit. A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change?
Answer» Electric flux linked with a surface is the number of electric lines of force cutting through the surface normally. It’s SI unit is Nm²C ^–1 or Vm on decreasing the radius of spherical surface to half there will be no effect on the electric flux.

Discussion

649.	Figure below show regular hexagons with charges at the vertices. In which of the following cases the electric field at the centre is not zero
Answer» Correct Answer - A Electric field at a point due to positive charges acts away from the charges acts away frm the charge and due to negative charge it acts towards the charge.

Discussion

650.	A parallel plate air capacitor consists of two circular plates of diameter `8 cm` . At what distance should the plates be held so as to have the same capacitance as that of a sphere of a diameter `20 cm` ?A. `4xx10^(-3)m`B. `1xx10^(-3)cm`C. `1xx10^(-2)cm`D. `1xx10^(-3)m`
Answer» Correct Answer - D `r_(1)=(4)/(2)=2xx10^(-2)cm` `r_(2)=(20)/(2)=10xx20^(-2)cm` Capacity of parallel plate capacitor = Capacitor of spherical capacitor `(epsilon_(0)A)/(d)=4piepsilon_(0)r_(2)` `"or "(epsilon_(0)pi r_(1)^(2))/(d)=4pi epsilon_(0)r_(2)` `d=(r_(1)^(2))/(4_(2))` `=((2xx10^(-2))^(2))/(4xx10xx10^(-2))=(4xx10^(-4))/(4xx10^(-1))` Distance between plates, `d=1xx10^(-3)m`

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