JEE Physics Electric Charges and Fields MCQs Set 09

Answer: (a) 16 μC

Explanation:
1. Charge removed from electrons = number × e = \( 10^{14}\times1.6\times10^{-19}=1.6\times10^{-5} \) C.
2. Removing electrons leaves the sphere positively charged, so the charge is +16 μC.

Teacher's Note:
1. Losing electrons makes a body positive; gaining electrons makes it negative.
2. Charge is always an integral multiple of e.

Answer: (b) 6.25 × 10¹⁸

Explanation:
1. \( n=\frac{Q}{e}=\frac{1}{1.6\times10^{-19}}=6.25\times10^{18} \).

Teacher's Note:
1. Quantisation of charge: Q = ne.
2. This number is worth remembering for quick estimates.

Answer: (b) K : 1

Explanation:
1. In air \( F_a=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2} \); in a medium \( F_m=\frac{F_a}{K} \).
2. \( F_a:F_m=K:1 \).

Teacher's Note:
1. A dielectric medium reduces the force between charges by the factor K.
2. For a fixed charge pair at fixed distance, only the medium changes the force.

Answer: (c) Along the diagonal BD

Explanation:
1. The equal charges +q at A and C are at the same distance from O, so their fields cancel.
2. The field of +2q at B points from B towards D; the field of −2q at D points towards D as well, so they add.
3. The net field (and force on a positive charge) is along diagonal BD.

Teacher's Note:
1. Look for pairs that cancel by symmetry first.
2. A positive charge and a negative charge on opposite corners give fields in the same direction at the centre.

Answer: (d) \( Q_1=\frac{Q}{2},\ Q_2=\frac{Q}{2} \)

Explanation:
1. \( F\propto Q_1(Q-Q_1) \) which is a parabola in \( Q_1 \).
2. Maximum when \( \frac{d}{dQ_1}[Q_1(Q-Q_1)]=Q-2Q_1=0 \), i.e. \( Q_1=Q_2=\frac{Q}{2} \).

Teacher's Note:
1. For a fixed sum, a product is largest when the two numbers are equal.
2. The same idea is used in many maximum-force or maximum-power problems.

Answer: (b) −Q/4

Explanation:
1. Consider an end charge Q at distance d from the middle charge and 2d from the other end charge.
2. Forces on it: \( \frac{kQq}{d^2}+\frac{kQ^2}{(2d)^2}=0 \).
3. \( q=-\frac{Q}{4} \).

Teacher's Note:
1. For equilibrium of the whole system, check one end charge; the middle one is balanced by symmetry.
2. The middle charge must be opposite in sign to the end charges.

Answer: (b) mg/e

Explanation:
1. Set \( eE=mg \).
2. \( E=\frac{mg}{e} \).

Teacher's Note:
1. Equate the electric force with the weight.
2. Check units: (kg·m/s²)/C gives N/C, which is the unit of E.

Answer: (a) At a distance of 5 cm from 25 μC

Explanation:
1. The field is zero between the charges (like charges). Let it be x from the 25 μC charge.
2. \( \frac{25}{x^2}=\frac{36}{(11-x)^2}\Rightarrow\frac{5}{x}=\frac{6}{11-x} \).
3. \( 55-5x=6x\Rightarrow x=5 \) cm from the 25 μC charge.

Teacher's Note:
1. Between two like charges the fields oppose, so the null point lies between them, closer to the smaller charge.
2. Take square roots of the ratio to avoid a quadratic.

Answer: (b) 260 N/C

Explanation:
1. Radius \( r=10^{-5} \) cm \( =10^{-7} \) m. Mass \( m=\frac43\pi r^3\rho=\frac43\pi(10^{-7})^3(1000)=4.19\times10^{-18} \) kg.
2. \( eE=mg\Rightarrow E=\frac{4.19\times10^{-18}\times10}{1.6\times10^{-19}}\approx262 \) N/C, about 260 N/C.

Teacher's Note:
1. Suspension means electric force equals weight.
2. Convert the radius to metres before cubing it; a slip of 10² in the radius changes the answer by 10⁶.

Answer: (d) Zero

Explanation:
1. Flux entering through one flat face is \( -E\pi R^2 \) and flux leaving through the other is \( +E\pi R^2 \).
2. The curved surface is parallel to the field, so it contributes nothing.
3. Net flux \( =0 \).

Teacher's Note:
1. The net flux through any closed surface is zero when it encloses no charge.
2. Count entering flux as negative and leaving flux as positive.

Answer: (b) \( \frac{100Q}{\varepsilon_0} \)

Explanation:
1. Length of wire enclosed = 1 m = 100 cm, so enclosed charge \( =100Q \).
2. By Gauss's law \( \phi=\frac{q_{enc}}{\varepsilon_0}=\frac{100Q}{\varepsilon_0} \).

Teacher's Note:
1. Flux depends only on the enclosed charge, not on the radius of the Gaussian surface.
2. Convert 1 m to cm because the charge density is given per cm.

Answer: (b) \( (\phi_2-\phi_1)\varepsilon_0 \)

Explanation:
1. Net outward flux \( =\phi_2-\phi_1 \).
2. Gauss's law: \( \phi_{net}=\frac{q}{\varepsilon_0}\Rightarrow q=(\phi_2-\phi_1)\varepsilon_0 \).

Teacher's Note:
1. Flux entering counts negative, so subtract it from the outgoing flux.
2. Multiply by ε₀, do not divide.

Answer: (b) \( \frac{\sigma}{\varepsilon_0} \)

Explanation:
1. For conducting sheets the charge sits on the facing surfaces, so each sheet effectively behaves as a one-sided charged surface.
2. Between the sheets both fields add: \( E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0} \), uniform everywhere between.

Teacher's Note:
1. Opposite charged parallel plates give a uniform field only between them.
2. Outside the plates the fields cancel and E = 0.

Answer: (c) 120 V/m

Explanation:
1. Outside: \( E=\frac{kQ}{r^2} \), so \( kQ=100\times(0.2)^2=4 \). At the surface (0.1 m): \( E_s=\frac{4}{0.01}=400 \) V/m.
2. Inside a uniformly charged sphere \( E\propto r \): \( E=400\times\frac{3}{10}=120 \) V/m.

Teacher's Note:
1. Outside a sphere the charge acts as a point charge at the centre; inside, field grows linearly with r.
2. Both expressions agree (E = 400 V/m) at r = R.

Answer: (b) Q/2

Explanation:
1. Balance: \( qE=mg\Rightarrow q\frac{V}{d}=mg\Rightarrow q\propto\frac{m}{V} \).
2. Mass \( m\propto r^3 \), so the second drop has \( m/8 \); voltage is \( \frac{1}{4} \) of the first.
3. \( q_2=Q\times\frac{1/8}{1/4}=\frac{Q}{2} \).

Teacher's Note:
1. Mass scales as the cube of the radius for a sphere.
2. Set up a ratio instead of calculating the actual values.

Answer: (c) \( \frac{\rho r}{3\varepsilon_0} \)

Explanation:
1. Charge inside radius r: \( q=\rho\cdot\frac43\pi r^3 \).
2. Gauss: \( E\cdot4\pi r^2=\frac{q}{\varepsilon_0}\Rightarrow E=\frac{\rho r}{3\varepsilon_0} \).

Teacher's Note:
1. Only the charge inside the Gaussian sphere matters.
2. The field is linear in r inside and falls as 1/r² outside.

Answer: (c) 0

Explanation:
1. \( U=-pE\cos\theta \).
2. U is minimum (−pE) when \( \cos\theta=1 \), i.e. \( \theta=0 \).

Teacher's Note:
1. Stable equilibrium of a dipole is when p is parallel to E.
2. At θ = π the potential energy is maximum and the equilibrium is unstable.

Answer: (d) A stationary charge produces both electric and magnetic fields

Explanation:
1. Charge is quantized (q = ne) and conserved in every process.
2. A charge at rest produces only an electric field; a magnetic field needs moving charge.
3. So statement (d) is wrong.

Teacher's Note:
1. Static charges: electric field only.
2. Moving charges (currents) are needed for magnetism.

Answer: (c) May be neutral

Explanation:
1. Repulsion needs both balls charged alike, so 2, 3, 4 and 5 are charged; 2 and 4 attract, so they carry opposite charges.
2. Ball 1 attracts both 2 and 4, which carry opposite charges. A charged ball 1 would repel one of them.
3. Hence ball 1 must be neutral (it is attracted by induction).

Teacher's Note:
1. Only repulsion proves that both bodies are charged; attraction may also occur between a charged and a neutral body.
2. Use the repelling pairs to fix the signs first.

Answer: (c) 180°

Explanation:
1. Without gravity the only forces on each ball are the string tension and the Coulomb repulsion.
2. Equilibrium needs the tension to balance the repulsion, which acts along the line joining the balls.
3. So both strings lie along that line: the angle is 180°.

Teacher's Note:
1. With gravity the angle is less than 180°; the weight is what pulls the balls down.
2. In weightlessness the strings become collinear.

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

Explanation:
1. Like charges: the null point lies between them. Let it be x from q.
2. \( \frac{kq}{x^2}=\frac{k\cdot4q}{(30-x)^2}\Rightarrow30-x=2x\Rightarrow x=10 \) cm.
3. So the point is 10 cm from q (20 cm from 4q), on the line between them.

Teacher's Note:
1. The null point is nearer the weaker charge.
2. The ratio of distances equals the square root of the ratio of charges.

Answer: (b) 10 μC, 10 μC

Explanation:
1. \( F\propto q(20-q) \).
2. This is maximum at \( q=10 \) μC, giving two equal parts of 10 μC.

Teacher's Note:
1. Equal parts maximise a product with fixed sum.
2. The maximum product here is 100 (μC)².

Answer: (b) \( \frac{E^2q^2t^2}{2m} \)

Explanation:
1. Acceleration \( a=\frac{qE}{m} \), speed \( v=at=\frac{qEt}{m} \).
2. \( KE=\frac12mv^2=\frac{E^2q^2t^2}{2m} \).

Teacher's Note:
1. Use v = at for uniform acceleration from rest.
2. Alternatively KE = work done = qE × (½at²), which gives the same result.

Answer: (c) \( \frac{1}{4\pi\varepsilon_0}\frac{2q}{3\sqrt3R^2} \)

Explanation:
1. On the axis \( E=\frac{kqx}{(R^2+x^2)^{3/2}} \). Setting \( dE/dx=0 \) gives \( x=\frac{R}{\sqrt2} \).
2. \( E_{max}=\frac{kq\,R/\sqrt2}{(3R^2/2)^{3/2}}=\frac{2kq}{3\sqrt3R^2} \).

Teacher's Note:
1. The field is zero at the centre and at infinity, so it has a maximum in between.
2. The maximum occurs at a distance R/√2 from the centre.

Answer: (c) 173.2

Explanation:
1. The area vector of a surface in the xy-plane is along z: \( \vec A=100\hat k \).
2. \( \phi=\vec E\cdot\vec A=\sqrt3\times100=173.2 \) V m.

Teacher's Note:
1. Only the component of E perpendicular to the surface counts.
2. Flux = E A cos θ, and here only E_z contributes.