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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
The sketech below show cross secion so equipotential surfaces between two charged.Conductors that are shwon in solid black.Some points on the wquipotenital surfaces .near the conductors are marked as `A,B,C…. .` The arrangements lies in air (Take `epsilon_(0)=8.85xx10^(-12)C^(2)//Nm^(2)]` A positive charge is placed at `B`,When it is released:A. no force will be exerted on it .B. it will move towards `A`C. it will move towards `C`D. it will move towards `E` |
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Answer» Correct Answer - B Direction of `E.F` at `B`is towards `A` will exert froce in this direction only, causing the positive charge to move .[`vec(E)` is perpendicular to equipotential surface and its direction is form high potential to low potential]. |
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| 152. |
Uniform electric field of magnitude `100 Vm^-1` in space is directed along the line `y = 3 + x`. Find the potential difference between points `A (3,1) and B (1,3)`.A. `100V`B. `200sqrt(2)V`C. `200V`D. `0` |
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Answer» Correct Answer - D No change in potenial for the wire placed on `X` axis.So `W=0` and for wires on `Y` &`Z` axes is `W=Deltar(lambda)/(2pioepsilon_(0))l n(r_(2))/(r_(1))W=W_(1)+W_(2)` `W=(4lambda)/(2piepsilon_(0))i n((sqrt(2))/(1))W=(lambda)/(piepsilon_(0)) l n2` |
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| 153. |
Charge cannot exist without mass but mass can exist without charge. `B`: charge is invariant but mass alone may velocity `C`: Charge is conserved but mass alone may not be conserved.A. `A,B,C` are trueB. `A,B,C` are not trueC. `A,B` are only trueD. `A,B` are false and `C`. |
| Answer» Correct Answer - 1 | |
| 154. |
A thin copper ring of radius `a` is charged with `q` units of electricity.An electron is placed at the centre of the copper ring .If the electron is displaced a little, it will have frequencyA. `(1)/(2pi)sqrt((eq)/(4piepsilon_(0)ma^(3)))`B. `(1)/(2pi)sqrt((eq)/(4piepsilon_(0)ema^(3)))`C. `sqrt((eq)/(4piepsilon_(0)ma))`D. `sqrt((q)/(4piepsilon_(0)ema^(3)))` |
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Answer» Correct Answer - 1 Range =`(u^(2)sin^(2)theta)/(2(g+-(EQ)/(m))` |
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| 155. |
Statement :-A:-Electrical potential may exist at a point where the electrical potential field is zero. Statement B:- Electrical field may exist at a point where the electircal potential is zero. Statement C:- The electric potential inside a charge conducting sphere is constant .A. `A,B` are trueB. `B,C` are trueC. `A,C` are trueD. `A,B,C` are true. |
| Answer» Correct Answer - 4 | |
| 156. |
A thin fixed of radius 1 metre has a positive charge `1xx10^-5` coulomb uniformly distributed over it. A particle of mass 0.9 gm and having a negative charge of `1xx10^-6` coulomb is placed on the axis at a distance of 1cm from the centre of the ring. Show that the motion of the negatively charged particle is approaximately simple harmonic. Calculate the time period of oscillations.A. `0.23s`B. `0.39s`C. `0.49s`D. `0.63s` |
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Answer» Correct Answer - 4 `E=(1)/(4piepsilon_(0))(qx)/((a^(2)+x^(2))^(3//2))=(qx)/(4piepsilon_(0)a^(3))` `:. m=(d^(2)x)/(dt^(2))=-(1)/(4piepsilon_(0))(qex)/(a^(3))` SO motion is `S.H.M` `omega^(2)=(1)/(4piepsilon_(0))(qe)/(ma^(3))` |
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| 157. |
Two point charges Q and ` - Q //4` are separated by a distance ` X`. Then .A. potential is zero at point on the axis which is `x//3` on the right side of the charge `-Q//4`B. potential is zero at point on the axis which is `x//5` on the right side of the charge `-Q//4`C. electric fiels is zero at a point on the axis which is at a distance `x` on the right side of the charge `-Q//4`D. there exist two points on the axis where electric field is zero . |
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Answer» Correct Answer - A,B,C Electric potential and electric field |
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| 158. |
A spheres carrying charge `0.01C` is kept at rest without falling down ,touching a wall by applying an electric filed `100N//C`.If the coefficient of friction between the sphere and the wall is `0.2` the weight of the sphere isA. `4N`B. `2N`C. `20N`D. `0.2N` |
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Answer» Correct Answer - 4 `F=(Qq)/(4piepsilon_(0))(x)/(R^(3))=-kx` and `T=2pisqrt((m)/(k))` |
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| 159. |
`N` fundamental charges each of charge `q` are to be distributed as two point charges separated by a fixed distance ,then the maximum to minimum force bears a ratio (N is even and greater than `2`)A. `(N-1)^(2)/(4N^(2))`B. `(4N^(2))/((N-1))`C. `(N^(2))/(4(N-1))`D. `(2N^(2))/((N-1))` |
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Answer» Correct Answer - 3 `(F_(max))/(F_(min))=((N//2)^(2))/((N-1)1)` |
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| 160. |
Two identical particles of charge `q` each are connected by a masless spring of force cosntant `k` .They are placed over a smoothe horizontal surface .They are released wheen unstretched .If maximum extneison of the spring is `r`,the value of `k` is :(neglect gravitational effect)A. `k=(q)/(r)sqrt((1)/(piepsilon_(0)r))`B. `k=(1)/(4piepsilon_(0))(q^(2))/(l^(2))xx(1)/(r)`C. `k=(2q)/(r)sqrt((1)/(piepsilon_(0)r))`D. `k=(q)/(r)sqrt((2)/(piepsilon_(0)r))` |
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Answer» Correct Answer - 2 `F_(C)=kx` |
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| 161. |
A particle A having a charge of `2.0xx 10^(-6)` C and a mass of 100 g is placed at the bottom of a smoth inclined plane of inclination `[email protected]` . Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?A. `27cm`B. `16cm`C. `30cm`D. `45cm` |
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Answer» Correct Answer - 1 `mg sintheta=(1)/(4piepsilon_(0))(q^(2))/(r^(2))` |
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| 162. |
A particle A having a charge of `2.0xx 10^(-6)` C and a mass of 100 g is placed at the bottom of a smoth inclined plane of inclination `[email protected]` . Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?A. `27cm`B. `16cm`C. `30cm`D. `45cm` |
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Answer» Correct Answer - 1 `mg sintheta=(1)/(4piepsilon_(0))(q^(2))/(r^(2))` |
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| 163. |
Two opposite and equal chrages `4 xx 10^(-8)` coulomb when placed `2 xx 10^(-2) cm` away, from a dipole. If dipole is placed in an ecternal electric field `4 xx 10^(8)` newton//coulomb, the value of maximum torque and the work done in rotating it through `180^(@)` will beA. `32xx10^(-4)Nm` and `32xx10^(-4)J`B. `64xx10^(-4)Nm` and `64xx10^(-4)J`C. `64xx10^(-4)Nm` and `32xx10^(-4)J`D. `16xx10^(-4)Nm` and `32xx10^(-4)J` |
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Answer» Correct Answer - 4 particles moves in a direction where potential energy of the system decreased. |
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| 164. |
`n` charges `Q,4Q,9Q,16Q….` are placed at distance of `1,2,3…..` metre form a point `0,`on the same straight line .The electric intensity at `0` isA. `(Q)/(4piepsilon_(0)n^(2))`B. `(Q)/(4piepsilon_(0)n)`C. InfinityD. `(nQ)/(4piepsilon_(0))` |
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Answer» Correct Answer - 4 `E=(1)/(4piepsilon_(0))[(Q_(1))/(x_(1)^(2))+Q_(2)/(x_(2)^(2))+….+(Q_(n))/(x_(n)^(2))]` |
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| 165. |
Two point charges `-q` and `+q` are located at points `(0,0-a)` and `(0,0,a)` resepctively.The electric potential at point `(0,0,z)`is `(Z gt a)`A. `(qa)/(4piepsilon_(0)z^(2))`B. `(q)/(4piepsilon_(0)a)`C. `(2qa)/(4piepsilon_(0)(z^(2)-a^(2)))`D. `(2qa)/(4piepsilon_(0)(z^(2)+a^(2)))` |
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Answer» Correct Answer - 3 the distance of point `P` from charge `+Q` is `r_(1)=z-a` and from, charge `-q` is `r_(2)=z+a` Potnetial at `P` is `(1)/(4piepsilon_(0))((q)/(r_(1))-(q)/(r_(2)))=(1)/(4piepsilon_(0))(2qa)/((z^(2)-a^(2)))` |
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| 166. |
An electric dipole is formed two particles fixed at the ends of a light rad of length `I`. The mass of each particle is `m` and charges are `-q` and `+q`The system is suspended by a torsionless thread in an electric field of intensity `E` that the dipole axis is parallel to the field if it is slightly displaced the period of angular motion isA. `(1)/(2pi)sqrt((2qE)/(ml))`B. `2pisqrt((2ml)/(qE))`C. `2pisqrt((ml)/(2qE))`D. `(1)/(2pi)sqrt((ml)/(4qE))` |
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Answer» Correct Answer - 3 `t=PEsintheta,tau=lalpha,Ialpha=PEsintheta` `I="moment of inertia" =`(ml^(2))/(2)` `:. "Time period"=`2pisqrt((I)/(PE))` |
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| 167. |
Two concentric spherical shells have radii `0.15m` and `0.3m`. The inner part is given a charge `-6xx10^(-2)mu C` and electron escapes from the inner sphere with negligible speed ,Assuming that the region between the sphere is vaccum ,the speed with which it wil strike the outer sphere is `x xx 10^(6)ms^(-1)` .Find the value of `x`? `(e)/(m)"for electron"=1.76xx10^(11)Ckg^(-1)` |
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Answer» Correct Answer - 8 Gain `K.E =e=(V_(2)-V_(1))` `rArr (1)/(2)mv^(2) =e(1)/(4piepsilon_(0))q((1)/(a)-(1)/(b))` `V^(2)=(2e)/(m)[(1)/(4piepsilon_(0))q(ab)/(b-a)]` ` rArr v=8xx10^(6)ms^(-1)=x xx10^(6) x=8` |
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| 168. |
A particle having a charge of `q=8.85muC` is placed on the axis of a circular ring of radius `R=30cm` at a point `P` at a distance of `a=40cm` from the centre of the ring.The electric flux passing through the ring is `x xx 10^(5)N//C` .Find the value of `x`? |
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Answer» Correct Answer - 1 `phi_(E)= phi_(E)=(q)/(2epsilon_(0))(1-(a)/(sqrt(R^(2)+a^(2))))` ` rArr phi_(E)=(8.85xx10^(-6))/(2(8.85xx10^(-12)))(1-(40)/(50))` `rArr phi_(E)=(10^(6))/(2)(1-(4)/(5))` `rArr phi_(E)=1xx10^(5)=x xx 10^(5)x=1` |
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| 169. |
Two equal negative charges `-q` are fixed at points `(0, -a)` and `(0,a)` on y-axis. A poistive charge Q is released from rest at point `(2a, 0)` on the x-axis. The charge Q willA. execute simple harmonic motion about the origin.B. move to the origin and remain at restC. move to infinityD. execute oscillartory but not simple harmonic motion. |
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Answer» Correct Answer - 4 `F=(1)/(4piepsilon_(0)epsilon_(r))(q_(1)q_(2))/(d^(2))` |
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| 170. |
The electric field on two sides of a large charged plate is shown in figure. The charge density on the plate in SI units is given by `(epsilon_0` is the permittivity of free space in SI units). A. `2epsilon_(0)`B. `4epsilon_(0)`C. `10epsilon_(0)`D. zero |
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Answer» Correct Answer - 2 Electric field due to plate =`(sigma)/(2epsilon_(0))` |
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| 171. |
Assertion. A metallic shield in the form of a hollow shell, can be built to block an electric field. Reason. In a hollow spherical shell, the electric field inside is not zero at every point.A. Both `A` and `R` are true and `R` is the correct explanaation of `A`B. Both `A` and `R` are true and `R` is not the correct explanation of `A`C. `A` is true and `R` is falseD. `A` is false and `R` is true. |
| Answer» Correct Answer - 1 | |
| 172. |
A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P a distance `R/2` from the centre of the shell isA. `(2Q)/(4piepsilon_(0)R)`B. `(2Q)/(4piepsilon_(0)R)-(2q)/(4piepsilon_(0)R)`C. `(2Q)/(4piepsilon_(0)R)+(2q)/(4piepsilon_(0)R)`D. `((q+Q))/(4piepsilon_(0))(2)/(R)` |
| Answer» Correct Answer - 3 | |
| 173. |
Three concentric spherical conductors A, B and C of radii R, 2R and 4R respectively. A and C is shorted and B is uniformly charged Charge on conductor A is A. `q_(A)=q_(C)=-Q//3`B. `q_(A)=(Q)/(3),q_(C)=(2Q)/(3)`C. `q_(A)=(Q)/(3),q_(C)=(Q)/(3)`D. `q_(A)=-(Q)/(3),q_(C)=(Q)/(3)` |
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Answer» Correct Answer - D `V_(A)=V_(C)` `(1)/(4piepsilon_(0))[(q_A)/(R)+(q_(B))/(2R)+q_(C)/(4R)]=(1)/(4piepsilon_(0))[(q_(A)+q_(B)+q_(C))/(4R)]` `4q_(A)+2q_(B)+q_(C)=q_(A)+q_(B)+q_(C)` `q_(A)=-(Q)/(3) q_(C)=(Q)/(3)` |
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| 174. |
The electrostatic potential inside a charged spherical ball is given by `phi = ar^(2) + b`, where r is distance from the center of the ball, a and b are constants. Calculate the charge density inside the ball. |
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Answer» Here ,`phi=ar^(2)+b,As phi=ar^(2)+b` `:.ointE.ds=(q)/(epsilon_(0)),-2ar.4pir^(2)=(q)/(epsilon_(0))rArr q=-89epsilon_(0)pir^(2)` `p=(q)/((4)/(3)pir^(3)) ,rArr p=-6aepsilon_(0)` |
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| 175. |
A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point `(0, 0, z_0)` where `z_0gt0`. Then the motion of P isA. Periodic ,for all value of `z_(0)`satisfying `0 lt z_(0) oo`B. Simple harmonic for all value of `z_(0)` satisfying `0 lt z_(0) le R`C. Approximately simple harmonic ,provided `z_(0) lt lt R`D. Such that `P` crosses `O` and continues to move along the negative "z-axis" towards `z-=-oo` |
| Answer» Correct Answer - 1 | |
| 176. |
The p.d between two plates separated by a distance of `1 mm is 100V`.The force on an electron placed in between the plates isA. `10^(5)N`B. `1.6xx10^(-24)N`C. `1.6xx10^(-14)N`D. `1.6xx10^(-19)N` |
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Answer» Correct Answer - 3 `F=Eq=(Vq)/(d)` |
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| 177. |
An electron travelling from infinity with velocity `v` into an electric field due to two stationary electrons separated by a distance of `2m`.if it comes to rest when it reaches the mid point of the line joining the stationary electrons.The initial velocity `v` of the electron isA. `16`B. `32`C. `16sqrt(2)`D. `32sqrt(2)` |
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Answer» Correct Answer - 2 `(1)/(2)mv^(2)=(1)/(4piepsilon_(0))(q_(1)q_(2))/(r )` |
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| 178. |
The distance between plates of a parallel plate capacitor is `5d`. Let the positively charged plate is at `x=0` and negatively charged plate is at `x=5d`. Two slabs one fo conductor and other of a dielectric of equal thickness `d` are inserted between the plates as shown in figure. Potential verus distance graph will look like: A. B. C. D. |
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Answer» Correct Answer - 2 `V_(1)-V_(2)=kq((1)/(r_(1))-(1)/(r_(2)))` `r_(2)-r_(1)=((V_(1)-V_(2))r_(1)r_(2))/(kq)`, but (r_(2)-r_(1))=t` `:.ralpha_(1)r_(2)` If P.D is constant then `(r_(2)-r_(1))=t` |
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| 179. |
An electron travelling from infinity with velocity `v` into an electric field to due to two stationsly electrons separated by a distance of `2m`.if it comes to rest when it reaches themid point of the line joinining the stationary electrons.The initial velocity `v` of the electron isA. `16m//s`B. `32m//s`C. `16sqrt(2)m//s`D. `32sqrt(2)m//s` |
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Answer» Correct Answer - 2 `E=-(dv)/(dx)E` inside the conductor is zero. |
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| 180. |
Two point charges q and -q are separated by a distance 2a. Evaluate the flux of the electric field strength vector across a circle of radius R. A. `(q)/(2epsilon_(0)){1-(a)/(sqrt(a^(2)+r^(2)))}`B. `(q)/(epsilon_(0)){1-(a)/(sqrt(a^(2)+r^(2)))}`C. `(2q)/(epsilon_(0)){1-(a)/(sqrt(a^(2)+r^(2)))}`D. zero |
| Answer» Correct Answer - 4 | |
| 181. |
A dielectric slab of thickness `d` is inserted in a parallel plate capacitor whose negative plate is at `x=0` and positive plate is at `x = 3d`. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from `0` to `3d(1998)`.A. the magnitude of the electric field remains the sameB. the direction of the filed remains the sameC. the electric potential increases continuouslyD. the electric potential increases at first,then increases and again increase. |
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Answer» Correct Answer - 2 Potential due to small element `p` at the centre `v=intdv=intk(lambda)/(r)dl=(1)/(4piepsilon_(0))(lambda)/(r)intdl` `dv=K.lambda(dl)/(r),(1)/(4piepsilon_(0))(lambda)/(r)pir rArr (lambda)/(4epsilon_(0))` |
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| 182. |
Choose the wrong statementA. work done in moving a charge on equipotential surface is zeroB. Electric lines of force are always normal to an equipotnetial surfaceC. when two like charges are brought nearer, then electrostatic potential energy of the system gets decreased.D. Electric lines of force diverge form positive charge and converage towards negative charge. |
| Answer» Correct Answer - 3 | |
| 183. |
A dielectric slab of thickness `d` is inserted in a parallel plate capacitor whose negative plate is at `x=0` and positive plate is at `x = 3d`. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from `0` to `3d(1998)`.A. the magnitude of the electric field remains the sameB. the direction of the filed remains the sameC. the electric potential increases continuouslyD. the electric potential decreases at first,then increases and again decrease. |
| Answer» Correct Answer - 2 | |
| 184. |
A point charge `q` is placed at orign Let ` vec E_A, vec E_B ` and `vec E_C ` be the electric field at theree points ` (A) ( 1, 2, 3 ), B (1,1,-1)` and ` C( 2,2,2) ` due to charge `q`. Then- [i] ` vec E_A _|_ vec E_B`[ii] `|vec E_B |=4|vecE_(c)|`select the correct alternative .A. `vec(E)_(A)botvec(E)_(B)`B. `vec(E)_(A)||vec(E)_(B)``C. `|vec(E)_(B)|=4|vec(E)_(C)|`D. `|vec(E)_(B)|=8|vec(E)_(C)|` |
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Answer» Correct Answer - C `bar(E)_(A)` is along `bar(OA)` and `bar(E)_(B)` is along bar(OB) `bar(OA)=hat(i)_2hat(j)+3hat(k),bar(OB)=hat(i)+hat(j)-hat(k)` since `vec(OA).vec(OA)=0` so `vec(E)_(A) _|_vec(E)_(B)` further `|vec(E)|lt (1)/(r^(2))|vec(OC)|=2|vec(OB)|,|vec(E)_(B)|=4|vec(E)_(C)|` |
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