JEE Physics Electric Charges and Fields MCQs Set 08

Answer: (a) \( >\frac{F}{q_0} \)

Explanation:
1. The test charge repels the ball, and the ball hangs on a thread, so it moves away from the point.
2. The distance to the point increases, so the field actually acting on the test charge is smaller than the original E.
3. \( \frac{F}{q_0}<E \), that is, E is greater than F/q0.

Teacher's Note:
1. The definition E = F/q0 holds only for a vanishingly small test charge that does not disturb the source.
2. Here the source can move, so a finite test charge changes the field.

Answer: (b) 8 in −x direction

Explanation:
1. \( E_x=-\frac{\partial V}{\partial x}=-8x \), and E_y = E_z = 0.
2. At x = 1: E = −8 V/m, which is 8 V/m in the −x direction.

Teacher's Note:
1. Electric field is the negative gradient of potential.
2. Only the x-derivative is non-zero because V does not depend on y or z.

Answer: (b) 100

Explanation:
1. Volume is conserved: \( \frac43\pi R^3=1000\times\frac43\pi r^3\Rightarrow R=10r \).
2. Charge on the big drop: Q = 1000q.
3. \( \frac{V_{big}}{V_{small}}=\frac{kQ/R}{kq/r}=\frac{1000q/10r}{q/r}=100 \)

Teacher's Note:
1. On coalescence the total volume (and hence R = n^{1/3} r) and the total charge are conserved.
2. Potential of a charged sphere is kQ/R.

Answer: (a) The kinetic energy of the electron is more than that of the proton

Explanation:
1. No external force acts, so total momentum stays zero: the two particles have equal and opposite momenta p.
2. \( K=\frac{p^2}{2m} \), so the lighter particle has the larger kinetic energy.
3. The electron is lighter, so its kinetic energy is greater at every separation.

Teacher's Note:
1. Equal forces act on both (Newton's third law) for equal times, so they gain equal momentum.
2. For equal momentum, kinetic energy is inversely proportional to mass.

Answer: (d) 1 : 1

Explanation:
1. A conductor in electrostatic equilibrium is an equipotential body.
2. The potential is the same at the corner, on the faces and at the centre, so the ratio is 1 : 1.

Teacher's Note:
1. The field inside a conductor is zero, so no work is needed to move a charge between two points of the conductor.
2. The surface charge density is larger at the corners, but the potential is the same everywhere.

Answer: (a) 10 cm away from q and between the charges

Explanation:
1. Like charges: the neutral point lies between them, nearer the smaller charge.
2. Let the point be x from q: \( \frac{kq}{x^2}=\frac{k\,4q}{(30-x)^2} \).
3. Taking square roots: \( 30-x=2x\Rightarrow x=10\ \text{cm} \).

Teacher's Note:
1. Between two like charges the fields oppose, so zero field is possible only there.
2. The neutral point divides the distance in the ratio of the square roots of the charges.

Answer: (b) zero

Explanation:
1. At an equatorial point the distances from +q and −q are equal.
2. So the potentials kq/r and −kq/r cancel and the potential is zero.

Teacher's Note:
1. The potential of a dipole is V = p cosθ/(4πε0 r²), and θ = 90° on the equatorial line.
2. The field there is not zero; it points opposite to the dipole moment.

Answer: (a) of the same frequency and with a shifted mean position

Explanation:
1. The electric force QE is constant, so it only adds a constant force to the spring force.
2. The new equilibrium position is displaced by QE/k.
3. The restoring force about the new mean position is still −kx, so the frequency \( \frac1{2\pi}\sqrt{\frac km} \) is unchanged.

Teacher's Note:
1. A constant force shifts the equilibrium position but does not change the force constant.
2. Frequency depends only on k and m.

Answer: (a) 0.1 V

Explanation:
1. \( V=\frac{W}{q}=\frac{2}{20}=0.1\ \text{V} \)
2. The distance (2 cm) is not needed.

Teacher's Note:
1. Potential difference is work done per unit charge.
2. The distance moved is extra information here.

Answer: (d) east-west

Explanation:
1. \( \frac{dV}{dl}=-E\cos\theta \), which vanishes when the direction makes 90° with E.
2. Perpendicular to north is the east-west direction (equipotential line).

Teacher's Note:
1. The rate of change of potential along a direction is the negative component of the field along it.
2. Equipotential surfaces are perpendicular to the field.

Answer: (d) 39.7 N/m²

Explanation:
1. Electrostatic pressure on a conductor: \( \frac FA=\frac{\sigma^2}{2\varepsilon_0} \).
2. \( =\frac{(26.5\times10^{-6})^2}{2\times8.85\times10^{-12}}=\frac{7.02\times10^{-10}}{1.77\times10^{-11}}\approx39.7\ \text{N/m}^2 \)

Teacher's Note:
1. The force per unit area is half the field outside multiplied by σ, because the surface charge feels the average of the fields inside (zero) and outside.
2. So the pressure is σ²/(2ε0).

Answer: (c) \( \frac1{d^3} \)

Explanation:
1. On the axis, \( E=\frac{1}{4\pi\varepsilon_0}\frac{2p}{d^3} \) for d much greater than the dipole length.

Teacher's Note:
1. A dipole is a pair of opposite charges whose fields partly cancel, so its field falls faster (1/d³) than that of a single charge (1/d²).
2. The same 1/d³ dependence holds on the equatorial line.

Answer: (c) the nuclei do not come very close due to Coulombic repulsion

Explanation:
1. Fusion needs the nuclei to come within about 10⁻¹⁵ m, where the nuclear force acts.
2. At room temperature the thermal energy is far too small to overcome the Coulomb repulsion between the positively charged nuclei.

Teacher's Note:
1. Nuclear fusion needs temperatures of millions of kelvin so that the nuclei can approach each other closely.
2. Coulomb repulsion between like charges is the barrier.

Answer: (c) \( \frac{\sqrt7K}{8} \)

Explanation:
1. General dipole field: \( \vec E=\frac{K}{r^3}\left[3(\vec p\cdot\hat r)\hat r-\vec p\right] \).
2. Here r = 2 along x, so \( \hat r=\hat i \) and \( \vec p\cdot\hat r=1 \).
3. \( \vec E=\frac{K}{8}\left[3\hat i-(\hat i+\sqrt3\hat j)\right]=\frac K8(2\hat i-\sqrt3\hat j) \).
4. \( |\vec E|=\frac K8\sqrt{4+3}=\frac{\sqrt7K}{8} \)

Teacher's Note:
1. Use the vector form of the dipole field when the point is neither on the axis nor on the equator.
2. Take the component of p along the line to the point first.

Answer: (b) 5500 î V m⁻¹

Explanation:
1. \( E_x=-\frac{dV}{dx}=\frac{1000}{x^2}+\frac{3000}{x^3}+\frac{1500}{x^4} \).
2. At x = 1: E = 1000 + 3000 + 1500 = 5500 V/m along +x.

Teacher's Note:
1. Differentiate each term and remember the minus sign.
2. V is given only along the x-axis, so the field has only the x-component.

Answer: (b) 1 and 3 are correct

Explanation:
1. With both fields uniform, the net force mg + qE is constant.
2. If the initial velocity is along the net force (or zero), the path is a straight line; otherwise the particle moves like a projectile along a parabola.
3. A circular path needs a force of constant magnitude always perpendicular to the velocity, which a constant force cannot give.

Teacher's Note:
1. A constant force always gives straight-line or parabolic motion.
2. Uniform circular motion needs a centripetal force that changes direction.

Answer: (a) \( T=\frac{\sigma^2r}{8\varepsilon_0} \)

Explanation:
1. Electrostatic pressure (outward): \( \frac{\sigma^2}{2\varepsilon_0} \).
2. A soap bubble has two surfaces, so the inward surface-tension pressure is \( \frac{4T}{r} \).
3. Equal inside and outside pressure means these balance: \( \frac{4T}{r}=\frac{\sigma^2}{2\varepsilon_0}\Rightarrow T=\frac{\sigma^2r}{8\varepsilon_0} \).

Teacher's Note:
1. A soap film has two free surfaces, so the excess pressure is 4T/r rather than 2T/r.
2. The electric repulsion tends to expand the bubble.

Answer: (d) \( \frac{q\sigma}{\varepsilon_0mg} \)

Explanation:
1. Field just outside a charged conductor: \( E=\frac{\sigma}{\varepsilon_0} \) (perpendicular to the plane).
2. Forces on the ball: weight mg downwards, electric force qE horizontally, and tension along the string.
3. The string makes angle θ with the vertical plane, so \( \tan\theta=\frac{qE}{mg}=\frac{q\sigma}{\varepsilon_0mg} \).

Teacher's Note:
1. For a conductor the field outside is σ/ε0, whereas for a thin non-conducting sheet it is σ/(2ε0).
2. The angle with the vertical is given by the ratio of horizontal to vertical force.

Answer: (b) \( \sqrt{\ln\left(\frac{r}{r_0}\right)} \)

Explanation:
1. Field of the line: \( E=\frac{\lambda}{2\pi\varepsilon_0 r} \).
2. Work done by the field: \( \frac12mv^2=q\int_{r_0}^{r}\frac{\lambda}{2\pi\varepsilon_0 x}dx=\frac{q\lambda}{2\pi\varepsilon_0}\ln\frac{r}{r_0} \).
3. \( v=\sqrt{\frac{q\lambda}{\pi\varepsilon_0m}\ln\frac r{r_0}}\ \propto\ \sqrt{\ln\frac{r}{r_0}} \)

Teacher's Note:
1. Use energy conservation with the logarithmic potential of a line charge.
2. The speed is proportional to the square root of the logarithm.

Answer: (c) 3q/2

Explanation:
1. Let the initial charge be q0. It is stationary in the field E: \( q_0E=mg \).
2. The drag at speed v0 equals mg (terminal velocity under gravity alone).
3. With charge 3q the drop moves up at v0, so drag acts downwards: \( 3qE=mg+mg=2mg \).
4. Dividing: \( \frac{3q}{q_0}=2\Rightarrow q_0=\frac{3q}{2} \).

Teacher's Note:
1. Drag always opposes motion: it points up when the drop falls and down when it rises.
2. Write the force balance for each situation separately.

Answer: (b) 640 C

Explanation:
1. Equal potentials: \( \frac{Q_E}{R_E}=\frac{q}{r} \).
2. \( Q_E=1\times10^{-6}\times\frac{6.4\times10^{6}}{10^{-2}}=640\ \text{C} \)

Teacher's Note:
1. Potential of a sphere is kQ/R, so equal potentials need charge proportional to radius.
2. Convert both radii to metres.

Answer: (c) \( \frac Q2 \)

Explanation:
1. \( F=\frac{k\,q(Q-q)}{r^2} \).
2. \( \frac{dF}{dq}\propto Q-2q=0\Rightarrow q=\frac Q2 \)

Teacher's Note:
1. The product q(Q − q) is largest when the two parts are equal.
2. The same result follows from the AM-GM inequality.

Answer: (a) \( \frac{\lambda}{3\sqrt3\,\varepsilon_0R} \)

Explanation:
1. Field on the axis of a ring: \( E=\frac{kQz}{(R^2+z^2)^{3/2}} \) with Q = 2πRλ.
2. For z = R/√2: \( R^2+z^2=\frac{3R^2}{2},\ \left(\frac{3R^2}{2}\right)^{3/2}=\frac{3\sqrt3}{2\sqrt2}R^3 \).
3. \( E=\frac{2\pi R\lambda}{4\pi\varepsilon_0}\cdot\frac{R/\sqrt2}{\frac{3\sqrt3}{2\sqrt2}R^3}=\frac{\lambda}{3\sqrt3\,\varepsilon_0R} \)

Teacher's Note:
1. Use the on-axis formula for a ring and substitute carefully.
2. The field is directed along +z for positive charge.

Answer: (a) zero

Explanation:
1. Between R1 and R2 the point lies inside the metal of the shell.
2. The electric field inside a conductor in equilibrium is zero, so the normal component is zero (the induced charge −Q on the inner surface cancels the field of Q).

Teacher's Note:
1. Gauss's law: the enclosed charge is Q + (−Q) = 0, so the flux through the Gaussian surface is zero.
2. The field in the cavity (r < R1) and outside (r > R2) is still non-zero.

Answer: (b) F

Explanation:
1. Let the charges be q and q at distance d, so \( F=\frac{kq^2}{d^2} \).
2. After touching, the first ball and the third ball each hold q/2; the second ball has q.
3. The third ball (q/2) is at d/2 from each: force from the first (q/2): \( \frac{k(q/2)(q/2)}{(d/2)^2}=F \), from the second (q): \( \frac{k(q)(q/2)}{(d/2)^2}=2F \).
4. They act in opposite directions, so the net force is 2F − F = F.

Teacher's Note:
1. Identical conducting balls share charge equally on contact.
2. Forces from the two neighbours push the middle ball in opposite directions.

Answer: (c) 4

Explanation:
1. Centres are 3R apart, so the point is 2R from the small centre and R from the large centre.
2. Small sphere (point is outside): \( E_1=\frac{\rho_1R^3}{3\varepsilon_0(2R)^2}=\frac{\rho_1R}{12\varepsilon_0} \).
3. Large sphere (point is inside at distance R): \( E_2=\frac{\rho_2R}{3\varepsilon_0} \), directed oppositely.
4. Equal magnitudes: \( \frac{\rho_1}{12}=\frac{\rho_2}{3}\Rightarrow\frac{\rho_1}{\rho_2}=4 \).

Teacher's Note:
1. Outside a uniformly charged sphere the field behaves as that of a point charge at the centre; inside it grows linearly with distance.
2. The two fields point in opposite directions between the centres when the charges have the same sign.

Answer: (d) \( -\frac{qQ}{6\pi\varepsilon_0L} \) and zero

Explanation:
1. Potential at C is zero (equal and opposite charges at equal distances).
2. At D: distance from A is 3L and from B is L, so \( V_D=k\left(\frac{q}{3L}-\frac qL\right)=-\frac{2kq}{3L}=-\frac{q}{6\pi\varepsilon_0L} \).
3. \( W_{C\to D}=Q(V_D-V_C)=-\frac{qQ}{6\pi\varepsilon_0L} \).
4. At infinity V = 0 as well, so the work from C to infinity is zero.

Teacher's Note:
1. The work done by the field depends only on the end points, not on the path (the semicircle is irrelevant).
2. Work done by an external agent equals Q times the potential difference.

Answer: (c) 2

Explanation:
1. Time in the field: t = L/v0, vertical velocity gained: \( v_y=\frac{qE}{m}\cdot\frac{L}{v_0} \).
2. \( \tan\theta=\frac{v_y}{v_0}=\frac{qEL}{mv_0^2}\propto\frac qm \).
3. For a proton q/m = e/m_p; for an α-particle q/m = 2e/(4m_p) = e/(2m_p).
4. \( \frac{\tan\theta_1}{\tan\theta_2}=\frac{e/m_p}{e/2m_p}=2 \)

Teacher's Note:
1. The path inside the plates is a parabola, like projectile motion, with acceleration qE/m.
2. The α-particle has twice the charge but four times the mass, so its charge-to-mass ratio is half of the proton's.

Answer: (d) r/4

Explanation:
1. At the closest point all the kinetic energy has become potential energy: \( \frac12mv^2=\frac{kqQ}{r} \).
2. Doubling v multiplies the kinetic energy by 4, so the closest distance becomes one quarter: r/4.

Teacher's Note:
1. Closest approach distance is inversely proportional to the initial kinetic energy.
2. Energy goes as v², so 2v gives a distance of r/4.

Answer: (d) the cube remains neutral

Explanation:
1. The cube is isolated, so its total charge cannot change.
2. Charges are only redistributed by induction: the near face gets negative charge and the far face gets an equal positive charge.
3. The net charge remains zero, and the interior stays neutral.

Teacher's Note:
1. Induction separates charge but does not create or remove any.
2. Total charge of an isolated body is conserved.