JEE Physics Electric Charges and Fields MCQs Set 24

Answer: (d) \( \dfrac{5q}{6\varepsilon_0} \)

Explanation:
1. The charge is at the centre of the cube, so by symmetry each of the six faces would receive one sixth of the total flux \( q/\varepsilon_0 \).
2. With one face open, the flux through the five faces is \( \tfrac56\cdot\dfrac q{\varepsilon_0}=\dfrac{5q}{6\varepsilon_0} \).

Teacher's Note:
1. Use symmetry before Gauss's law. A cube has six identical faces seen from its centre.

Answer: (a) 2E

Explanation:
1. Non-conducting sheet: \( E=\sigma/2\varepsilon_0 \) with \( \sigma \) the surface density of the sheet.
2. Conducting sheet with the same total charge: the charge sits on both faces so each face carries \( \sigma/2 \) per unit area of one side... for a conductor the field near the surface is \( \sigma_{face}/\varepsilon_0 \) where the face density equals the full charge density of the sheet, giving \( \sigma/\varepsilon_0=2E \).

Teacher's Note:
1. A conductor places its charge on the outer surfaces. The field just outside a conductor is surface density over epsilon naught.

Answer: (b) 0

Explanation:
1. The charge Q sits in the cavity of the conductor. The induced charges on the shell make the field due to everything outside the cavity (including q) zero inside the cavity.
2. So the net force on Q due to q is zero (electrostatic shielding), even though q feels a force of 10 N.

Teacher's Note:
1. A conductor shields its cavity from external fields. Newton's third law does not hold between Q and q because the shell's induced charges also act.

Answer: (c) \( -\sigma \)

Explanation:
1. The force on one hemisphere due to the other = \( \dfrac{\sigma^2}{4\varepsilon_0}\pi R^2 \) (outward).
2. The inner shell of density \( \sigma' \) gives a field \( \dfrac{\sigma'(R/2)^2}{\varepsilon_0R^2}=\dfrac{\sigma'}{4\varepsilon_0} \) at radius R, so the force on the hemisphere is \( \sigma\cdot\dfrac{\sigma'}{4\varepsilon_0}\pi R^2 \).
3. Equilibrium: \( \sigma^2+\sigma\sigma'=0\Rightarrow\sigma'=-\sigma \).

Teacher's Note:
1. Force on a hemispherical shell equals the field at its surface times charge times projected area. Balance the repulsion from the other hemisphere against the attraction of the inner shell.

Answer: (c) \( \dfrac{1}{4\pi\varepsilon_0}\dfrac{3Q}{r^2} \)

Explanation:
1. A Gaussian sphere of radius r between a and b encloses only the charge +3Q of the inner sphere.
2. \( E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{3Q}{r^2} \).

Teacher's Note:
1. Only the charge inside the Gaussian surface counts. The outer shell does not affect the field inside it.

Answer: (a) \( \text{decreasing as }1/r^2 \)

Explanation:
1. The induced charge on the inner surface of the shell is -q, spread uniformly because q is at the centre. This is independent of Q' and Q.
2. Inside the cavity the field is therefore only that of q: \( E=\dfrac{q}{4\pi\varepsilon_0r^2} \), decreasing as \( 1/r^2 \).

Teacher's Note:
1. The cavity field depends only on the charge in the cavity and the shape of the cavity. Outside charges cannot change it.

Answer: (a) \( \text{decreasing as }1/r^2 \)

Explanation:
1. Removing Q' changes only the distribution on the outer surface of the shell.
2. The field inside is still that of the central charge q, \( E\propto1/r^2 \).

Teacher's Note:
1. Same logic as before: a conducting shell isolates its cavity. The field inside depends only on the charge at the centre.

Answer: (c) \( x^{-1/2} \)

Explanation:
1. Equilibrium for small angle: \( \dfrac{x}{2l}=\dfrac{q^2}{4\pi\varepsilon_0x^2mg} \), so \( x^3\propto q^2 \) and \( q\propto x^{3/2} \).
2. Constant leak rate: \( \dfrac{dq}{dt}=\text{const}\Rightarrow x^{1/2}\dfrac{dx}{dt}=\text{const} \).
3. So \( v=\dfrac{dx}{dt}\propto x^{-1/2} \).

Teacher's Note:
1. Find how the equilibrium separation depends on charge. Then differentiate with respect to time.

Answer: (a) 3

Explanation:
1. When the discharged ball touches the other, the charge is shared equally, so each ball has half of the original charge: \( q'=q/2 \).
2. For small angles \( x^3\propto q^2 \), so \( b^3=a^3/4=108/4=27 \) and \( b=3 \) cm.

Teacher's Note:
1. Charge sharing between identical conductors halves the charge. Then use the cube-law relation between separation and charge.

Answer: (a) 6

Explanation:
1. \( \tan\theta=\dfrac{F}{W_{eff}} \). In air: \( F_a/W \); in liquid: \( F_a/(K\,W') \) where \( W'=W\left(1-\dfrac{0.8}{2.4}\right)=\dfrac23W \).
2. Unchanged angle: \( \dfrac{F_a}{W}=\dfrac{F_a}{K\cdot\frac23W}\Rightarrow K=\dfrac32 \), so 4K = 6.

Teacher's Note:
1. Two effects change in the liquid: the weight is reduced by buoyancy and the electric force is reduced by the dielectric constant. Equate the ratios.

Answer: (a) 2

Explanation:
1. Downward force on each ball: \( 3m\,g+QE=3.5\,mg \) and \( 2m\,g+QE=2.5\,mg \).
2. \( a=\dfrac{3.5mg-2.5mg}{3m+2m}=\dfrac g5=2\ \text{m/s}^2 \).

Teacher's Note:
1. The electric force adds to the weight for both positive charges. Apply Newton's second law to the whole system.

Answer: (a) 9

Explanation:
1. For a point on the perpendicular bisector: \( E=\dfrac{kq}{b\sqrt{b^2+(l/2)^2}} \).
2. \( E=\dfrac{9\times10^9\times5\times10^{-9}}{2\sqrt{4+2.25}}=\dfrac{45}{2\times2.5}=9\ \text{V/m} \).

Teacher's Note:
1. Integrate the field of small elements of the wire along the bisector, or use the standard result. The root of 6.25 is 2.5.

Answer: (a) 4

Explanation:
1. Field of the long string at distance r: \( \dfrac{\lambda_1}{2\pi\varepsilon_0r} \).
2. Force on the segment: \( F=\int_{r_0}^{r_0+l}\dfrac{\lambda_1\lambda_2}{2\pi\varepsilon_0r}dr=\dfrac{\lambda_1\lambda_2}{2\pi\varepsilon_0}\ln\dfrac{r_0+l}{r_0} \).
3. For \( l=r_0 \): \( F=\dfrac{\lambda_1\lambda_2}{4\pi\varepsilon_0}\cdot2\ln2=\dfrac{\lambda_1\lambda_2}{4\pi\varepsilon_0}\ln4 \), so x = 4.

Teacher's Note:
1. Integrate the force on each element of the segment. Rewrite 2 ln 2 as ln 4 to match the given form.

Answer: (a) 8

Explanation:
1. Flux through the face at x = 0 is zero (E = 0). At x = 2 m, \( E=2E_0 \) and the area is 4 m², so flux = \( 8E_0 \). Other faces have no flux (field along x).
2. By Gauss's law the enclosed charge is \( \varepsilon_0\Phi=8\varepsilon_0E_0 \), so n = 8.

Teacher's Note:
1. Only the faces perpendicular to the field contribute. The face at x = 0 has zero field.

Answer: (a) 9

Explanation:
1. Charge per unit cross-sectional area = area under the density curve = trapezoid: \( \tfrac12(1+0.5)\times12\rho_0=9\rho_0 \).
2. Cross-section is 1 m², so \( Q=9\rho_0 \) and \( \Phi=\dfrac{Q}{\varepsilon_0}=\dfrac{9\times8.85\times10^{-12}}{8.85\times10^{-12}}=9 \).

Teacher's Note:
1. Total flux out of a closed surface is the enclosed charge over epsilon naught. Use the area of the trapezoid for the charge.

Answer: (a) 8

Explanation:
1. An element dq = \( \lambda dx \) at distance x on the axis gives flux \( \dfrac{dq}{2\varepsilon_0}\left(1-\dfrac{x}{\sqrt{x^2+R^2}}\right) \) through the disc.
2. Integrating from 0 to infinity: \( \Phi=\dfrac{\lambda}{2\varepsilon_0}\left[x-\sqrt{x^2+R^2}\right]_0^\infty=\dfrac{\lambda R}{2\varepsilon_0}=8 \).

Teacher's Note:
1. The flux through a disc from a point charge on its axis uses the solid angle. Add up the contributions of all elements.

Answer: (a) 16

Explanation:
1. At r = 2 m, E = 4 V/m. By Gauss's law \( Q=4\pi\varepsilon_0a^2E=4\pi\varepsilon_0\times4\times4=64\pi\varepsilon_0 \).
2. Setting this equal to \( 4x\pi\varepsilon_0 \) gives x = 16.

Teacher's Note:
1. Gauss's law for a radial field is E times 4 pi r squared equals Q over epsilon naught. Evaluate the field at the surface.

Answer: (a) 32

Explanation:
1. Surface density \( \sigma=\dfrac{Q}{4\pi R^2} \). Outward force per unit area on a conductor surface = \( \dfrac{\sigma^2}{2\varepsilon_0} \).
2. The net force on the smaller part is this pressure times the area of the cutting plane: \( \pi(R^2-h^2)=\tfrac34\pi R^2 \).
3. \( F=\dfrac{\sigma^2}{2\varepsilon_0}\cdot\dfrac34\pi R^2=\dfrac{3Q^2}{128\pi\varepsilon_0R^2}=\dfrac{3kQ^2}{32R^2} \), so p = 32.

Teacher's Note:
1. Electrostatic pressure on a conductor is sigma squared over two epsilon naught. Use the projected area of the cut.

Answer: (c) (C) and (D) only

Explanation:
1. (A) is true only for identical bodies, not in general, so it is false.
2. (B) is false: charge is invariant and does not depend on speed.
3. (C) is true, and (D) is true: attraction can occur with an uncharged body, but repulsion needs both to carry charge of the same sign.

Teacher's Note:
1. Charge is invariant and always attached to matter. Repulsion is the sure test because attraction can be due to induction.

Answer: (b) (B) and (C)

Explanation:
1. Falling at constant speed: resistance upward = \( mg \). Rising at the same speed: resistance downward, so the electric force \( qE=mg+mg=2mg \) must be upward.
2. The charge is negative, so the field is directed downward. \( E=\dfrac{2mg}{q}=\dfrac{2\times1.6\times10^{-18}\times10}{9.6\times10^{-19}}=\dfrac{100}{3}\ \text{N/C} \).

Teacher's Note:
1. Resistance flips direction with the velocity. Remember the mass must be converted from grams to kilograms.

Answer: (a) (A) and (C)

Explanation:
1. Inside, \( E=\dfrac{kQr}{R^3} \) increases with r: (A) true, (B) false.
2. Outside, \( E=\dfrac{kQ}{r^2} \) decreases: (C) true.
3. Both expressions equal \( kQ/R^2 \) at r = R, so the field is continuous: (D) false.

Teacher's Note:
1. Inside a uniformly charged solid sphere the field grows linearly. The field is continuous at the surface for a volume charge.

Answer: (a) Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation of Statement 1

Explanation:
1. The electrostatic force is conservative, so work is path independent: Statement 1 is true.
2. Statement 2 is the defining property of a conservative force, and path independence follows from it, so it explains Statement 1.

Teacher's Note:
1. Path independence and zero work on a closed loop are two forms of the same property. That is why one explains the other.