Identify the incorrect statement among the following
Under electrostatic conditions, the electric field at the surface of a charged conductor must be perpendicular (normal) to the surface at every point. It cannot be tangential, otherwise charges would flow along the surface.
The number of electrons that should be removed from a metal coin such that coin acquires a positive charge of \(10^{-10}\text{ C}\) is
Using quantization of charge, \(q = ne ⇒ n = \frac{q}{e} = \frac{10^{-10}}{1.6 \times 10^{-19}} = 6.25 \times 10^8\).
Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is
When \(N = 64\) drops combine, the new radius is \(R = N^{1/3}r = 4r\). The new charge is \(Q = Nq = 64q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 \left(\frac{kq}{r}\right) = 16V\).
Match column I and Column II.
**Column I**
(A) Coulomb’s law
(B) Surface charge density
(C) Quantisation of charge
(D) Electric flux
**Column II**
(P) Charge/Area
(Q) \(\oint \vec{E} \cdot d\vec{s}\)
(R) \(q = ne\)
(S) Force is inversely proportional to square of distance
A matches with S (inverse square law), B with P (charge/area), C with R (\(q=ne\)), and D with Q (flux).
The electric field in a region is directed outward and is proportional to the distance \(r\) from the origin. If we take a spherical volume of radius \(r\) taking centre at the origin, then the charge contained inside the volume is proportional to
From Gauss's Law, the enclosed charge is proportional to the electric flux: \(q \propto E \cdot A\). Since \(E \propto r\) and area \(A \propto r^2\), we get \(q \propto r \cdot r^2 = r^3\).
A neutral body is charged positively by rubbing it, its weight
When a neutral body is charged positively, it loses electrons. Since electrons have a finite mass \((m_e \approx 9.1 \times 10^{-31}\text{ kg})\), the loss of electrons results in a slight decrease in its mass and weight.
The number of electrons that should be removed from a metal coin such that coin acquires a positive charge of \(10^{-10}\text{ C}\) is
By quantization of charge, \(q = ne\). Solving for \(n\), we get \(n = \frac{q}{e} = \frac{10^{-10}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 6.25 \times 10^8\).
Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is
By conserving volume, \(R = n^{1/3}r = 64^{1/3}r = 4r\). The total charge is \(Q = 64q\). Thus, potential of the big drop is \(V' = \frac{kQ}{R} = \frac{64kq}{4r} = 16V\).
Match column I and Column II.
**Column I**(A) Coulomb’s lawn(B) Surface charge density (C) Quantisation of charge (D) Electric flux
**Column II**(P) Charge/Area (Q) \( \oint \vec{E} \cdot d\vec{s} \) (R) \( q = ne \) (S) Force is inversely proportional to square of distance
Coulomb's law is an inverse square law (A-S). Surface charge density is charge per unit area (B-P). Charge is quantised as \( q=ne \) (C-R). Electric flux is defined by Gauss's integral (D-Q).
Assertion (A): The tyres of aircrafts are slightly conducting.
Reason (R): If a conductor is connected to ground, the extra charge induced on conductor will flow to ground.
Assertion (A) is true: Aircraft tyres are made slightly conducting to discharge static electricity accumulated during flight.nReason (R) is true: Grounding allows excess charge to flow to Earth.
(R) correctly explains (A) as the purpose of conducting tyres is to discharge static charge to the ground safely.