1638 + Interview Questions in ELECTROSTATICS IN GENERAL AWARENESS Page 4 InterviewSolution

151.	If 100 joules of work must be done to move electric charge equal to 4 C from a place, where potential is -10 volt to another place where potential is V volt, find the value of V.
Answer» Given: q₀ = 4 C, V_A = -10 volt, V_B = V volt, W_AB = 100 J To Find: Potential (V) Formula: V_B – V_A= \(\frac{W_{AB}}{q_0}\) Calculation: From formula, V – (-10) = \(\frac{100}{4}\) = 25 ∴ V + 10 = 25 ∴ V = 15 volt

151.

If 100 joules of work must be done to move electric charge equal to 4 C from a place, where potential is -10 volt to another place where potential is V volt, find the value of V.

Answer»

Given: q₀ = 4 C,

V_A = -10 volt,

V_B = V volt,

W_AB = 100 J

To Find: Potential (V)

Formula: V_B – V_A= \(\frac{W_{AB}}{q_0}\)

Calculation: From formula,

V – (-10) = \(\frac{100}{4}\) = 25

∴ V + 10 = 25

∴ V = 15 volt

Discussion

152.	A potential difference of 250 Volt is applied across the plate of a capacitor of 10 pF. Calculate the charge on the plates of the capacitor.
Answer» Here, `V = 250V , C = 10 pF = 10xx10^(-12) F = 10^(-11) F`. `:. Q = CV = 10^(-11)xx250 = 2.5xx10^(-9)C`.

Discussion

153.	Three capacities of respective capacities 6, 4 and `2muF` are connected in parallel and P.D. of 5 volt applied across it. Its total capacity and the charge on each condenser in order will beA. `12muF, 30muC, 10muC, 20muC`B. `12muF, 10muC, 30muC, 20muC`C. `12muF, 10muC, 20muC, 30muC`D. `12muF, 30muC, 20muC, 10muC`
Answer» Correct Answer - D `C_(P)=C_(1)+C_(2)+C_(3)=6+4+2=12muF` `Q_(1)=C_(1)V=6xx5=30muC` `Q_(2)=C_(2)V=4xx5=20muC` `Q_(3)=C_(3)V=2xx5=10muC`

Discussion

154.	A charge of `4xx10^(-9)C` is distributed uniformly over the circumference of a conducting ring of radius 0.3m. Calculate the field intensity at a point on the axis of the ring at 0.4m from its centre, and also at the centre.
Answer» Correct Answer - `115.2 N//C` ; Zero `q = 4xx10^(-9)C, a = 0.3m, r = 0.4m, E = ?` `E = (q r)/(4pi in_(0) (r^(2) + a^(2))^(3//2))` `E = (14.4)/((0.5)^(3)) = 115.2 N//C` At the center of ring, `r = 0, :. E = 0`

Discussion

155.	Write an expression for potential at a point P `vec(( r))` due to two point charges `q_(1) and q_(2)` at `vec(r_(1)) and vec(r_(2))` respectivley.
Answer» `V vec((r )) = (q_(1))/(4pi in_(0) \| vec(r ) - vec(r_(1)) \|) + (q_(2))/(4pi in_(0) \| vec(r ) - vec(r_(2)) \|)`

Discussion

156.	What work must be done in carrying an `alpha -` pariclae across a potential difference of 1 volt?
Answer» As `W = q xx DeltaV = 3.2xx10^(-19)xx1` `= 3.2xx10^(-19) J`

Discussion

157.	what is the SI unit of line interfral of electric field ?
Answer» volt or joule/coulomb, as the line intergal of electric field represents potential difference.

Discussion

158.	The figure shows the electric lines of force emerging from a charged body. If the electric fields at A and `E_(A)` and `E_(B)` respectively and if the the distance between A anf B is r, then A. `E_(A)gtE_(B)`B. `E_(A)ltE_(B)`C. `E_(A)=E_(B)`D. cannot be predicted
Answer» Correct Answer - A

Discussion

159.	Figure shows the electric lines of force emerging from a charged body. If the electric field at`A` and `B` are `E_(A)` and `E_(B)` respectively and if the displacement between `A` and `B` is `r` then A. `E_(A) gt E_(B)`B. `E_(A) lt E_(B)`C. `E_(A) = (E_(B))/(r)`D. `E_(A) = (E_(B))/(r^(2))`
Answer» Correct Answer - A

Discussion

160.	The potential difference points a and B in the given uniform electric field is : A. EaB. `Esqrt((q^(2)+b^(2)))`C. EbD. `(Eb//sqrt(2))`
Answer» Correct Answer - C

Discussion

161.	A particle of charges Q and mass m `m` travels through a potential difference V from rest. The final momentum of the particle isA. `(mV)/Q`B. `2Q sqrt(mV)`C. `sqrt(2mQV)`D. `sqrt((2QV)/m)`
Answer» Correct Answer - C

Discussion

162.	Can two equipotential surfaces cut each other?
Answer» No. An equipotentiol surface is normal to the electric field intensity. If two equipotential surfaces intersect, then at the point of intersection, there will be two directions of electric field intensity, which is not possible.

Discussion

163.	The figure shows the electric lines of force emerging from a charged body. If the electric fields at A and `E_(A)` and `E_(B)` respectively and if the the distance between A anf B is r, then A. `E_(A) lt E_(B)`B. `E_(A) gt E_(B)`C. `E_(A)=E_(B)/r`D. `E_(A)=E_(B)/r^(2)`
Answer» Correct Answer - B

Discussion

164.	An equipotential line and a line of force areA. never intersect each otherB. intersect at `45^(º)`C. intersect at `60^(º)`D. intersect at `90^(º)`
Answer» Correct Answer - D

Discussion

165.	The force between two electrons separated by a distance r is proportional toA. `r^(3)`B. `r^(1//2)`C. `r^(-1//3)`D. `r^(-2)`
Answer» Correct Answer - D

Discussion

166.	At a point on the axis (but not inside the dipole and not at infinity) of an electric dipoleA. The electric field is zeroB. The electric potential is zeroC. Neither the electric field not the electric potential is zeroD. The electric field is directed perpendicular to the axis of the dipole
Answer» Correct Answer - C

Discussion

167.	A plastic rod has been formed into a circle of radius R. It has a positive charge `+Q` uniformly distributed along one-quarter of its circumference and a negative charge of ` -6Q` uniformly distributed along the rest of the circumference (figure). With V = 0 at infinity, what is the electric potential `-6Q`(a) at the centre C of the circle and(b) at point P, which is on the central axis of the circle at distance z from the centre?
Answer» Correct Answer - D `a. V_C=(k.q_("net"))/R=(1/(4piepsilon_0)((-5Q)/R)` `b. V_p=(kq_("net"))/r` where r=distance of P from any point on circumfrence `=(1/(4piepsilon_0))((-5Q)/(sqrt(R^2+z^2)))`

Discussion

168.	Lines of force cut equipotential surfaceA. obliquelyB. normallyC. tangentiallyD. at `45^(@)`
Answer» Correct Answer - B

Discussion

169.	If a unit positive charge is taken from one point to another over an equipotential surface ,thenA. work is done on the chargeB. no work is doneC. work done is constantD. work is done by the charge
Answer» Correct Answer - B

Discussion

170.	An equipotential surface is that surface on which each and every point hasA. same potentialB. zero potentialC. negative potentialD. different potential
Answer» Correct Answer - A

Discussion

171.	If electric field is uniform, then the electric lines of forces are :A. DivergentB. ConvergentC. CircularD. Parallel
Answer» Correct Answer - D

Discussion

172.	The force between two electrons separated by a distance r is proportional toA. `r^(2)`B. `r^(4)`C. `r^(-2)`D. `r^(-4)`
Answer» Correct Answer - D

Discussion

173.	A ring of radius r has a uniformly spread charge + q on quarter of its circumference. The opposite quarter of the ring carries a charge – q uniformly spread over it. Find the electric potential at a point A shown in the figure. Point A is at a distance R(gtgt r) from the centre of the ring.
Answer» Correct Answer - `V=(qr)/(pi^(2) in_(0)R^(2))`

Discussion

174.	Which of the following is correct regarding equipotential surface ?A. Equipotential surfaces can never cross each other otherwise potential at a point will have two values which is absurdB. Equitential surfaces are always perpendicular to lines of forceC. If a charge is moved from one point to the other over an equipotential surface, work done will be zero as `W_(AB) = - U_(AB) = Q(V_(B) - V_(A) = 0 (as (V_(B) = V_(A))`D. All options are correct
Answer» Correct Answer - 4

Discussion

175.	Two identical electric dipoles are arranged parallel to each other with separation between them large compared to the length of individual dipole. The electrostatic energy of interaction of the two dipoles in this position is U. (a) Find work done in slowly rotating one of the dipoles by `90^(@)` so as to bring it to position shown in Fig. (b). (b) Find work done in rotating one of the dipoles by `180^(@)` so as to bring it to the position shown in Fig. (c).`O_1` and `O_2` are centers of the dipoles.
Answer» Correct Answer - (a). `-(KP^(2))/(r^(3))` (b) `-(2KP^(2))/(r^(3))`

Discussion

176.	Three dipoles each of dipole moment of magnitude p are placed tangentially on a circle of radius R in its plae positioned at equal angle from each other as shown in the figure. Then the magnitude of electric field intensity at the centre of the circle will be : A. `(4 kp)/R^(3)`B. `(2kp)/R^(3)`C. `(kp)/R^(3)`D. `0`
Answer» Correct Answer - B

Discussion

177.	Tow short electric dipoles are placed as shown The energy or electric interaction between these dipoles will be ` .A. ` (2 k P_1p_2 cos theta)/r^2 `B. ` (-2 kP1p_2 cos theta)/r^3`C. `(-2 P_1p2 sin theta)/r^3`D. ` (-4 kP_1P_2 cos theta)/r^3`
Answer» Correct Answer - B Potenial energy of ` P_2` in the field of ` P_1` ` =- P_2 E_1 cos theta ` ` =- P_2 (2 KP_1 cosq)/r^3` ` U=- (2 KP_1OP_2)/r^3 cos theta` .

Discussion

178.	Two short dipoles , each of dipole moment `P` are placed at a large separation `r`. The force between themA. is proportional to product of dipole momentaB. is inversely proportional to `r^(4)`C. the force is attractive , if direction of dipole momenta are same , repulsive if oppositeD. all options are correct
Answer» Correct Answer - 4

Discussion

179.	A solid sphere of radius R has charge q uniformly distributed over its volume. The distance from it surfce at which the electrostatic potential is equal to half of the potential at the centre isA. RB. 2RC. `R/3`D. `R/2`
Answer» Correct Answer - C `(kq)/r=3/2((kq)/R)/2or r=4/3R` `:.` Distance from surface =`r=R` `=R/3`

Discussion

180.	A massless rod of length L has two equal charges (q) tied to its ends. The rod is free to rotate in horizontal plane about a vertical axis passing through a point at a distance `(L)/(4)` from one of its ends. A uniform horizontal electric field (E) exists in the region. (a) Draw diagrams showing the stable and unstable equilibrium positions of the rod in the field. (b) Calculate the change in electric potential energy of the rod when it is rotated by an angle `Deltatheta` from its stable equilibrium position. (c) Calculate the time period of small oscillations of the rod about its stable equilibrium position. Take the mass of each charge to be m.
Answer» Correct Answer - (b). `DeltaU=(EqL)/(2)(1-cosDeltatheta)` (c) `T=pisqrt((5mL)/(Eq))`

Discussion

181.	(a). Use the method in part (b) of the previous problem of calculate the electrostatic self energy of a uniformly charegd sphere of radius R having charge Q. (b). Divide the above sphere (mentally) into two regions-spherical concentric part having radius `(R)/(2)` and the remaining annular part (between `(R)/(2)` and R). Denote the point charges in sphere of radius `R//2` by `q_(1),q_(2),q_(3)` .. . .etc. The charges in annular part be denoted by `Q_(1),Q_(2),Q_(3)` . . . etc. Calculate the electrostatic interaction energy for all pairs like `[(Q_(i),Q_(j))+(q_(i)+q_(j))]`.
Answer» Correct Answer - (a). `(3)/(5)(Q^(2))/(4pi epsi_(0)R)` (b). `(147)/(320)(Q^(2))/(4pi epsi_(0)R)`

Discussion

182.	Consider a uniformly charged spherical shell of radius R having charge Q charge Q can be through to be made up of number of point charges `q_(1),q_(2),q_(3). . . ` etc. the electrostatic energy of the charged shell is sum of interaction energies of all possible pairs of charges. `U=sum(q_(i)q_(j))/(4piepsilon_(0)r_(ij))` Where `r_(ij)` is distance between `q_(i)` and `q_(j)`. for continuous charg on the shell, the summation has to be carried through intergration. (a). Calculate the electrostatic energy of the shell. We can term this energy as self energy of the shell. (b). Calculate the work done in assembling a spherical shell of uniform charge Q and radius R by bringing charges in small installments from infinity and putting them of the shell. Do you ding the answers in (a) and (b) to be same?
Answer» Correct Answer - (a). `(Q^(2))/(8 pi in_(0)R)` (b). Same as in (a)

Discussion

183.	A point charge `q_(0)` is placed at the centre of uniformly charges ring of total charge Q and radius R. If the point charge is slightly displaced with negligible force along axis of the ring then find out its speed when it reaches a large distance.
Answer» Only electric force is acting on `q_(0)` `:. W_(el)=DeltaK" "=1/2 mv^(2)-0" "implies" "`Now `W_(el)")"_(c rarr oo) =q_(0) V_(c) =q_(0) (KQ)/R`. `:. (Kq_(0)Q)/R=1/2 mv^(2)" "implies" "v=sqrt((2Kq_(0)Q)/(mR))`

Discussion

184.	The magnitude of electrostatic unit of charge in S.I. system isA. `1.602xx10^(-19)C`B. `9.1xx10^(-31)C`C. `3xx10^(8)C`D. `3xx10^(-19)C`
Answer» Correct Answer - A

Discussion

185.	The smallest quantity of electricity called as electrostatic unit of charge is present onA. photonB. `alpha` - particleC. electron and protonD. none of these
Answer» Correct Answer - C

Discussion

186.	A positively charged sphere of mass m = 5kg is attached by a spring of force constant `K = 10^(4)` N/m. The sphere is tied with a thread so that spring is in its natural length. Another identical, negatively charged sphere is fixed with floor, vertically below the positively charged sphere as shown in fignre. If initial separation between sphere is `r_(0)` = 50cm and magnitnde ofcharge on each sphere is `q = 100muC`, calculate maximum elongation of spring when the thread is burnt. Take `g= 10 m//s^(2)` :
Answer» Correct Answer - [10cm]

Discussion

187.	Two charges `+- 10 muC` are placed `5xx10^(-3) m` apart. Determine the electric field at a point `Q, 0*15m` away from O, on a line passing through O and normal to the axis of the diople.
Answer» Here, `q = +- 10 muC = +- 10xx10^(-6) C`, `2a = 5xx10^(-3) m, r = 015m` The point Q lies on equatorial line of electric dipole, `As r gt gt` a So, `E = (p)/(4pi in_(0) r^(3)) = (q(2a))/(4pi in_(0) r^(3))` `E = (9xx10^(9)xx10xx10^(-6)xx5xx10^(-3))/((015)^(3))` `= 1*33xx10^(5) NC^(-1)`

Discussion

188.	Two point charges of `+16 muC and -9 mu C` are placed 8 cm apart in air. Determine the position of the point at which the resultant electric field is zero.
Answer» Correct Answer - 24cm to the right of `-9 muC` charge Let E = 0 at a distance x to the right of `-9 mu C` charge , Use `k (q_(1) xx1)/((8+x)^(2)) = (k q_(2)xx1)/(x^(2))`

Discussion

189.	Two charges `+30 muC and -30 muC` are placed 1cm apart. Calculate electric field at a point on the axial line at a distance of 20cm from the centre of dipole.
Answer» Correct Answer - `6.25xx10^(5) N//C` As `r gt gt a`, so `E = (2 vec(p))/(4pi in_(0) r^(3))` `= (9xx10^(9)xx2xx30xx10^(-6)xx10^(-2))/((20xx10^(-2))^(3)) = 6.25xx10^(5) N//C`

Discussion

190.	Two point charges of same magnitude and opposite sign are fixed at point a and B. A third small point charge is to be balanced at point P by the electrostatic force due to these two charges. The point P : A. lies on the perpendicular bisector of line ABB. is at mid point of line ABC. lies to the left of AD. none of these
Answer» Correct Answer - D

Discussion

191.	Two point charges each of `5 muC` but opposite in sign are placed 4cm from the mid point on the axial line of dipole.
Answer» Correct Answer - `10^(8) NC^(-1)` Here, `q = 5 muC = 5xx10^(-6)C` `2a = 4cm = 4xx10^(-2)m, E = ?` `r = 4 cm = 4xx10^(-2) m`. `E_("axial") = (2 pr)/(4 pi in_(0) (r^(2) - a^(2))^(2)) = (2 (q) (2a) r)/(4pi in_(0) (r^(2) - a^(2))^(2))` `= (9xx10^(9)xx2(5xx10^(-6))(4xx10^(-2)) xx4xx10^(-2))/([0.04^(2) - 0.02^(2)]^(2))` `E_("axial") = 10^(8) NC^(-1)`

Discussion

192.	Draw electric lines of forces due to an electic dipole.
Answer» The electric lines of force in repect of an electric dipole are as shown in Fig.

Discussion

193.	What is nature of sysmmetry of field due to a point charge ?
Answer» The field due to point charge has sphereical symmetry.

Discussion

194.	Charges `+q` and `-q` are placed at points `A` and `B` respectively which are a distance `2L` apart, `C` is the midpoint between `A` and `B`. The work done in moving a charge `+Q` along the semicircle `CRD` is A. `(q Q)/(2pi in_(0) L)`B. `(q Q)/(6pi in_(0) L)`C. `(-q Q)/(6pi in_(0) L)`D. `(q Q)/(4pi in_(0) L)`
Answer» Correct Answer - C As is clear from Fig, `V_(C) = 0`, `V_(D) = (q)/(4pi in_(0) (3L) - (q)/(4pi in_(0) L)) = (-q)/(6pi in_(0) L)` `W_(CDR) = Q (V_(D) - V_(C)) = (-qQ)/(6pi in_(0) L)`

Discussion

195.	Two point charges of `+3 muC` each are `100 cm` apart. At what point on the line joining the charges will the electric intensity be zero ?
Answer» At the center fo the two charges.

Discussion

196.	The point charges `Q` and `-2Q` are placed at some distance apart. If the electirc field at the location of `Q` is `E`, the electric field at the location of `-2Q` will beA. 3 EB. `E//2`C. ED. None of these
Answer» Correct Answer - D

Discussion

197.	Two point charges of `10^(-8)C and -10^(-8)C` are placed `0*1 m` apart. Calculate electric field intensity at A, B and C shown in Fig.
Answer» The charges `+ 10^(-8)C and -10^(-8)C and -10^(-8)C` are held at P and Q respectively, `PQ = 01m` Also, `PA = PB = 005 m` `QA = 005 and CP = CQ = 01m` At A, field intensity due to charge `q_(1)` `= (1)/(4pi in_(0)) (q_(1)xx1)/(AP^(2))` , along PA `= (9xx10^(9)x10^(-8))/((005)^(2)) = 36000 N//C` Field intensity due to charge `q_(2)` `= (9xx10^(9)x10^(-8))/((005)^(2))` along `AQ = 36000 N//C` `:.` Net field intensity at `A = 36000 + 36000` `= 72000 N//C = 72xx10^(4) N//C` along AQ At B. Field intensity due to charge `q_(1)` `= (1)/(4pi in_(0)) (q_(1)xx1)/(PB^(2))` along PB producd `(9xx10^(9)xx10^(-8))/((005)^(2)) = 4000 N//C` `:.` Net field intensity at `B = 36000 - 4000` `= 32000 N//c` `32xx10^(4) N//C` , along PB produced At C, Field intensity due to charge `q_(1)` , `E_(1) = (1)/(4pi in_(0)) (q)/(PC^(2))` , along PC. `E_(1) = (9xx10^(9)xx10^(-8))/((0-1)^(2)) = 9xx10^(3) N//C` Field intensity due to charge `q_(2)` `E_(2) = (1)/(4pi in_(0)) (q_(2))/(QC^(2))` along CQ, `E_(2) = (9xx10^(9)xx10^(-8))/((01)^(2)) = 9xx10^(3) N//C` Both, `E_(1) and E_(2)` are equal in magnitude, acting at `60^(@)` to CR parallel to BA. `:.` Resulant intensity at R `= E_(1) cos 60^(@) + E_(2) cos 60^(@) = 2 E_(1) cos 60^(@)` `= 2xx9xx10^(3)xx(1)/(2) = 9xx10^(3) N//C.` It is represented by CR parallel to BA. The components of `vec(E_(1))` and `vec(E_(2))` in directions `_\|_` to CR cancel out.

Discussion

198.	An inclined plane of length `5.60` m making an angle of `45^(@)` with the horizontal is placed in a uniform electric field `E=100 Vm^(-1)`. A particle of mass `1 kg` and charge `10^(-2)C` is allowed to slide down from rest position from maximum height of slope. If the coefficient of friction is 0.1, then the time taken by the particle to reach the bottom isA. 1 sB. 1.41 sC. 2 sD. None of these
Answer» Correct Answer - B

Discussion

199.	Answer the following : (a) The top fo the atmosphere is at about 400 kv with respect to the surface of earth, the field is about `100 Vm^(-1)`. Why then do we not get an electric shock, as we step out of out house into the open ? (Assume the house to be a steel cage so that there is no field inside). (b) A man fixes outside his house one evening a two meter high insulating slab carrying on its top, a large aluminium sheet of area `1 m^(2)`. Will he get an electric shock if he touches the metal sheet next morning ? (c) The discharging current in the atmosphere due to the small conducity of air is known to be 1800 A on an average over the globe. Why then does the atomsphere not discharge itself complety in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ? (d) What are teh forms of energy into which the electrical energy fo the atmosphere is dissipated during a lighting ?
Answer» (a) Since our body and the surface of earth, both are conducting, therefore, our body and the ground form an equipotential surface. As we step out into the open from our house , the orignal equipotential surfaces of open air charge, keeping our body and the ground at the same potential. That is why we donot get an electric shock. (b) Yes, the man will get the shock. This is because the steady discharging current of the atmostphere charegs up the aluminium sheet gradually adn raises its voltage to an extent depending on the capacitance of the condenser formed by the aluminium sheet, the ground and the insulating slab. (c) The atomosphere is being charged continously be thunderstoms and lighting all over the globe. It is also discharging due to the small conductivity of air. The two opposing process, on an average,are in equilibrium. Therefore the atmosphere keeps charged. (d) During a lighting the electrical energy of the atmosphere is dissipated in the form of light heat and sound.

Discussion

200.	A mass `m=20 g` has a charge `q= 3.0 mC`. It moves with a velocity of 20 m/s and enters a region of electric field of 80 N/C in the same direction as the velocity of the mass. The velocity of the mass after 3 s in this region isA. `80 ms^(-1)`B. `56 ms^(-1)`C. `44 ms^(-1)`D. `40 ms^(-1)`
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

If 100 joules of work must be done to move electric charge equal to 4 C from a place, where potential is -10 volt to another place where potential is V volt, find the value of V.

A potential difference of 250 Volt is applied across the plate of a capacitor of 10 pF. Calculate the charge on the plates of the capacitor.

A charge of `4xx10^(-9)C` is distributed uniformly over the circumference of a conducting ring of radius 0.3m. Calculate the field intensity at a point on the axis of the ring at 0.4m from its centre, and also at the centre.

Write an expression for potential at a point P `vec(( r))` due to two point charges `q_(1) and q_(2)` at `vec(r_(1)) and vec(r_(2))` respectivley.

What work must be done in carrying an `alpha -` pariclae across a potential difference of 1 volt?

what is the SI unit of line interfral of electric field ?

The figure shows the electric lines of force emerging from a charged body. If the electric fields at A and `E_(A)` and `E_(B)` respectively and if the the distance between A anf B is r, then A. `E_(A)gtE_(B)`B. `E_(A)ltE_(B)`C. `E_(A)=E_(B)`D. cannot be predicted

Figure shows the electric lines of force emerging from a charged body. If the electric field at`A` and `B` are `E_(A)` and `E_(B)` respectively and if the displacement between `A` and `B` is `r` then A. `E_(A) gt E_(B)`B. `E_(A) lt E_(B)`C. `E_(A) = (E_(B))/(r)`D. `E_(A) = (E_(B))/(r^(2))`

The potential difference points a and B in the given uniform electric field is : A. EaB. `Esqrt((q^(2)+b^(2)))`C. EbD. `(Eb//sqrt(2))`

A particle of charges Q and mass m `m` travels through a potential difference V from rest. The final momentum of the particle isA. `(mV)/Q`B. `2Q sqrt(mV)`C. `sqrt(2mQV)`D. `sqrt((2QV)/m)`

Can two equipotential surfaces cut each other?

The figure shows the electric lines of force emerging from a charged body. If the electric fields at A and `E_(A)` and `E_(B)` respectively and if the the distance between A anf B is r, then A. `E_(A) lt E_(B)`B. `E_(A) gt E_(B)`C. `E_(A)=E_(B)/r`D. `E_(A)=E_(B)/r^(2)`

An equipotential line and a line of force areA. never intersect each otherB. intersect at `45^(º)`C. intersect at `60^(º)`D. intersect at `90^(º)`

The force between two electrons separated by a distance r is proportional toA. `r^(3)`B. `r^(1//2)`C. `r^(-1//3)`D. `r^(-2)`

At a point on the axis (but not inside the dipole and not at infinity) of an electric dipoleA. The electric field is zeroB. The electric potential is zeroC. Neither the electric field not the electric potential is zeroD. The electric field is directed perpendicular to the axis of the dipole

Lines of force cut equipotential surfaceA. obliquelyB. normallyC. tangentiallyD. at `45^(@)`

If a unit positive charge is taken from one point to another over an equipotential surface ,thenA. work is done on the chargeB. no work is doneC. work done is constantD. work is done by the charge

An equipotential surface is that surface on which each and every point hasA. same potentialB. zero potentialC. negative potentialD. different potential

If electric field is uniform, then the electric lines of forces are :A. DivergentB. ConvergentC. CircularD. Parallel

The force between two electrons separated by a distance r is proportional toA. `r^(2)`B. `r^(4)`C. `r^(-2)`D. `r^(-4)`

A ring of radius r has a uniformly spread charge + q on quarter of its circumference. The opposite quarter of the ring carries a charge – q uniformly spread over it. Find the electric potential at a point A shown in the figure. Point A is at a distance R(gtgt r) from the centre of the ring.

Tow short electric dipoles are placed as shown The energy or electric interaction between these dipoles will be ` .A. ` (2 k P_1p_2 cos theta)/r^2 `B. ` (-2 kP1p_2 cos theta)/r^3`C. `(-2 P_1p2 sin theta)/r^3`D. ` (-4 kP_1P_2 cos theta)/r^3`

A solid sphere of radius R has charge q uniformly distributed over its volume. The distance from it surfce at which the electrostatic potential is equal to half of the potential at the centre isA. RB. 2RC. `R/3`D. `R/2`

A point charge `q_(0)` is placed at the centre of uniformly charges ring of total charge Q and radius R. If the point charge is slightly displaced with negligible force along axis of the ring then find out its speed when it reaches a large distance.

The magnitude of electrostatic unit of charge in S.I. system isA. `1.602xx10^(-19)C`B. `9.1xx10^(-31)C`C. `3xx10^(8)C`D. `3xx10^(-19)C`

The smallest quantity of electricity called as electrostatic unit of charge is present onA. photonB. `alpha` - particleC. electron and protonD. none of these

Two charges `+- 10 muC` are placed `5xx10^(-3) m` apart. Determine the electric field at a point `Q, 0*15m` away from O, on a line passing through O and normal to the axis of the diople.

Two point charges of `+16 muC and -9 mu C` are placed 8 cm apart in air. Determine the position of the point at which the resultant electric field is zero.

Two charges `+30 muC and -30 muC` are placed 1cm apart. Calculate electric field at a point on the axial line at a distance of 20cm from the centre of dipole.

Two point charges each of `5 muC` but opposite in sign are placed 4cm from the mid point on the axial line of dipole.

Draw electric lines of forces due to an electic dipole.

What is nature of sysmmetry of field due to a point charge ?

Two point charges of `+3 muC` each are `100 cm` apart. At what point on the line joining the charges will the electric intensity be zero ?

The point charges `Q` and `-2Q` are placed at some distance apart. If the electirc field at the location of `Q` is `E`, the electric field at the location of `-2Q` will beA. 3 EB. `E//2`C. ED. None of these

Two point charges of `10^(-8)C and -10^(-8)C` are placed `0*1 m` apart. Calculate electric field intensity at A, B and C shown in Fig.

A mass `m=20 g` has a charge `q= 3.0 mC`. It moves with a velocity of 20 m/s and enters a region of electric field of 80 N/C in the same direction as the velocity of the mass. The velocity of the mass after 3 s in this region isA. `80 ms^(-1)`B. `56 ms^(-1)`C. `44 ms^(-1)`D. `40 ms^(-1)`