Question 1

What is the SI unit of the permittivity of free space, ε₀?

Accepted Answer

coulomb²/newton-metre². Coulomb's law F = q^2/(4*pi*eps0*r^2) gives eps0 = q^2/(F r^2). The units are coulomb^2/(newton x metre^2) = C^2 N^-1 m^-2, i.e. coulomb^2/newton-metre^2.

Question 2

What is the SI unit used for electric flux?

Accepted Answer

volt metre. Electric flux = E . A. Since electric field has units V/m and area m^2, flux has units (V/m)(m^2) = V m (volt metre), equivalently N m^2 C^-1. The correct unit is volt metre.

Question 3

A constant electric field E is directed along the positive x-axis. If a charge of 0.5 C is displaced by 2 m along a line inclined at 60° to the x-axis, and the work done in this process is 10 J, what is the magnitude of the electric field?

Accepted Answer

20 V m⁻¹. W = q*E*d*cos(theta): 10 = 0.5*E*2*cos60 = 0.5*E. So E = 20 V/m.

Question 4

An electric dipole is located at the origin O along the x-axis. A point P lies 20 cm from O, and the line OP makes an angle π/3 with the x-axis. If the electric field at P is inclined at an angle θ to the x-axis, what is θ?

Accepted Answer

π/3 + tan⁻¹(√3/2). At polar angle theta the dipole field tilts from the radius vector by beta with tan(beta) = (1/2)tan(theta) = (1/2)tan(60) = sqrt(3)/2. The field's angle with the x-axis is theta + beta = pi/3 + arctan(sqrt(3)/2).

Question 5

A uniformly charged semicircular wire has radius a and linear charge density λ. The magnitude of the electric field at the centre of the semicircle is

Accepted Answer

λ/(2π ε₀ a). The electric field at the center of a uniformly charged semicircular wire can be derived by integrating the contributions from each infinitesimal charge element along the semicircle. Due to symmetry, the horizontal components of the electric field cancel out, while the vertical components add up, resulting in a total electric field magnitude of λ/(2π ε₀ a).

Question 6

A point Q is located on the perpendicular bisector of an electric dipole whose dipole moment is p. If Q is at a distance r from the dipole and r is much greater than the dipole size, the electric field at Q varies as

Accepted Answer

p and r⁻³. The electric field due to an electric dipole at a point far away (where r is much greater than the dipole size) decreases with the cube of the distance (r⁻³) and is directly proportional to the dipole moment (p), which is why the correct option is p and r⁻³.

Question 7

A solid sphere of radius R carries charge q spread throughout its volume with volume charge density given by ρ = k r^a, where k and a are constants and r is measured from the centre. If the electric field at the point r = R/2 is one-eighth of the field at the surface r = R, determine the v

Accepted Answer

2. Q_enc = integral of k r^a *4 pi r^2 dr gives E(r) = k r^(a+1)/(eps0 (a+3)), so E ~ r^(a+1). E(R/2)/E(R) = (1/2)^(a+1) = 1/8 = (1/2)^3, so a+1=3, a=2.

Question 8

An oil droplet of radius r and density ρ remains at rest in a uniform electric field E directed vertically upward. If the density of air is ρ0, with ρ0 > ρ, and e is the elementary charge, the droplet must have—

Accepted Answer

an excess of 4πr³(ρ−ρ0)g / 3Ee electrons. For equilibrium the electric force balances net gravity minus buoyancy: |q|E = (4/3)pi r^3 (rho0 - rho) g. Since rho0 > rho the charge must be negative, so the drop has an EXCESS of n = 4 pi r^3 (rho0-rho) g /(3 E e) electrons.

Question 9

A plane surface is associated with the area vector A = (2î + 3ĵ) m². If the electric field acting on it is E = 4î V/m, what is the electric flux through the surface?

Accepted Answer

8 V-m. The electric flux through a surface is calculated by taking the dot product of the electric field vector and the area vector. In this case, the dot product of E = 4î V/m and A = (2î + 3ĵ) m² results in 8 V-m, as only the î components contribute to the flux.

Question 10

Two identical small metal spheres A and B carry charges +4q and −4q, respectively. If they are kept at a fixed separation, the electrostatic force between them is F. Now a third identical sphere C, initially neutral, is first brought into contact with A, then with B, and finally taken away

Accepted Answer

F/8. When sphere C, initially neutral, comes into contact with sphere A, it acquires a charge of +2q, and when it contacts sphere B, it acquires a charge of -2q. After these interactions, sphere A has a charge of +2q and sphere B has a charge of -2q, resulting in a new electrostatic force that is one-fourth of the original force F, leading to a final interaction force of F/8.

Question 11

Two ideal electric dipoles with moments p1 and p2 are separated by a distance x, and their dipole moments are parallel to each other. The magnitude of the force between them is

Accepted Answer

(1/4πϵ0) · 6p1p2/x⁴. The correct option is based on the derived formula for the force between two aligned electric dipoles, which shows that the force is proportional to the product of their dipole moments and inversely proportional to the fourth power of the distance between them, resulting in the factor of 6 in the numerator.

Question 12

Identify the statement that is incorrect.

Accepted Answer

Charged particles cannot be created or annihilated.. The incorrect statement is 'Charged particles cannot be created or annihilated' - they can be (e.g., pair production/annihilation) provided net charge is conserved. The other three statements are all true.

Question 13

Two positive ions each have charge q and are a distance d apart. If the repulsive force between them is F, then the number of electrons absent from each ion, taking e as the charge of one electron, is

Accepted Answer

√(4πϵ0 Fd²/e²). The correct option relates the repulsive force between two positive ions to the number of electrons missing from each ion by using Coulomb's law. The formula shows that the force is proportional to the product of the charges (which are related to the number of missing electrons) and inversely proportional to the square of the distance between them, allowing us to derive the number of electrons in terms of the given variables.

Question 14

Choose the incorrect statement about a conductor in electrostatic equilibrium.

Accepted Answer

At the surface of a charged conductor, the electrostatic field must be tangent to the surface at every point.. At the surface of a charged conductor the electrostatic field is perpendicular (normal) to the surface, not tangent to it. That statement is the incorrect one; the others are all true.

Question 15

A charge q is kept at the midpoint of the line segment joining two identical charges Q. If the arrangement is to remain in equilibrium, what must be the value of q?

Accepted Answer

-Q/4. For the system to be in equilibrium, the net force acting on the charge q must be zero. The attractive forces from the two identical charges Q must balance the repulsive force from q, which occurs when q is negative and equal to -Q/4, allowing the forces to counteract each other.

Question 16

If an enclosed surface has electric flux ϕ₁ entering it and ϕ₂ leaving it, what is the net charge enclosed by the surface?

Accepted Answer

(ϕ₂ − ϕ₁)/ε₀. According to Gauss's law, the net electric flux through a closed surface is directly proportional to the net charge enclosed within that surface. The net charge is calculated by subtracting the total flux entering the surface from the total flux leaving it, which is represented by (ϕ₂ − ϕ₁)/ε₀.

Question 17

Two point charges, +8q at x = 0 and −2q at x = L, are placed on the x-axis. At which position on the x-axis does the resultant electric field due to these charges become zero?

Accepted Answer

2L. The zero-field point is outside the smaller (-2q) charge: 8q/x^2 = 2q/(x-L)^2 -> 2(x-L) = x -> x = 2L.

Question 18

An electric dipole is kept making an angle of 30° with a non-uniform electric field. What effect will it experience?

Accepted Answer

Both a torque and a translational force. In a non-uniform field the two charges of the dipole feel unequal forces, giving a net translational force; since the dipole is at 30 deg to the field it also feels a torque p x E. So it experiences both a torque and a translational force.

Question 19

A square has a charge Q at each pair of diagonally opposite vertices, and a charge q at the remaining two vertices. If the resultant electrostatic force on any one of the charges Q is zero, what is the value of Q/q?

Accepted Answer

−2√2. For charge Q at a corner: the two q charges at the adjacent corners (distance L) give a resultant sqrt(2)*kQq/L^2 along the diagonal, and the opposite Q (distance L*sqrt(2)) gives kQ^2/(2L^2). Setting the net to zero: sqrt(2)*q + Q/2 = 0 -> Q/q = -2*sqrt(2).

Question 20

A solid sphere of radius R carries a non-uniform volume charge density given by ρ(r) = (Q/π R⁴) r, where r is the distance from the centre and Q is the total charge. If point P lies inside the sphere at a distance r1 from the centre, what is the magnitude of the electric field at P?

Accepted Answer

Qr1²/(4π ε₀ R⁴). Enclosed charge q(r1) = integral of (Q/pi R^4) r * 4 pi r^2 dr from 0 to r1 = Q r1^4 / R^4. Then E = q/(4 pi e0 r1^2) = Q r1^2 / (4 pi e0 R^4).

Question 21

The potential inside a charged spherical ball is given by ϕ = ar² + b where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is:

Accepted Answer

−6aϵ0. phi = a r^2 + b -> E = -dphi/dr = -2 a r. Using Gauss in spherical form, rho = e0*(1/r^2) d/dr(r^2 E) = e0*(1/r^2)(-6 a r^2) = -6 a e0. The charge density is -6 a e0.

Question 22

Two charges, each equal to q, are kept at x = −a and x = a on the x-axis. A particle of mass m and charge q0 = q/2 is placed at the origin. If charge q0 is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to

Accepted Answer

y. The net force acting on the particle is due to the electric fields created by the two charges, which are symmetric about the y-axis. When the particle is displaced slightly along the y-axis, the forces from the two charges will create a restoring force that is proportional to the displacement y, indicating that the force acts in the direction of the displacement.

Question 23

Three point charges, +Q, q, and +Q, are located on the x-axis at x = d/2 and x = d from the origin as specified. If the resultant force on a +Q charge placed at x = 0 is zero, what must be the value of q?

Accepted Answer

−Q/4. On the +Q at x=0: the +Q at x=d repels it, while q at x=d/2 must pull it back. Balancing kQ^2/d^2 = k Q|q|/(d/2)^2 = 4kQ|q|/d^2 gives |q| = Q/4, and q must be negative, so q = -Q/4.

Question 24

A ring of radius R carries charge uniformly. Along the line passing through its centre and perpendicular to its plane, the electric field attains its maximum magnitude at a point located a distance h from the centre. What is h?

Accepted Answer

R/√2. The axial field of a ring is E = kQx/(x^2+R^2)^(3/2). Setting dE/dx = 0 gives x^2 = R^2/2, so the field peaks at h = R/sqrt(2).

Question 25

The potential at a point x (measured in μm) due to two charges situated on the x-axis is given by V(x) = 20/(x² − 4) volt. The electric field E at x = 4 μm is given by

Accepted Answer

(10/9) volt/μ m and in the +ve x direction. V = 20/(x^2-4) -> dV/dx = -40x/(x^2-4)^2. At x=4: dV/dx = -160/144 = -10/9. So E = -dV/dx = +10/9 V/um, directed along +x.

Question 26

This is a two-statement question. After reading the statements, select the option that most accurately describes them.

A non-conducting solid sphere of radius R carries a uniform positive volume charge density ρ. Because of this charge distribution, the electric potential is finite at the

Accepted Answer

Statement 1 is incorrect and Statement 2 is correct.. Statement 1 is incorrect because the change in potential energy when moving a charge from the center to the surface of the sphere does not equal qρ/(3ε0). However, Statement 2 is correct as it accurately describes the electric field inside the sphere, which is derived from the uniform charge distribution.

Question 27

Assume that an electric field E = 30x²î exists in space. Then the potential difference V_A - V_O, where V_O is the potential at the original and V_A the potential at x = 2 m is -

Accepted Answer

-80 J. The potential difference between two points in an electric field can be calculated using the formula V_A - V_O = - ∫_(O)^(A) E · dr. Here, integrating the electric field ext{E} = 30x² î from 0 to 2 meters results in a potential difference of -80 J, indicating that the potential at point A is lower than at point O.

Question 28

Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result,

Accepted Answer

3F/8. Initially A = B = q, F = kq^2/d^2. C (0) touches A -> both q/2. Then C (q/2) touches B (q) -> each (q + q/2)/2 = 3q/4. New force = k(q/2)(3q/4)/d^2 = (3/8)kq^2/d^2 = 3F/8.

Question 29

A solid ball of radius R has a charge density ρ given by ρ = ρ₀ (1 − r/R) for 0 ≤ r ≤ R. The electric field outside the ball is -

Accepted Answer

ρ₀R³/(12ε₀r²). The correct option is derived from applying Gauss's law to the charge distribution of the solid ball. The charge density decreases linearly with radius, leading to a total charge that can be calculated and results in the electric field outside the ball being proportional to the total charge divided by the square of the distance from the center, yielding the specific form of the electric field as ρ₀R³/(12ε₀r²).

Question 30

A particle of mass m and charge q is in an electric and magnetic field given by E = 2i + 3j; B = 4j + 6k. The charged particle is shifted from he origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is:

Accepted Answer

5q. The magnetic force is always perpendicular to velocity and does no work, so only the electric force contributes: W = qE.d = q(2i+3j).(i+j) = q(2+3) = 5q.

Question 31

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is:

Accepted Answer

R/√(2). On the axis E ∝ x/(x^2 + R^2)^(3/2). Setting dE/dx = 0 gives x^2 = R^2/2, so the field is maximum at h = R/sqrt(2).

Question 32

The electric field in a region is given by E = (Ax + B)î, where E is in N C⁻¹ and x is in metres. The value of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and at x = -5 is V2, then V1 - V2 is

Accepted Answer

180 V. E = 20x+10. V1 - V2 = -integral_{-5}^{1}(20x+10)dx = -[10x^2+10x]_{-5}^{1} = -(20 - 200) = 180 V.

Question 33

Four point charges -q, +q, +q and -q are placed on y-axis at y = -2d, y = -d, y = +d, y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as:

Accepted Answer

E ∝ 1/D⁴. Total charge = 0 and dipole moment Sum(q*y) = 0, but the quadrupole moment Sum(q*y^2) = -6qd^2 is nonzero. A quadrupole field falls off as 1/D^4, so E ~ 1/D^4.

Question 34

Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have charge (−q) each. The potential V and the electric field E at the centre of the

Accepted Answer

V = 0, E = 0. All 10 charges are equidistant R from the centre, so V = (k/R)(5(+q)+5(-q)) = 0. For E, summing the field vectors: the cosine/sine sums over the five +q positions (0,72,144,216,288 deg) and over the five -q positions (36,108,180,252,324 deg) each vanish (sum of 5 equally-spaced unit vectors = 0), so E_x = E_y = 0. Hence V = 0 and E = 0.

Question 35

Two identical electric point dipoles have dipole moments p₁ = p î and p₂ = −p î and are held on the x axis at distance 'a' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their spe

Accepted Answer

(3) (p)/(a)√((1)/(2πε₀ m a)). For collinear dipoles U = (1/4*pi*e0)*(2p^2/a^3). All of this converts to KE = m v^2 (both masses), giving v^2 = p^2/(2*pi*e0*m*a^3), so v = (p/a)*sqrt(1/(2*pi*e0*m*a)).

Question 36

In finding the electric field using Gauss law the formula |E| = q_enc/(ε0 |A|) is applicable. In the formula ε0 is permittivity of free space, A is the area of Gaussian surface and q_enc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situati

Accepted Answer

Only when the Gaussian surface is an equipotential surface and |E| is constant on the surface.. The simple form |E| = q_enc/(eps0|A|) requires that E be perpendicular to the surface (so the surface is equipotential) and that |E| be constant over it, so that the flux integral reduces to |E||A|. Both conditions must hold.

Question 37

Find out the surface charge density at the intersection of point x = 3 m plane and x-axis, in the region of uniform line charge of 8 nC/m lying along the z-axis in free space.

Accepted Answer

0.424 nC m⁻². The surface charge density at the intersection of the specified plane and the x-axis can be calculated using the formula that relates line charge density to surface charge density, taking into account the geometry of the charge distribution. Given the uniform line charge of 8 nC/m, the resulting surface charge density at the specified point is 0.424 nC/m².

Question 38

The electric field in a region is given by E = (2/5)E0 i + (3/5)E0 j with E0 = 4.0×10³ N/C. The flux of this field through a rectangular surface area 0.4 m² parallel to the Y – Z plane is ________ Nm²C⁻¹.

Accepted Answer

640. Flux = Ex * A = (2/5)(4.0x10^3)(0.4) = 1600 x 0.4 = 640 Nm^2/C.

Question 39

An oil drop of radius 2 mm with a density 3 g cm⁻³ is held stationary under a constant electric field 3.55 × 10⁵ V m⁻¹ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? [consider g = 9.81 m/s²]

Accepted Answer

1.73 × 10¹⁰. m = (4/3)pi(2e-3)^3(3000) ~ 1.005e-4 kg; q = mg/E = (1.005e-4 x 9.81)/(3.55e5) ~ 2.78e-9 C; n = q/e ~ 2.78e-9/1.6e-19 ~ 1.73x10^10 electrons.

Question 40

A certain charge Q is divided into two parts q and (Q − q). How should the charges Q and q be divided so that (Q − q) placed at a certain distance apart experience maximum electrostatic repulsion ?

Accepted Answer

Q = 2q. To achieve maximum electrostatic repulsion between the two charges, the force between them, which is proportional to the product of the charges, should be maximized. The optimal division occurs when one charge is twice the other, leading to the relationship Q = 2q, which maximizes the product Q(Q - q) at a fixed total charge.

Question 41

An electric dipole is placed on x-axis in proximity to a line charge of linear charge density 3.0 × 10⁻⁶ C/m. Line charge is placed on z-axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm respectively from its origin. If total force of 4 N is exerted on the

Accepted Answer

4.44 μC. Line-charge field E = lambda/(2*pi*eps0*r) = 2k*lambda/r. Net force on dipole = q*2k*lambda*(1/r1 - 1/r2) with r1=0.010 m, r2=0.012 m: 2k*lambda = 2*9e9*3e-6 = 5.4e4, and (1/0.010 - 1/0.012) = 16.67 /m, so F = q*5.4e4*16.67 = q*9.0e5. Setting F = 4 N gives q = 4/9.0e5 = 4.44 microC.

Question 42

Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: (m = mass of charged pa

Accepted Answer

(q²/(2πϵ0md³))^(1/2). The correct option represents the angular frequency of simple harmonic motion for the proton, derived from the net force acting on it due to the two fixed electrons. The force is proportional to the displacement and inversely proportional to the cube of the distance, leading to the specific form of the frequency given by the equation.

Question 43

The electric field in a region is given E = ((3)/(5) î + (4)/(5) ĵ) N/C. The ratio of flux of reported field through the rectangular surface of area 0.2 m² (parallel to y-z plane) to that of surface of area 0.3 m² (parallel to x-z plane) is a: b, where a = _____. [Here, î, ĵ and k̂ are

Accepted Answer

3. The electric flux through a surface is given by the product of the electric field and the area of the surface projected in the direction of the field. For the rectangular surface parallel to the y-z plane, the area is 0.2 m² and the electric field has a component of rac{4}{5} N/C in the j direction, resulting in a flux of 0.2 * rac{4}{5} = 0.16 N·m²/C. For the surface parallel to the x-z plane, the area is 0.3 m² and the electric field has a component of rac{3}{5} N/C in the i direction, resulting in a flux of 0.3 * rac{3}{5} = 0.18 N·m²/C. The ratio of these two fluxes is 0.16:0.18, which

Question 44

Given below are two statements:
Statement I: An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.
Statement II: If R is the radius of a solid metallic sphere and Q be

Accepted Answer

Statement I is true but Statement II is false. Statement I is correct because the electric field due to an electric dipole can exist within the sphere, but the net electric flux through a closed surface surrounding it is zero due to symmetry. Statement II is incorrect because, for a solid metallic sphere, the electric field inside is zero, and thus the electric flux through a closed surface inside the sphere must also be zero.

Question 45

A uniformly charged disc of radius R having surface charge density σ is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin:-

Accepted Answer

E = σ/(2ε0) (1 - Z/(Z² + R²)^(1/2)). The correct option accurately describes the electric field intensity along the z-axis due to a uniformly charged disc by incorporating the effects of both the surface charge density and the geometric relationship between the distance from the disc and its radius, leading to the correct expression for the electric field.

Question 46

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical ?

Accepted Answer

x = ((q² l)/(2πε₀ mg))^(1/3). For small theta, separation x = 2*l*sin(theta) ~ 2*l*theta, and tan(theta) ~ x/(2l) equals the Coulomb-to-weight ratio k*q^2/(x^2*m*g). Setting x/(2l) = q^2/(4*pi*e0*x^2*m*g) gives x^3 = q^2*l/(2*pi*e0*m*g), so x = (q^2*l/(2*pi*e0*m*g))^(1/3).

Question 47

Two particles A and B having charges 20 μC and -5 μC respectively are held fixed with a separation of 5 cm. At what position should a third charged particle be placed so that it does not experience a net electric force?
20μC A —— 5 cm —— B -5μC

Accepted Answer

At 5 cm from -5 μC on the right side of system. The third charged particle must be placed where the electric forces from both charges balance each other. At 5 cm from the -5 μC charge on the right side, the attractive force from the -5 μC charge and the repulsive force from the 20 μC charge will equalize, resulting in no net force on the third particle.

Question 48

Which of the following physical quantities have the same dimensions ?

Accepted Answer

Electric displacement (D̄) and surface charge density. Electric displacement and surface charge density both represent the distribution of electric charge per unit area, thus they share the same dimensional formula, which is essential for understanding their relationship in electromagnetism.

Question 49

Two point charges A and B of magnitude +8 × 10⁻⁶ C and -8 × 10⁻⁶ C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 × 10⁴ NC⁻¹. The distance 'd' between the point charges A and B is:

Accepted Answer

3.0 m. The electric field at point O, which is equidistant from both charges, is the result of the superposition of the fields created by each charge. Given the magnitudes of the charges and the calculated electric field strength, the distance 'd' can be derived using the formula for the electric field due to point charges, confirming that the correct distance is 3.0 m.

Question 50

Two identical metallic spheres A and B placed at certain distance in air repel each other with a force of F. Another identical uncharged sphere C is first placed in contact with A and then in contact with B and finally placed at midpoint between spheres A and B. The force experienced by sp

Accepted Answer

3F/4. C (uncharged) touching A gives each q/2; C then touching B gives both 3q/4. At the midpoint (r/2 from each), force from B = k(3q/4)(3q/4)/(r/2)^2 = (9/4)F and from A = (3/2)F act oppositely, so net force on C = (9/4 - 3/2)F = 3F/4, option (3F/4).

JEE Main Physics: Electric Charges and Fields questions with solutions

Questions