Q: A charge q0 is distributed uniformly on a ring of radius R. A sphere of equal radius R is constructed with its centre on the circumference of the ring. Find the electric flux through the surface of the sphere.

Charge on arcPQ of ring is given asVPQ=3q0This is also the charge enclosed in the sphere so flex through surface of sphere is given asϕ=ε0qin ⇒ϕ=ε0(q0/3)=3ε0q0

Question 1

(i) Find flux through the hemispherical cup due to the charge q placed as shown in figure.

Accepted Answer

Half of the flow originated by charge q will come out of hemispherical surface of the cup.[2ϵ0​q​]

Question 2

Two point charges q and −q are separated by a distance 2l. Find the flux of the electric field strength vector across a circle of radius R located symmetrically in a plane normal to line joining the charges.

Accepted Answer

Electric field strength due to the point charges at the location of elemental ring is given as
E=(l2+r2)2Kq​
⇒Ecosθ=2πϵ0​1​⋅(l2+r2)3/2ql​
Total flux through circle is given as
ϕ=∫0R​E×2πrdr
⇒ϕ=4πϵ0​ql​∫0R​(r2+l2)3/22πrdr​
Substituting, r2+l2=z
also 2rdr=dz
⇒ϕ=4∈0​ql​∫0R​z−3/2dz
⇒ϕ=4∈0​ql​∣∣​−21​z−1/2​∣∣​0R​
⇒ϕ=2∈0​−ql​∣∣​(r2+l2)−1/2∣∣​0R​
⇒ϕ=2ϵ0​−ql​∣∣​r2+l2​1​∣∣​0R​
⇒ϕ=2∈0​−ql​(R2+l2​1​−l1​)
⇒ϕ=ϵ0​ql​(l1​−R2+l2​1​)
Above result can be directly calculated by using the concept of solid angle. Students are advised to verify this result by using that method also.

Question 3

A charge q0​ is distributed uniformly on a ring of radius R. A sphere of equal radius R is constructed with its centre on the circumference of the ring. Find the electric flux through the surface of the sphere.

Accepted Answer

Charge on arcPQ of ring is given asVPQ​=3q0​​This is also the charge enclosed in the sphere so flex through surface of sphere is given asϕ=ε0​qin ​​⇒ϕ=ε0​(q0​/3)​=3ε0​q0​​

Question 4

An electric field given by E=4i^+3(y2+2)j^​ pierces a gaussian cube of side 1 m placed at origin such that one of its corners is at origin and rest of sides are along positive side of coordinate axis. Find the magnitude of net charge enclosed within the cube

Accepted Answer

We calculate the total electric flux coming out of the cube, which is given as
Along z axis as E2​=0 no flux is these
Along x axis as E= constant
⇒ϕin x​=ϕout x​
⇒ϕx​=0
For y=0 plane electric flux getting in the cube is
ϕin ​=∫E⋅dA=∫3(0+2)j^​⋅dA(−j^​)=6∫dA=−6
For y=1 plane flux coming out from the cube is
ϕout ​=∫E⋅dA=∫3(1+2)j^​⋅dA(j^​)=9∫dA=9
⇒ Using Gauss's laws for cube, we have
ϕnet ​=ϵ0​qenclosed ​​=3
⇒qenclosed ​=3∈0​

Question 5

A point charge q is located on the axis of a disc of radius R at a distance b from the plane of the disc as shown in figure-1.295. What should be the radius of the disc if onefourth of the total electric flux from the charge passes through the disc.

Accepted Answer

Using concept of solid angle the flux of Q which passes through disc is given as
ϕ=2ε0​Q​(1−cosθ)
Given that
ϕ=41​ϕtotal ​=4ε0​Q​
⇒4ε0​Q​=2ε0​Q​(1−cosθ)
⇒cosθ=21​
⇒θ=60∘
We use tan60∘=bR​
⇒R=btan60∘
⇒R=3​b

Question 6

A cube has sides of length L=0.2 m. It is placed with one corner at the origin as shown in figure-1.297. The electric field is uniform and given by E=(2.5i^−4.2j^​)N/C. Find the electric flux through the entire cube.

Accepted Answer

Given electric field is uniform in space we know that net flux from any closed surface in uniform electric field is zero.

Question 7

Find the electric flux through a cubical surface due to a point charge q placed (a) at centre of one face (b) corner of the cubical box as shown in figures below, in case (b) find the flux through each face of the cube.

Accepted Answer

(vi) If charge is kept at the centre of a face first we should enclose the charge by assuming an identical imaginary cube on other side as shown in figure below.
The total flux through both the cube surfaces is given as
ϕtotal ​=ϵ0​q​
Thus the flux through given cube is given as
ϕ=2ϵ0​q​

(b) If the charge is placed at the corner of a cube, seven more identical cubes are required to enclose the charge completely and symmetrically as shown in figure below.

The total flux through all the faces of 8 cubes is
ϕtotal ​=ϵ0​q​
Flux through each cube at one corner this charge is located is given as
ϕ=8ϵ0​q​
This flux emerging equally through three front faces of the cube. Other three faces; junction of which charge is placed have no flux. Therefore the flux through one of the faces is given as
=3q/8ϵ0​​=24ϵ0​q​

Question 8

For a spherically symmetrical charge distribution, electric field at a distance r from the centre of sphere is E=kr7r^, where k is a constant. What will be the volume charge density at a distance r from the centre of sphere?

Accepted Answer

We consider an elemental shell in the region of radius x and width dx as shown in figure.

If volume charge density in the shell is p(x) then we use Gauss's law for the surfaces of shell as
E2​(4π(x+dx)2)−E1​(4πx2)=ϵ0​ρ(x)4πx2dx​
⇒k(x+dx)7[4p(x+dx)2]−kx7(4πx2)=ϵ0​ρ(x)4πx2dx​
⇒4πk[(x+dx)9−x9]=ϵ0​ρ(x)4πx2dx​
⇒k[x9(1+x9dx​)−x9]=ϵ0​ρ(x)x2dx​
⇒9kx8dx=ϵ0​ρ(x)x2dx​
⇒ρ(x)=9k∈0​x6
⇒ρ(r)=9k∈0​r6

Question 9

A non-conducting spherical ball of radius R contains a spherically symmetric charge with volume charge density ρ=krn, where r is the distance from the centre of the ball and n is a constant. What should be n such that the electric field inside the ball is directly proportional to square of distance from the centre?

Accepted Answer

Using Gauss's law for a symmetric spherical surface of radius r we have
∮E⋅dA=ϵ0​∫0R​ρ4πx2dx​
⇒E×4πr2=ϵ0​k∫rn×4πr2dr​=ϵ0​4πk​n+3rk+3​
⇒E=(n+3)ϵ0​k​(rn+1)
Given that Eαr2 so we use
n+1=2
⇒n=1

Question 10

An infinite, uniformly charged sheet with surface charge density σ cuts through a spherical Gaussian surface of radius R at a distance x from its center, as shown in the figure-1.298. The electric flux ϕ through the Gaussian surface is :

Accepted Answer

(x) The enclosed section of sheet within the spherical surface is a disc of radius r which in given as
r=R2−x2​
Thus flux coming out from spherical surface is given as
ϕ=∈0​σ[π(R2−x2)]​

Question 11

A charge Q is placed at each of two opposite corners of a square and a charge −q is placed at each of the remaining two corners. If the resultant force on Q is zero, how are Q and q related.

Accepted Answer

The different forces on C due to other charges are shown in figure. Let a be the side of the square. The force on due to A is given asFAC​=2a2KQ2​..(1)If the resultant force on C is to be zero, the force on C due to D and B must be along CD and CB respectively so charge q must be negative.The resultant force FCB​ and FCD​ along CA are given asFC​=a2KQ2​⋅2​1​+a2KQq​⋅2​1​⇒FC​=2​⋅a2KqQ​...(2)As resultant force on C is zero, we have from equations-(1) and (2)2​⋅a2KqQ​=2a2KQ2​⇒Q=−22​q

Question 12

Two balls of the same radius and weight are suspended on threads so that their surface are in contact. A charge of q0​=4×10−7C is given to the balls which makes them repel each other and diverge to an angle of 60∘. Find the mass of the balls if the distance of balls from the point of suspension to the centre of ball is 20 cm. Find the density of the material of the balls if the angle of divergence becomes 54∘ when the balls are immersed in kerosene of density 800 kg m−3. dielectric constant of kerosene is εr​=2.

Accepted Answer

Let 2α be the angle between the two threads as shown in figure, F is the resultant force, W is weight of the ball and F1​ is Coulomb's force of repulsion.From above figure we haver=2lsinαFor equilibrium of ball we haveF1​=Wtanα=(4πε0​1​)(4l2sin2α)(q02​/4)​⇒W=(4πε0​1​)16l2sin2αtanαq02​​...(1)⇒W=0.0156 NThus mass of ball is given asm=9.80.0156​=1.592 gWhen the ball is immersed in kerosene, we have if FB​ is the buoyant force on ball, we useW−FB​=(4πε0​K1​)16l2sin2α2​tanα2​q02​​...(2)and W−FB​=(ρ1​−ρ2​)Vg...(3),Where ρ1​ and ρ2​ are the densities of ball and kerosene respectively. From equations-(1), (2) and (3) we haveWW−FB​​=Ksin2α2​tanα2​sin2αtanα​=ρ1​ρ1​−ρ2​​=1−ρ1​ρ2​​⇒ρ1​=Ksin2α2​tanα2​−sin2αtanαρ2​Ksin2α2​tanα2​​Substituting the values we haveρ1​=2559 kg/m3

Question 13

Two equal positive point charges are separated by a distance 2a. A point test charge is located in a plane which is normal to the line joining these charges and midway between them.(a) Calculate the radius r of the circle of symmetry in this plane for which the force on the test charge has a maximum value.(b) What is the direction of this force, assuming a positive test charge ?

Accepted Answer

Let C be the point on the circle of symmetry of radius r where a test charge q0​ is placed as shown in figure. Now the force F exerted on this test charge by either charge Q is given byF=(a2+r2)KQq0​​The resultant force on the test charge will be given asFR​=2Fcosθ⇒FR​=(a2+r2)2KQq0​​⋅cosθ⇒FR​=(a2+r2)2KQq0​​⋅(a2+r2)1/2r​⇒FR​=(a2+r2)3/22KQq0​r​...(1)For the force to be maximum, we havedrdFR​​=0⇒drdFR​​=2KQq[(a2+r2)3(a2+r2)3/2⋅1−23​(a2+r2)1/2⋅2r⋅r​]=0⇒(a2+r2)3/2=23​(a2+r2)1/2⋅2r2⇒(a2+r2)=3r2⇒a2=2r2⇒r=a/2​As shown in figure, the force is radial and away form the centre of the circle.

Question 14

Consider a fixed charge Q and another charge q is placed at a distance x0​ from Q on a smooth plane surface. Find the velocity of charge q as a function of x.

Accepted Answer

Let any instant the separation between them is x. The force between them isF=4πϵ0​1​x2Qq​Acceleration of charge q is given asa=4πϵ0​m1​x2Qq​⇒dxvdv​=4πϵ0​m1​Qqx−2⇒∫0v​vdv=4πϵ0​m1​Qq∫x0​x​x−2dx⇒2v2​=4πϵ0​m1​Qq∣∣​−x1​∣∣​x0​x​⇒v=[2πϵ0​mQq​{x0​1​−x1​}]1/2

Question 15

A positive point charge 50μC is located in the plane xy at the position vector r0​=2i^+3j^​, where i^ and j^​ are the unit vectors of the x and y axis. Find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i^−5j^​. Here r0​ and r are expressed in meter.

Accepted Answer

We know that electric field vector due to charge q is given asE=4πϵ0​1​∣r−r0​∣3q(r−r0​)​⇒E=9×109×(50×10−6)∣(8i^−5j^​)−(2i^+3j^​)∣3[(8i^−5j^​)−(2i^+3j^​)]​⇒E=9×109×50×10−6∣(6i^−8j^​)∣3(6i^−8j^​)​⇒E=4.5×105103(6i^−8j^​)​⇒E=4.5×102(6i^−8j^​)⇒⇒​E=2.7i^−3.6j^​kV/m∣E∣=4.5kV/m​

Question 16

Four point charges, each of charge +q, are rigidly fixed at the four corners of a square planar soap film of side ' a '. The surface tension of the soap film is σ. If the he system of charges and planar film are in equilibrium, then side of square is given as a=k[σq2​]1/N, find the values of k and N.

Accepted Answer

F2​=4πε0​(2​a)2q2​=4πε0​×2a2q2​Net force on side AB of the film isF=2F1​+2F2​cos45⇒F=4πε0​a22q2​+4πε0​22​a22q2​⇒F=4πε0​a2q2​(2+2​1​)Force on AB due to surface tension =2σa.4πε0​a2q2​(2+2​1​)=2σa⇒a=[4πϵ0​1​(1+22​1​)⋅σq2​]1/3...(1)Given that a=k(σq2​)1/N...(2)Comparing equation (1) and (2), we have N=3and k=[4πϵ0​1​(1+22​1​)]1/3

Question 17

Two identical beads each having a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, non-conducting walls, the beads move, and at equilibrium they are a distance R apart (figure-1.42). Determine the charge on each bead.

Accepted Answer

Figure shows FBD of ball B.For equilibrium of ball B we use Lami's theorem which gives.  sin(90+30∘)mg​=sin(90∘+60∘)F​⇒F=3​mg​⇒4πε0​1​R2(q)(q)​=3​mg​⇒q=3​4πε0​mgR2​​

Question 18

A block of mass m with a positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure. The distance of the block from the wall is d. A horizontal electric field E is switched on in the space in rightward direction. Consider elastic collisions with the wall, find the time period of resulting oscillatory motion of the block. Analyse the motion and state if it is a simple harmonic motion.

Accepted Answer

Here acceleration of block is given asa=mqE​Time taken by block to reach wall is given as t=qE2dm​​Velocity at the time of impact isv=2ad​⇒v=m2qEd​​When the block will rebound time taken by block in coming to rest is same is the time it takes in reading the wall as accelerations and initial velocity is same. Thus time period of oscillations of block is twice the time it takes in reading the wall which is given asT=2t=2qE2md​​Since the restoring force is independent of x, the displacement from mean position, this is not a simple harmonic motion.

Question 19

Two horizontal parallel conducting plates are kept at a separation d=1.5×10−2 m apart one above the other in air as shown in figure-1.61. The upper plate is maintained at a positive potential of 1.5kV while the other plate is earthed which maintains it at zero potential. Calculate the number of electrons which must be attached to a small oil drop of mass m=4.9×10−15 kg between the plates to maintain it at rest. Consider density of air is negligible in comparison with that of oil. If the potential of above plate is suddenly changed to −1.5kV, what will be the initial acceleration of the charged drop? Also calculate the terminal velocity of the drop if its radius is r=5.0×10−6 m and the coefficient of viscosity of air is η=1.8×10−5 N−s/m2.

Accepted Answer

For the equilibrium, the drop must have negative charge q, so that its weight is balanced by the electric force given as.qE=mg..(1)⇒q=Emg​The electric field is given asE=dV​=1.5×10−21.5×103​=105 V/m⇒q=105(4.9×1015)×(9.8)​⇒q=4.8×10−19CIf n is the number of electrons on the oil drop, then we haven=eq​=1.6×10−194.8×10−19​⇒n=3When polarity of plates is reversed, the electric force also acts in downward direction, and so net force on drop isF=mg+qE...(2)From equations-(1) and (2), we getF=2mgInitial acceleration of the dropa=mF​=m2mg​=2 gAs drop accelerates in downward direction, its velocity increases and hence viscous force increases in upward direction. At certain moment the net force on the drop becomes zero and thereafter drop will move with constant velocity. If v is the terminal velocity, then by Stoke's rule we have2mg=6πηvv⇒v=3πηrmg​⇒v=3π×(1.8×10−5)×(5.0×10−6)(4.9×10−15)×9.8​⇒v=5.7×10−5 m/s

Question 20

A uniform electric field of intensity E=106 V/m exist in vertically downwards direction in a region. A particle of mass m=0.01 kg and charge q=10−6C is suspended by an inextensible thread of length l=1 m. The particle is displaced slightly from its mean position and released. Calculate the time period of its oscillation. What minimum velocity should be given to the particle at rest from its equilibrium position so that it completes a full circle in vertical plane? Calculate the maximum and minimum tension in the thread in its circular motion in vertical plane.

Accepted Answer

The situation described in question is shown in figure.Acceleration a due to electric field is given asa=mqE​=0.0110−6×106​=100 m/s2Thus effective acceleration would be given asg′=g+mqE​=9.8+100=109.8 m/s2The time period of particle would be given asT=2π(g′l​)​=2π(109.81​)​⇒T=0.6 sFrom figure-127, at point A forces on particle are given aslmvA2​​=TA​−mg′=TA​−(qE+mg)...(1)At point B, we havelmvB2​​=TB​+(qE+mg)With minimums speed at A, to complete the circle at point B tension TB​ should be zero, thus we havelmvBmin2​​=qE+mg...(2)Using work energy theorem at points A and B, we have21​mvA2​−2mgl−2gEl=21​mvB2​+UB​...(3)From equation-(2) and (3), we getvA2​=m5(qE+mg)l​=5g′l=5×109.8×1⇒vA​=23.42 m/s and TA​=(qE+mg)+lmvA2​​⇒TA​=6(qE+mg)=6(1+0.098)⇒TA​=6.588 N

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Solutions for Physics Galaxy Electrostatics and Current Electricity Vol 3 A by Ashish Arora

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