JEE Physics Electric Charges and Fields MCQs Set 12

Answer: (a) \( d\sqrt K \)

Explanation:
1. Force in the medium \( F=\frac{q_1q_2}{4\pi\varepsilon_0Kd^2} \).
2. In air at distance d': \( \frac{q_1q_2}{4\pi\varepsilon_0d'^2} \). Equating, \( d'^2=Kd^2\Rightarrow d'=d\sqrt K \).

Teacher's Note:
1. A dielectric medium reduces the force by the factor K.
2. A medium weakens the force, so in air the same force needs the charges to be farther apart: d' = d√K.

Answer: (a) 2 mm

Explanation:
1. Time in the field \( t=\frac{0.1}{3\times10^7}=3.33\times10^{-9}\ \text{s} \).
2. Acceleration \( a=\frac qmE=2\times10^{11}\times1800=3.6\times10^{14}\ \text{m/s}^2 \).
3. \( y=\frac12at^2=\frac12\times3.6\times10^{14}\times(3.33\times10^{-9})^2=2\times10^{-3}\ \text{m}=2\ \text{mm} \).

Teacher's Note:
1. Horizontal motion is uniform; vertical motion is uniformly accelerated.
2. 1.8 kV/m = 1800 V/m.

Answer: (a) x = 6 cm

Explanation:
1. At x = 2 cm the field of the 10 \( \mu \)C charge points along +x: \( E_1=\frac{k\times10}{2^2} \).
2. The 40 \( \mu \)C charge must produce an equal field along -x, so it lies beyond x = 2 cm at distance d: \( \frac{40}{d^2}=\frac{10}{4}\Rightarrow d=4\ \text{cm} \).
3. Its position is \( x=2+4=6\ \text{cm} \).

Teacher's Note:
1. The two fields must be equal and opposite at the point.
2. The larger charge sits farther from the point.

Answer: (a) \( \frac{q}{12\varepsilon_0} \)

Explanation:
1. Join a mirror image of the cuboid on the other side of S to get a cube of side 2L with the charge q at its centre.
2. The flux through each face of the cube is \( \frac{q}{6\varepsilon_0} \).
3. Only half of the flux from the charge enters the original cuboid, so the flux through the face opposite to S is \( \frac12\times\frac{q}{6\varepsilon_0}=\frac q{12\varepsilon_0} \).

Teacher's Note:
1. The charge lies on the surface S, so only half of its field lines enter the cuboid.
2. Symmetry makes the flux through the six faces of the cube equal.

Answer: (a) 24 cm

Explanation:
1. The forces can cancel only outside the charges, on the side of the smaller charge (beyond x = 12 cm).
2. With the proton at distance x from the origin: \( \frac{4}{x^2}=\frac{1}{(x-12)^2}\Rightarrow2(x-12)=x \).
3. \( x=24\ \text{cm} \).

Teacher's Note:
1. Between the charges the two forces add, so cancellation is impossible there.
2. The point must be nearer to the smaller charge.

Answer: (a) 12

Explanation:
1. Electric field \( E=-\frac{dV}{dr}=-4ar \).
2. Gauss's law for a uniformly charged sphere: \( E\cdot4\pi r^2=\frac{\rho}{\varepsilon}\cdot\frac43\pi r^3\Rightarrow E=\frac{\rho r}{3\varepsilon} \).
3. \( \frac{\rho r}{3\varepsilon}=-4ar\Rightarrow\rho=-12a\varepsilon \), so \( \lambda=12 \).

Teacher's Note:
1. Find the field from the potential, then apply Gauss's law with a spherical surface of radius r.
2. The negative sign shows that the charge is negative.

Answer: (b) \( E_I\ne0,\ E_{II}=0,\ E_{III}\ne0 \)

Explanation:
1. Region I: the point charge produces a field \( \frac{kQ}{r^2}\ne0 \).
2. Region II is inside the conducting material, so the field is zero; charge -Q is induced on the inner surface and +Q on the outer surface.
3. Region III: the total enclosed charge is Q, so the field is \( \frac{kQ}{r^2}\ne0 \).

Teacher's Note:
1. The field inside the metal of a conductor in equilibrium is always zero.
2. The induced charges rearrange so that the field outside depends only on the total charge.

Answer: (a) 12

Explanation:
1. Faces perpendicular to x: at x = 1, \( E_x=2 \): flux \( 2\times(2\times3)=12 \) (outward); at x = 0, flux 0.
2. Faces perpendicular to y: at y = 2, \( E_y=-8 \): flux \( -8\times(1\times3)=-24 \); at y = 0, flux 0.
3. Faces perpendicular to z: at z = 3 flux \( 6\times2=12 \) outward; at z = 0 flux \( -6\times2=-12 \).
4. Net flux \( =12-24+12-12=-12 \), so \( q=\varepsilon_0\Phi=-12\varepsilon_0 \) and n = 12 (in magnitude).

Teacher's Note:
1. Add the outward flux through all six faces, taking care with the signs.
2. The net flux through the closed surface is q/ε0 by Gauss's law.

Answer: (a) 640 V cm

Explanation:
1. The field is along x, so only the faces perpendicular to x contribute: at x = 0 the field is 0, and at x = 0.2 m \( E=4000\times0.04=160\ \text{V/m} \).
2. \( \Phi=160\times(0.2)^2=6.4\ \text{V m}=640\ \text{V cm} \).

Teacher's Note:
1. Faces parallel to the field have zero flux.
2. Convert 1 V m = 100 V cm.

Answer: (a) The net potential of the system cannot be zero at a point but the net electric field can be zero at that point.

Explanation:
1. Potential is a scalar, and each positive charge contributes a positive potential, so the sum cannot be zero.
2. The field is a vector: for example at the midpoint of two equal positive charges the fields cancel, so the net field can be zero.

Teacher's Note:
1. Scalars add with sign; vectors add with direction.
2. Electric field zero does not imply zero potential.

Answer: (d) \( \vec E_I=-\frac\sigma{\varepsilon_0}\hat n,\ \vec E_{II}=0,\ \vec E_{III}=\frac\sigma{\varepsilon_0}\hat n \)

Explanation:
1. Each sheet produces a field of magnitude \( \frac{\sigma}{2\varepsilon_0} \) pointing away from it.
2. Between the sheets the two fields are opposite and cancel: \( E_{II}=0 \).
3. In region I both fields point left and in region III both point right: \( E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac\sigma{\varepsilon_0} \), so \( \vec E_I=-\frac\sigma{\varepsilon_0}\hat n,\ \vec E_{III}=+\frac\sigma{\varepsilon_0}\hat n \).

Teacher's Note:
1. Superpose the fields of the two sheets with their directions.
2. The field of an infinite sheet is independent of the distance.

Answer: (a) 2

Explanation:
1. At distance y on the bisector: \( E=\frac{2kqy}{(a^2+y^2)^{3/2}} \).
2. Maximise: \( \frac{dE}{dy}=0\Rightarrow(a^2+y^2)-3y^2=0\Rightarrow y=\frac{a}{\sqrt2} \), so x = 2.

Teacher's Note:
1. Only the components along the bisector add up.
2. Differentiate E with respect to y and set the derivative to zero.

Answer: (a) 288

Explanation:
1. Flux only through the faces at x = a (x = 0 has zero field): \( \Phi=E_0a\cdot a^2=E_0a^3=4\times10^4\times(0.02)^3=0.32\ \text{V m} \).
2. \( q=\varepsilon_0\Phi=9\times10^{-12}\times0.32=2.88\times10^{-12}\ \text{C}=288\times10^{-14}\ \text{C} \).

Teacher's Note:
1. Apply Gauss's law to the cube.
2. Convert 2 cm to 0.02 m.