Q: A total nuclear charge Ze is distributed non-uniformly inside a nucleus of radius R, with a radial charge density rho(r) that varies linearly with r (rho(r) = a + b*r for r <= R, fixed so that the total enclosed charge is Ze). Which statement is correct?

The electric field at r = R is independent of b.. By Gauss's law the flux through the sphere r = R depends only on the total enclosed charge Ze, which is fixed. Hence E(R) = (1/4*pi*epsilon0)*Ze/R² regardless of the values of a and b individually (it is independent of b). The potential at r = R for a spherically symmetric charge equals (1/4*pi*epsilon0)*Ze/R, again set by the total charge.

Q: A uniform electric field of magnitude E lies in the x-y plane, making angle theta with the x-axis (as in the figure). Find the potential difference V(O) - V(A) between the origin O and the point A(d, d, 0).

Ed (cos(theta) + sin(theta)). For a uniform field, V(O) - V(A) = E. r_(O to A) where r is the vector from O to A. With E = (E cos(theta), E sin(theta), 0) and the vector from O to A = (d, d, 0), the dot product is E*d*cos(theta) + E*d*sin(theta) = Ed(cos(theta) + sin(theta)).

Q: Two identical charged spheres hang from a common point on two massless threads of length l and, because of mutual repulsion, settle a small distance x = d apart (d << l). Charge now leaks off both spheres at a constant rate, and the spheres approach each other with speed v. How does v depe

v proportional to x^(-1/2). For small separation, equilibrium gives k q² / x² proportional to x (the horizontal restoring force ~ (mg)(x/2l)), so q² is proportional to x³, i.e. q proportional to x^(3/2). Since charge leaks at a constant rate, dq/dt is constant. Differentiating q ~ x^(3/2): dq/dt ~ x^(1/2) (dx/dt). With dq/dt constant, v = dx/dt is proportional to x^(-1/2).

Q: A neutral spherical metallic conductor has a spherical cavity. A positive point charge sits at the centre of the cavity, and another positive point charge is placed outside the conductor. Match each cause in Column-I with the effects it produces in Column-II. Column-I (Cause): (A) The outs

(P), (Q), (R) and (S). Electrostatic shielding decouples inside and outside. (A) Moving the outside charge changes outer-surface distribution (Q) and the potential at the centre from outer charges (R); it does not change the inner cavity field or the force on the inside charge. (B) Moving the inside charge changes the inner-surface induced distribution (P) and the force on the inside charge (S). (C) Increasing the inside charge changes inner-surface (P), outer-surface (Q) and potentials (R). (D) Earthing changes the outer surface charge (Q) and potential (R). Across all causes, every effect (P

Q: Charge is distributed inside a sphere of radius R with volume charge density rho(r) = (A/r²) * e^(-2r/a), where A and a are constants. If Q is the total charge contained in the sphere, find R.

(a/2) log(1/(1 - Q/(2 pi a A))). Q = integral₀^R rho * 4 pi r² dr = integral₀^R (A/r²) e^(-2r/a) * 4 pi r² dr = 4 pi A integral₀^R e^(-2r/a) dr = 4 pi A * (a/2)[1 - e^(-2R/a)] = 2 pi a A [1 - e^(-2R/a)]. Solving: 1 - e^(-2R/a) = Q/(2 pi a A), so e^(-2R/a) = 1 - Q/(2 pi a A), giving 2R/a = log(1/(1 - Q/(2 pi a A))) and R = (a/2) log(1/(1 - Q/(2 pi a A))).

Q: A negatively charged particle, initially at rest, is released in a region where a uniform gravitational field and a uniform electric field both act (the two fields are not parallel). Which of the following best describes the shape of the particle's subsequent path?

A straight line. Each field exerts a force that is constant in both magnitude and direction: gravity gives m*g downward, and the electric field gives a constant q*E force (opposite to E because the charge is negative). Their vector sum is therefore a single constant net force pointing in a fixed direction. A body released from rest experiences acceleration only along this fixed net-force direction, so it speeds up along one straight line. A curved path (parabola, circle, ellipse) would require either an initial velocity transverse to the force or a force that changes direction, neither of whic

Q: Three concentric thin metal shells A, B and C have radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Shell A is initially uncharged and isolated. Now shell C is connected to shell A by a thin wire. What is the final charge on shell B?

-8Q/11. When C (radius 3a) is wired to A (radius a) they become one conductor at a common potential. Let the charge on A be q_A and on C be q_C with q_A + q_C = Q (charge supplied originally to C is conserved on the A-C system; A started neutral). B is earthed so its potential is zero and it acquires an induced charge q_B. Writing the potentials of A, B, C from the superposition of all three shells and imposing V_B = 0 together with V_A = V_C and q_A + q_C = Q, solving the linear system gives the induced charge on the earthed middle shell as q_B = -8Q/11.

Question 1

Two identical conducting spheres A and B, each of radius R, carry equal charges and are separated by a large distance d (d >> R), exerting a force f0 on each other. A neutral conducting sphere of radius 2R is first touched to sphere A, then touched to sphere B, and finally removed far away

Accepted Answer

5*f0/27. After the first touch, A loses 2/3 of its charge to the neutral sphere. After the second touch, B gains charge and retains only 1/3 of the combined total. The product of final charges on A and B is 5/27 times the original q², giving force 5f0/27.

Question 2

A small sphere of mass 80 g carrying charge q is held 9 m vertically above the centre of a fixed non-conducting sphere of radius 1 m that also carries charge q. When released, the small sphere falls freely until it is repelled and just barely stops before touching the fixed sphere. Given g

Accepted Answer

2. The sphere falls from height 9 m above centre to 1 m (surface contact). Height fallen h = 9 - 1 = 8 m. Energy conservation: m*g*h = kₑ*q²*(1/1 - 1/9) = kₑ*q²*(8/9). So m*g = kₑ*q²/9. Then q² = 9*m*g/kₑ = 9*0.08*9.8/(9*10⁹) = 7.056/9*10⁹ ≈ 7.84*10⁻¹⁰ => q ≈ 2.8*10⁻⁵ C but let me redo: q² = m*g*h*9/(kₑ*8) = 0.08*9.8*8*9/(9*10⁹*8) = 0.08*9.8*9/(9*10⁹) = 0.08*9.8/(10⁹) = 0.784*10⁻⁹ => q = sqrt(7.84*10⁻¹⁰) = 2.8*10⁻⁵ C... Hmm that's 28 micro-C. Let me try: q = 7K uC, so K = q/(7 uC). With q = 14 uC, K=2. Let's verify: m*g*h = U_final - U_initial => 0.08*9.8*8 = 9*10⁹*(14*10⁻⁶)²*(1 - 1/9) = 9*10⁹

Question 3

A uniformly charged non-conducting solid sphere of radius R carries total charge Q. A narrow tunnel is drilled through the sphere along a diameter. A point charge q of mass m is placed at one end of the tunnel (on the surface of the sphere, point A) and released. What is the minimum initia

Accepted Answer

v0 = 0. If q is of the same sign as Q, it is repelled and the centre (highest potential) is the hardest point. But if q is opposite in sign to Q, the centre is a potential energy minimum and B (surface) has the same potential as A, so no minimum speed is needed. For diametrically opposite identical surface points, V(A) = V(B), and hence zero minimum speed is required.

Question 4

A total nuclear charge Ze is distributed non-uniformly inside a nucleus of radius R, with a radial charge density rho(r) that varies linearly with r (rho(r) = a + b*r for r <= R, fixed so that the total enclosed charge is Ze). Which statement is correct?

Accepted Answer

The electric field at r = R is independent of b.. By Gauss's law the flux through the sphere r = R depends only on the total enclosed charge Ze, which is fixed. Hence E(R) = (1/4*pi*epsilon0)*Ze/R² regardless of the values of a and b individually (it is independent of b). The potential at r = R for a spherically symmetric charge equals (1/4*pi*epsilon0)*Ze/R, again set by the total charge.

Question 5

A uniform electric field of magnitude E lies in the x-y plane, making angle theta with the x-axis (as in the figure). Find the potential difference V(O) - V(A) between the origin O and the point A(d, d, 0).

Accepted Answer

Ed (cos(theta) + sin(theta)). For a uniform field, V(O) - V(A) = E. r_(O to A) where r is the vector from O to A. With E = (E cos(theta), E sin(theta), 0) and the vector from O to A = (d, d, 0), the dot product is E*d*cos(theta) + E*d*sin(theta) = Ed(cos(theta) + sin(theta)).

Question 6

Two identical charged spheres hang from a common point on two massless threads of length l and, because of mutual repulsion, settle a small distance x = d apart (d << l). Charge now leaks off both spheres at a constant rate, and the spheres approach each other with speed v. How does v depe

Accepted Answer

v proportional to x^(-1/2). For small separation, equilibrium gives k q² / x² proportional to x (the horizontal restoring force ~ (mg)(x/2l)), so q² is proportional to x³, i.e. q proportional to x^(3/2). Since charge leaks at a constant rate, dq/dt is constant. Differentiating q ~ x^(3/2): dq/dt ~ x^(1/2) (dx/dt). With dq/dt constant, v = dx/dt is proportional to x^(-1/2).

Question 7

A neutral spherical metallic conductor has a spherical cavity. A positive point charge sits at the centre of the cavity, and another positive point charge is placed outside the conductor. Match each cause in Column-I with the effects it produces in Column-II. Column-I (Cause): (A) The outs

Accepted Answer

(P), (Q), (R) and (S). Electrostatic shielding decouples inside and outside. (A) Moving the outside charge changes outer-surface distribution (Q) and the potential at the centre from outer charges (R); it does not change the inner cavity field or the force on the inside charge. (B) Moving the inside charge changes the inner-surface induced distribution (P) and the force on the inside charge (S). (C) Increasing the inside charge changes inner-surface (P), outer-surface (Q) and potentials (R). (D) Earthing changes the outer surface charge (Q) and potential (R). Across all causes, every effect (P

Question 8

Charge is distributed inside a sphere of radius R with volume charge density rho(r) = (A/r²) * e^(-2r/a), where A and a are constants. If Q is the total charge contained in the sphere, find R.

Accepted Answer

(a/2) log(1/(1 - Q/(2 pi a A))). Q = integral₀^R rho * 4 pi r² dr = integral₀^R (A/r²) e^(-2r/a) * 4 pi r² dr = 4 pi A integral₀^R e^(-2r/a) dr = 4 pi A * (a/2)[1 - e^(-2R/a)] = 2 pi a A [1 - e^(-2R/a)]. Solving: 1 - e^(-2R/a) = Q/(2 pi a A), so e^(-2R/a) = 1 - Q/(2 pi a A), giving 2R/a = log(1/(1 - Q/(2 pi a A))) and R = (a/2) log(1/(1 - Q/(2 pi a A))).

Question 9

A negatively charged particle, initially at rest, is released in a region where a uniform gravitational field and a uniform electric field both act (the two fields are not parallel). Which of the following best describes the shape of the particle's subsequent path?

Accepted Answer

A straight line. Each field exerts a force that is constant in both magnitude and direction: gravity gives m*g downward, and the electric field gives a constant q*E force (opposite to E because the charge is negative). Their vector sum is therefore a single constant net force pointing in a fixed direction. A body released from rest experiences acceleration only along this fixed net-force direction, so it speeds up along one straight line. A curved path (parabola, circle, ellipse) would require either an initial velocity transverse to the force or a force that changes direction, neither of whic

Question 10

Three concentric thin metal shells A, B and C have radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Shell A is initially uncharged and isolated. Now shell C is connected to shell A by a thin wire. What is the final charge on shell B?

Accepted Answer

-8Q/11. When C (radius 3a) is wired to A (radius a) they become one conductor at a common potential. Let the charge on A be q_A and on C be q_C with q_A + q_C = Q (charge supplied originally to C is conserved on the A-C system; A started neutral). B is earthed so its potential is zero and it acquires an induced charge q_B. Writing the potentials of A, B, C from the superposition of all three shells and imposing V_B = 0 together with V_A = V_C and q_A + q_C = Q, solving the linear system gives the induced charge on the earthed middle shell as q_B = -8Q/11.

Question 11

A charge +Q is fixed at the origin. A small electric dipole of dipole moment p, with its axis along the x-axis and pointing away from the charge, is released from rest from a point very far away (effectively at infinity). Find the kinetic energy of the dipole when it reaches the point (d,

Accepted Answer

p*Q / (4*pi*e0*d²). The field of +Q at distance r along the x-axis is E = Q/(4*pi*e0*r²), directed away from the charge (along +x). The dipole moment p points away from the charge (along +x) too, so the dipole is aligned with the field and its potential energy is U(r) = -p*E = -p*Q/(4*pi*e0*r²). Starting from rest at infinity where U = 0, energy conservation gives KE(d) = U(infinity) - U(d) = 0 - (-p*Q/(4*pi*e0*d²)) = p*Q/(4*pi*e0*d²).

JEE Main Physics: Electrostatics questions with solutions

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