1.
Two identical conducting spheres A and B carry charges +8Q and −2Q respectively. They are first brought into contact and separated. Then sphere A is connected momentarily to earth and separated. If the final charge on B remains unchanged after earthing A, find the final charge on A and explain the physical reason.
When identical spheres touch, total charge = +8Q − 2Q = +6Q. Each gets +3Q.
After separation, B has +3Q. Sphere A is earthed, so its excess charge flows to earth and its final charge becomes 0.
B remains +3Q because it is isolated.
Final charge on A = 0.
2.
A charge +q is placed at the centre of a cube. The electric flux through one face of the cube is Φ. If the same charge is shifted slightly away from the centre but remains inside the cube, what happens to the total flux through the cube and the flux through one face?
Total flux through the cube remains q/ε0 by Gauss law because charge is still inside the closed surface.
But flux through one face will not remain equal to Φ because symmetry is broken.
Only total flux is fixed; individual face flux changes.
3.
Two point charges +Q and +4Q are fixed at points A and B separated by distance d. At what point on the line AB should a charge q be placed so that the net electric force on q is zero?
Let the point be at distance x from +Q. Then distance from +4Q is d − x.
For zero force:
kQq/x² = k(4Q)q/(d − x)²
1/x² = 4/(d − x)²
d − x = 2x
3x = d
x = d/3.
So the charge must be placed between the charges, at distance d/3 from +Q.
4.
A uniformly charged thin ring of radius R carries total charge Q. A charge q is placed on its axis at distance x from the centre. If x ≪ R, show that the force on q is approximately proportional to x.
Electric field on axis of ring:
E = kQx/(R² + x²)3/2
Since x ≪ R, R² + x² ≈ R².
Therefore E ≈ kQx/R³.
Force F = qE = kQqx/R³.
Hence F ∝ x.
5.
Three charges +q, +q and −q are placed at the vertices of an equilateral triangle of side a. Find the magnitude of net electric force on the negative charge.
The negative charge is attracted by both positive charges.
Each force has magnitude F = kq²/a².
Angle between the two forces is 60°.
Resultant:
Fnet = √(F² + F² + 2F²cos60°)
= √(2F² + F²)
= √3F
= √3 kq²/a².
6.
Two charges +q and −q are placed at x = −a and x = +a respectively. Find the electric field at a point P on the y-axis at distance y from the origin.
Distance of P from each charge is r = √(a² + y²). Vertical components cancel and horizontal components add.
E = 2kqa / (a² + y²)3/2, directed from +q towards −q.
7.
A charge Q is uniformly distributed over a thin spherical shell of radius R. Find electric field at a point at distance r from the centre for r < R and r > R.
For r < R, charge enclosed = 0, so E = 0.
For r > R, shell behaves like point charge at centre.
E = kQ/r².
8.
A point charge q is placed at the centre of a conducting spherical shell. What is the induced charge on the inner and outer surfaces of the shell?
To make electric field zero inside the conducting material, inner surface gets −q.
If shell was initially neutral, outer surface gets +q.
Therefore inner charge = −q and outer charge = +q.
9.
Two charges +Q and −4Q are separated by distance d. Find the point on the line joining them where electric field is zero.
For unlike charges, zero field cannot lie between them. It lies outside on the side of smaller charge +Q.
Let distance from +Q be x away from −4Q.
kQ/x² = k4Q/(d + x)²
d + x = 2x
x = d.
So point is at distance d from +Q on the side opposite to −4Q.
10.
A dipole of moment p is placed in a uniform electric field E making angle θ with the field. Find torque and potential energy of the dipole.
Torque on dipole: τ = pE sinθ.
Potential energy: U = −pE cosθ.
Stable equilibrium occurs at θ = 0° and unstable at θ = 180°.
11.
Electric field due to an electric dipole at an axial point is Ea and at an equatorial point at the same distance is Ee. Find Ea : Ee for large distance.
For large distance r:
Eaxial = 2kp/r³
Eequatorial = kp/r³
Therefore Ea : Ee = 2 : 1.
12.
A charged particle of mass m and charge q is released from rest in a uniform electric field E. Find its velocity after travelling distance s.
Force F = qE.
Acceleration a = qE/m.
Using v² = 2as:
v = √(2qEs/m).
13.
A charge q is placed at the corner of a cube. Find the electric flux through the cube.
Imagine 8 identical cubes joined so that charge q becomes at centre of a larger cube.
Total flux through large cube = q/ε0.
Flux through one small cube = q/(8ε0).
14.
A charge q is placed at the centre of one face of a cube. Find the flux through the cube.
Put another identical cube sharing that face. Then charge becomes inside the combined closed surface.
Total flux = q/ε0.
Flux through one cube = q/(2ε0).
15.
An infinite line charge has linear charge density λ. Find electric field at distance r from the line using Gauss law.
Choose cylindrical Gaussian surface of radius r and length L.
Flux = E(2πrL).
Charge enclosed = λL.
E(2πrL) = λL/ε0.
E = λ/(2πε0r).
16.
An infinite plane sheet has uniform surface charge density σ. Find electric field on either side of the sheet.
Use a pillbox Gaussian surface.
Flux = 2EA.
Charge enclosed = σA.
2EA = σA/ε0.
E = σ/(2ε0).
17.
Two infinite parallel sheets carry surface charge densities +σ and −σ. Find electric field between the sheets and outside them.
Field due to each sheet = σ/(2ε0).
Between sheets, fields are in same direction: E = σ/ε0.
Outside sheets, fields cancel: E = 0.
18.
A uniformly charged solid sphere of radius R has total charge Q. Find electric field at distance r from centre for r < R.
Charge enclosed inside radius r: qenc = Q(r³/R³).
By Gauss law: E(4πr²) = qenc/ε0.
E = kQr/R³.
Field inside is proportional to r.
19.
Four equal charges +q are placed at the corners of a square of side a. Find the electric field at the centre.
Due to symmetry, fields due to opposite charges cancel pairwise.
Therefore net electric field at centre = 0.
20.
Three charges +q, +q and +q are placed at three corners of a square of side a. Find direction of electric field at the empty corner.
The two adjacent charges produce fields along perpendicular directions, each kq/a².
The diagonal charge produces field along diagonal with magnitude kq/(2a²).
Resultant is along the diagonal away from the square because of symmetry of adjacent charges plus diagonal contribution.
21.
A charge q is divided into two parts q₁ and q₂. For fixed separation r, what division gives maximum repulsive force between them?
Let q₁ = x and q₂ = q − x.
Force F = kx(q − x)/r².
Product x(q − x) is maximum when x = q/2.
Hence q₁ = q₂ = q/2.
22.
A small charge q of mass m is suspended by a string in a uniform horizontal electric field E. Find the angle made by the string with vertical.
Forces: weight mg downward, electric force qE horizontal, tension along string.
tanθ = qE/mg.
Therefore θ = tan−1(qE/mg).
23.
A point charge q is placed at distance d from the centre of a grounded conducting sphere. What is the qualitative nature of force on q?
Grounded conductor develops induced charge of opposite sign nearer to q.
Attraction from nearer opposite charge dominates.
Therefore force on q is attractive towards the sphere.
24.
Two identical pith balls each of mass m and charge q are suspended from same point by strings of length l. If separation between balls is x and x ≪ l, find approximate x.
For each ball: tanθ = F/mg.
For small angle, tanθ ≈ sinθ ≈ x/(2l).
Coulomb force F = kq²/x².
x/(2l) = kq²/(mgx²).
x³ = 2lkq²/mg.
x = (2lkq²/mg)1/3.
25.
A dipole is placed in a non-uniform electric field. Does it experience force, torque, or both?
In uniform electric field, dipole experiences torque but no net force.
In non-uniform electric field, forces on +q and −q are unequal.
Therefore it can experience both net force and torque.
26.
Electric field in a region is given by E = αx î. Find charge density ρ in the region.
Using differential form of Gauss law: ∇·E = ρ/ε0.
Ex = αx.
∂Ex/∂x = α.
Therefore ρ = ε0α.
27.
Electric field in space is E = A(x î + y ĵ + z k̂). Find the charge enclosed in a cube of side a whose one corner is at origin and edges are along axes.
∇·E = A + A + A = 3A.
Charge density ρ = ε03A.
Volume of cube = a³.
Charge enclosed = 3ε0Aa³.
28.
A charge +Q is placed at the centre of a regular tetrahedron. Find flux through one face.
Total flux through tetrahedron = Q/ε0.
Since charge is at centre, symmetry distributes flux equally through 4 faces.
Flux through one face = Q/(4ε0).
29.
A point charge q is placed inside an irregular closed surface. If the surface is deformed without crossing the charge, what happens to total electric flux?
Total flux depends only on enclosed charge, not on shape or size of surface.
Since charge remains inside, flux remains q/ε0.
30.
Two charges +q and +q are fixed at x = −a and x = +a. Find the nature of equilibrium of a charge +Q placed at origin for small displacement along x-axis.
At origin net force is zero.
If +Q is displaced slightly towards +a, repulsion from right charge increases and from left decreases, giving net force towards left, opposite displacement.
Hence equilibrium is stable along x-axis.
But in perpendicular direction it is unstable.
31.
Two identical charges q are placed at distance 2a. A third charge Q is placed at the midpoint. What should be the sign of Q for the system to be in equilibrium?
The two end charges repel each other.
To balance this repulsion, the middle charge must attract each end charge.
Therefore Q must have opposite sign to q.
32.
Find the value of Q in the previous problem if the end charges are +q and separation between them is 2a.
Force on left +q due to right +q: kq²/(2a)² = kq²/(4a²), leftward.
Force due to middle Q at distance a must be rightward: kq|Q|/a².
Equate: kq|Q|/a² = kq²/(4a²).
|Q| = q/4.
Since Q is opposite sign, Q = −q/4.
33.
A charge Q is uniformly distributed over a ring of radius R. Find the point on the axis where electric field is maximum.
E = kQx/(R² + x²)3/2.
For maximum, differentiate with respect to x.
dE/dx = 0 gives R² − 2x² = 0.
x = R/√2.
Field is maximum at distance R/√2 from centre.
34.
A uniformly charged semicircular arc of radius R has total charge Q. Find electric field at the centre.
Linear charge density λ = Q/(πR).
For element dq, dE = kdq/R².
Horizontal components cancel, vertical components add.
E = 2kλ/R = 2kQ/(πR²).
Direction is along symmetry axis away from the arc for positive charge.
35.
A thin circular disc has uniform surface charge density σ and radius R. Find electric field near its centre on axis at distance x where x ≪ R.
Field on axis of disc:
E = σ/(2ε0) [1 − x/√(x² + R²)].
For x ≪ R, √(x² + R²) ≈ R.
E ≈ σ/(2ε0) [1 − x/R].
36.
A charge q moves in a circular path around a fixed charge Q due to electrostatic attraction. Find its speed if radius of orbit is r.
Electrostatic force provides centripetal force:
kQq/r² = mv²/r.
v² = kQq/(mr).
v = √(kQq/mr).
37.
Two small spheres of charges q and 4q are separated by distance r. They are touched and separated again to same distance. Find ratio of final force to initial force.
Initial force Fi = k(q)(4q)/r² = 4kq²/r².
Total charge = 5q. Identical spheres after contact each get 5q/2.
Final force Ff = k(5q/2)²/r² = 25kq²/(4r²).
Ratio Ff/Fi = 25/16.
38.
Two charges +q and −2q are placed at distance d. Find the location where electric potential is zero on the line joining them.
Potential zero means kq/r₁ − k2q/r₂ = 0.
So r₂ = 2r₁.
Between charges, if distance from +q is x, distance from −2q is d − x.
d − x = 2x → x = d/3.
Another point also exists outside on side of +q.
39.
A charge q is placed at origin. A spherical Gaussian surface of radius R is drawn around it. If radius is doubled, how does electric flux change?
Electric field decreases as 1/R², but area increases as R².
Total flux remains q/ε0.
Therefore flux does not change.
40.
A dipole is placed in uniform electric field. For which angle is torque maximum and for which angle is potential energy minimum?
Torque τ = pE sinθ, maximum when sinθ = 1, so θ = 90°.
Potential energy U = −pE cosθ, minimum when cosθ = 1, so θ = 0°.
41.
A charge q is placed at centre of a sphere. The electric flux through a circular cap of the sphere depends on what geometrical quantity?
Flux from a point charge through any open surface depends on solid angle Ω subtended by that surface.
Φ = qΩ/(4πε0).
Therefore it depends on solid angle of the cap at the charge.
42.
A charged particle enters a uniform electric field perpendicular to its initial velocity. What type of path does it follow?
Initial horizontal velocity remains constant.
Electric field produces constant acceleration perpendicular to velocity.
This is like projectile motion.
Therefore path is parabolic.
43.
A proton and an electron are placed in the same uniform electric field. Compare magnitudes of force and acceleration.
Force magnitude F = eE for both proton and electron.
But acceleration a = F/m.
Since electron mass is much smaller, electron acceleration is much larger.
Directions of force are opposite.
44.
An electric dipole has charges ±q separated by 2a. At a point on axial line far away at distance r from centre, find approximate field.
Dipole moment p = q(2a).
At far axial point:
E = 2kp/r³.
Therefore E = 2k(2qa)/r³ = 4kqa/r³.
45.
At an equatorial point of a dipole, why is electric field opposite to dipole moment?
Dipole moment points from −q to +q.
At equatorial point, resultant of fields due to +q and −q has components along direction from +q to −q.
Hence field is opposite to dipole moment.
46.
A closed surface encloses charges +q, −2q, +3q and −q. Find total electric flux through the surface.
Net enclosed charge = q − 2q + 3q − q = q.
By Gauss law, flux = q/ε0.
47.
If electric field lines are closer in region A than in region B, what can be concluded about electric field strength?
Density of electric field lines represents magnitude of electric field.
Since lines are closer in region A, electric field is stronger in region A than in region B.
48.
Why can two electric field lines never intersect?
Tangent to an electric field line gives direction of electric field.
If two lines intersect, electric field at that point would have two directions, which is impossible.
Hence field lines never intersect.
49.
A positive charge is released from rest in a non-uniform electric field. Will it always move along an electric field line?
Initially acceleration is along electric field.
But as charge moves, field direction may change. Due to inertia, velocity need not remain tangent to field line.
Therefore it does not always follow an electric field line exactly.
50.
A charge q is placed inside a neutral conducting cavity. If the conductor is isolated, what is the charge induced on inner and outer surfaces?
To keep electric field zero inside conducting material, inner surface gets −q.
Since conductor is neutral overall, outer surface gets +q.
Therefore inner surface charge = −q and outer surface charge = +q.