JEE Physics Electric Charges and Fields MCQs Set 19

Answer: (c) only in (iii)

Explanation:
1. (i) Induction pushes positive charge from the knob to the leaves, and the leaves, initially closed, diverge.
2. (ii) Touching transfers positive charge, so the leaves carry more charge and diverge further.
3. (iii) The electroscope now has a net positive charge. A nearby negative rod attracts positive charge towards the knob, away from the leaves, so the leaves collapse towards each other (converge).

Teacher's Note:
1. Bringing an opposite charge near the knob draws charge out of the leaves, so they close. A like charge pushes more charge into the leaves.

Answer: (a) for no range of \( r \)

Explanation:
1. Because the flux is non-zero in every range (\( +2 \), \( -4 \), \( +6 \) in units of \( 10^5 \) N m\( ^2 \) C\( ^{-1} \)), the field cannot be zero on the whole Gaussian sphere in any of the regions.
2. By spherical symmetry the field is radial with constant magnitude on each Gaussian sphere, so \( E=\frac{\phi}{4\pi r^2}\ne0 \) everywhere.

Teacher's Note:
1. With spherical symmetry the field on a Gaussian sphere is \( \frac{\phi}{4\pi r^2} \). It vanishes only where the flux is zero.

Answer: (d) \( 30 \)

Explanation:
1. Let the radii be \( a \) (sphere), \( b \) and \( c \) (inner and outer surfaces of the shell). The shell is a conductor, so its surfaces are at the same potential.
2. The potential difference between the sphere and the shell depends only on the charge \( Q \) on the sphere: \( \frac{Q}{4\pi\epsilon_0}\left(\frac1a-\frac1b\right) \).
3. Charge given to the shell changes both potentials equally, so the difference stays \( 30 \) V.

Teacher's Note:
1. A uniform shift of charge on an outer conductor only adds a constant to the potentials inside it, not to differences.

Answer: (d) \( v\propto\sqrt{\ln\left(\frac{r}{r_0}\right)} \)

Explanation:
1. The potential difference between \( r_0 \) and \( r \) is \( \frac{\lambda}{2\pi\epsilon_0}\ln\frac{r}{r_0} \) (field \( \frac\lambda{2\pi\epsilon_0r} \)).
2. Energy conservation: \( \frac12mv^2=\frac{q\lambda}{2\pi\epsilon_0}\ln\frac r{r_0} \).
3. Hence \( v\propto\sqrt{\ln\frac{r}{r_0}} \).

Teacher's Note:
1. The potential of a line charge is logarithmic in distance. Use work-energy for the speed.

Answer: (d) Statement 1 is false and statement 2 is true

Explanation:
1. Statement 2 is true: by Gauss law \( E=\frac{\rho r}{3\epsilon_0} \) inside.
2. The potential difference from centre to surface is \( \int_0^R\frac{\rho r}{3\epsilon_0}dr=\frac{\rho R^2}{6\epsilon_0} \), so \( \Delta U=\frac{q\rho R^2}{6\epsilon_0} \), not \( \frac{q\rho R^2}{3\epsilon_0} \). Statement 1 is false.

Teacher's Note:
1. Integrate the field from the centre to the surface for the potential difference; do not stop at the field value.

Answer: (b) (i), (ii), (iii) and (iv)

Explanation:
1. An uncharged conductor in the field of the sheet is polarised: the face nearer the sheet gets charge opposite to the sheet and the far face gets charge of the same sign as the sheet.
2. The nearer opposite charge is closer to the source, so the net force is attractive: \( M \) attracts \( A \) and \( B \).
3. Similarly the far face of \( A \) and the near face of \( B \) have opposite signs, so \( A \) and \( B \) attract each other (Newton's third law gives \( B \) attracts \( A \)). All four are correct.

Teacher's Note:
1. Polarisation of a neutral conductor always gives net attraction because the opposite charge is on the nearer side.

Answer: (c) (ii), (iii) and (iv) only

Explanation:
1. At \( (2,0) \) the field is along \( +x \), so the charge is on the x-axis to the left of that point (ii true).
2. The field at \( (3,3) \) points along \( (4,3) \): from the charge \( (x_0,0) \), \( \frac{3}{3-x_0}=\frac34 \) gives \( x_0=-1 \) cm, so (i) is false.
3. At \( (3,3) \) the distance is \( 5 \) cm and \( E=7.2\times5=36 \) N C\( ^{-1} \): \( q=\frac{36\times(0.05)^2}{9\times10^9}=10^{-11} \) C (iii true). At the origin \( V=\frac{9\times10^9\times10^{-11}}{0.01}=9 \) V (iv true).

Teacher's Note:
1. The direction of the field gives the line from the source charge. Use \( E=\frac{kq}{r^2} \) for the charge and \( V=\frac{kq}{r} \) for the potential.

Answer: (a) \( a=8^{-1/4}R \)

Explanation:
1. Total charge: \( \int_0^R4\pi kr^3dr=\pi kR^4=2Q \), so \( k=\frac{2Q}{\pi R^4} \).
2. Charge enclosed within \( a \): \( \pi ka^4=\frac{2Qa^4}{R^4} \). Force on \( A \) from the sphere (attractive): \( \frac{1}{4\pi\epsilon_0}\frac{Q\cdot2Qa^4/R^4}{a^2}=\frac{1}{4\pi\epsilon_0}\frac{2Q^2a^2}{R^4} \) inwards.
3. Repulsion from \( B \): \( \frac{1}{4\pi\epsilon_0}\frac{Q^2}{(2a)^2} \) outwards. Balance: \( \frac{2a^2}{R^4}=\frac{1}{4a^2} \), so \( a^4=\frac{R^4}{8} \) and \( a=8^{-1/4}R \).

Teacher's Note:
1. The charge \( A \) feels the field of the sphere (only the charge enclosed within radius \( a \)) and the repulsion from \( B \).

Answer: (b) \( R_1=0 \) and \( R_2 \lt R_4-R_3 \), and also \( 2R \lt R_4 \)

Explanation:
1. Inside: \( V(r)=\frac{V_0}{2}\left(3-\frac{r^2}{R^2}\right) \). Outside: \( V(r)=\frac{V_0R}{r} \).
2. \( \frac{3V_0}{2} \) is the central value, so \( R_1=0 \). \( \frac{5V_0}{4} \) gives \( r^2=\frac{R^2}{2} \), so \( R_2=\frac{R}{\sqrt2}\approx0.71R \).
3. \( \frac{3V_0}{4} \) gives \( R_3=\frac{4R}{3} \) and \( \frac{V_0}{4} \) gives \( R_4=4R \). Then \( R_4-R_3=\frac{8R}{3} \gt R_2 \), and \( R_4=4R \gt 2R \). So \( R_1=0 \), \( R_2 \lt R_4-R_3 \) and \( 2R \lt R_4 \) all hold.

Teacher's Note:
1. Use the potential inside, \( \frac{V_0}{2}(3-\frac{r^2}{R^2}) \), and outside, \( \frac{V_0R}{r} \), to find each radius.

Answer: (c) \( \frac{a^2V}{2r^2} \)

Explanation:
1. Each drop carries charge \( q=4\pi\epsilon_0rV \).
2. Half the volume is filled: \( N\cdot\frac43\pi r^3=\frac12\cdot\frac43\pi a^3 \), so \( N=\frac{a^3}{2r^3} \).
3. Potential of the sphere \( =\frac{Nq}{4\pi\epsilon_0a}=\frac{N rV}{a}=\frac{a^2V}{2r^2} \).

Teacher's Note:
1. The charge on a drop follows from \( V=\frac{q}{4\pi\epsilon_0r} \). Count the drops from the volume ratio.

Answer: (c) \( -2\sigma \)

Explanation:
1. Let the inner shell have density \( \sigma' \), so its charge is \( 4\pi\left(\frac R2\right)^2\sigma'=\pi R^2\sigma' \).
2. Just inside the outer shell the field is \( E_{in}=\frac{\sigma'}{4\epsilon_0} \); just outside it is \( E_{out}=\frac{\sigma}{\epsilon_0}+\frac{\sigma'}{4\epsilon_0} \). The field acting on the surface charge is the average, \( \frac{\sigma}{2\epsilon_0}+\frac{\sigma'}{4\epsilon_0} \).
3. The force pushing a hemisphere away is \( \sigma\,(\pi R^2)\,E_{avg} \). It vanishes when \( \frac{\sigma}{2}+\frac{\sigma'}{4}=0 \), so \( \sigma'=-2\sigma \).

Teacher's Note:
1. The field on a surface charge layer is the average of the fields on its two sides. The force along the axis acts on the projected area \( \pi R^2 \).

Answer: (a) \( v\propto x^{-1/2} \)

Explanation:
1. For small angles the equilibrium gives \( \frac{x}{2l}=\frac{kq^2}{x^2mg} \), so \( q^2\propto x^3 \) and \( q\propto x^{3/2} \).
2. Constant leak rate: \( \frac{dq}{dt}= \) constant. Differentiating, \( \frac32x^{1/2}\frac{dx}{dt}= \) constant.
3. So the speed of approach \( v=\left|\frac{dx}{dt}\right|\propto x^{-1/2} \).

Teacher's Note:
1. Find \( q \) as a function of \( x \) from the equilibrium condition. Then use the constant rate of change of \( q \).

Answer: (a) the work done by the electrostatic field is independent of the path from \( P \) to \( Q \), because the net work by a conservative force around a closed loop is zero

Explanation:
1. The electrostatic force is conservative.
2. For a conservative force the work around a closed loop is zero, which implies that the work from \( P \) to \( Q \) is the same for every path joining them.

Teacher's Note:
1. Path independence of work and zero work on a closed loop are equivalent statements of a conservative force.

Answer: (a) \( 4 \)

Explanation:
1. Force on a small element \( \lambda_2dr \) at distance \( r \): \( \lambda_2dr\cdot\frac{\lambda_1}{2\pi\epsilon_0r} \).
2. \( F=\frac{\lambda_1\lambda_2}{2\pi\epsilon_0}\ln\frac{r_0+\ell}{r_0}=\frac{\lambda_1\lambda_2}{4\pi\epsilon_0}\ln\left(\frac{r_0+\ell}{r_0}\right)^2 \).
3. With \( \ell=r_0 \): \( x=2^2=4 \).

Teacher's Note:
1. Write the force on a small element using the field of the line charge and integrate. Bring the factor \( 2 \) inside the logarithm as a square.

Answer: (d) \( 3 \)

Explanation:
1. Inside the cavity \( E=0 \), so \( V_A=V_{r=a} \).
2. For \( a \lt r \lt b \): \( E=\frac{\rho\left(r^3-a^3\right)}{3\epsilon_0r^2} \).
3. \( V_A-V_B=\int_a^{2a}\frac{\rho}{3\epsilon_0}\left(r-\frac{a^3}{r^2}\right)dr=\frac{\rho}{3\epsilon_0}\left[\frac{r^2}{2}+\frac{a^3}{r}\right]_a^{2a}=\frac{\rho a^2}{3\epsilon_0} \), so \( n=3 \).

Teacher's Note:
1. Use Gauss law for the enclosed charge of the shell portion, then integrate the field from \( a \) to \( b \).

Answer: (a) \( 2 \)

Explanation:
1. The first ball gains kinetic energy \( U_1=kq^2\sum_{j\ne1}\frac1{r_{1j}}=kq^2S \), where \( S \) is the same sum for every vertex by symmetry.
2. The second ball, released after the first has gone, interacts only with the remaining \( 2017 \) balls: \( U_2=kq^2\left(S-\frac1a\right) \).
3. \( K=U_1-U_2=\frac{kq^2}{a} \), so \( q^2=\frac{Ka}{k}=\frac{9\times10^9\times4\times10^{-6}}{9\times10^9}=4\times10^{-6} \) and \( q=2\times10^{-3} \) C \( =2 \) mC.

Teacher's Note:
1. By symmetry every vertex has the same potential energy sum. The difference arises only from the missing neighbour.

Answer: (c) (i) and (iii) only

Explanation:
1. Inside, \( E=\frac{\rho r}{3\epsilon_0} \) increases linearly with \( r \) (i true, ii false).
2. Outside, \( E=\frac{Q}{4\pi\epsilon_0r^2} \) decreases with \( r \) (iii true).
3. At \( r=R \) both expressions give \( \frac{Q}{4\pi\epsilon_0R^2} \), so \( E \) is continuous (iv false).

Teacher's Note:
1. For a volume distribution the field is continuous. A discontinuity occurs only across a surface charge.

Answer: (b) (i) and (iii) only

Explanation:
1. \( E=-\frac{dV}{dr} \), so its magnitude is the magnitude of the potential gradient (i true).
2. By definition it is the force per unit positive test charge (iii true).
3. The force on an electron is \( eE \) (not \( E \)), and there is no charge 'at' the point in general (ii and iv false).

Teacher's Note:
1. Field intensity is the force per unit charge and the negative gradient of the potential.

Answer: (b) (i) and (iv) only

Explanation:
1. Gaussian cylinder of radius \( x \lt R \) and length \( \ell \): \( E\,2\pi x\ell=\frac{\rho\pi x^2\ell}{\epsilon_0} \), so \( E=\frac{\rho x}{2\epsilon_0} \) (i).
2. For \( x \gt R \) the enclosed charge is \( \rho\pi R^2\ell \): \( E=\frac{\rho R^2}{2\epsilon_0x} \) (iv).
3. The other two expressions are off by a factor of \( 2 \).

Teacher's Note:
1. Use a coaxial cylinder as the Gaussian surface. The enclosed charge is \( \rho\times \) (area \( \times\ell \)).

Answer: (b) \( \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} \), directed radially, and unchanged if \( Q' \) is removed

Explanation:
1. The conducting shell screens its interior from \( Q' \) and from the charge \( Q \) on the shell; the inner surface carries \( -q \) spread uniformly.
2. The field inside the cavity is then due to \( q \) alone: \( E=\frac{q}{4\pi\epsilon_0r^2} \), which decreases with \( r \) as an inverse square and does not depend on \( Q' \).
3. Removing \( Q' \) therefore leaves the field for \( r \lt R \) unchanged.

Teacher's Note:
1. Inside a closed conducting shell the field depends only on the charges inside it. Outside charges are screened.

Answer: (d) \( 4 \)

Explanation:
1. At the closest approach both particles move with the same velocity \( \frac V2 \) (momentum conservation).
2. Energy conservation: \( \frac12mV^2=2\cdot\frac12m\left(\frac V2\right)^2+\frac{Q^2}{4\pi\epsilon_0d} \), so \( \frac{Q^2}{4\pi\epsilon_0d}=\frac14mV^2 \).
3. \( d=\frac{4Q^2}{4\pi\epsilon_0mV^2} \), so \( x=4 \).

Teacher's Note:
1. At minimum separation the relative velocity is zero. Use conservation of momentum and energy together.