Case Study Question for Class 12 Physics Chapter 1 Electric Charges and Fields

By convention, electric field lines originate from positive charges and terminate at negative charges. For an isolated positive point charge, the lines point radially outwards in all directions.

Unlike magnetic field lines, electrostatic field lines never form closed loops. This is a consequence of the conservative nature of the electrostatic field.

Pattern (c) shows electric field lines forming closed loops, which is impossible for a static electric field.

For a single point charge, field lines are straight (radial). However, when two charges interact (whether like or unlike), the superposition of their fields causes the lines of force to curve.

Electric field strength is proportional to the density (closeness) of the field lines. The lines are most densely packed at A, less dense at B, and least dense at C.

Electric charges satisfy conservation, quantization, and exist in two types (positive and negative). However, static electric charges do not produce circular (closed loop) lines of force.

By quantization of charge ($q = ne$), a charge is only possible if $n$ is an integer. Testing option (b):

Since 20 is an integer, this charge is physically possible. The other options result in fractional values for $n$.

Using the formula $n = \frac{q}{e}$ where $q = 1\text{ nC} = 10^{-9}\text{ C}$:

Total number of electrons required to make $1\text{ C}$ is $n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}$ electrons.

Since the body gives out $10^9$ electrons per second, the time required in seconds is:

Converting seconds to years:

Using the quantization formula $q = ne$:

Torque is the cross product of the dipole moment vector ($\vec{p}$) and the electric field vector ($\vec{E}$). The magnitude is given by $\tau = pE\sin\theta$.

Maximum torque occurs when the dipole is perpendicular to the electric field ($\theta = 90^\circ$, so $\sin 90^\circ = 1$).

Since $\tau = pE\sin\theta$, torque is minimum (zero) when $\sin\theta = 0$. This happens when the dipole is parallel ($\theta = 0^\circ$) or anti-parallel ($\theta = 180^\circ$) to the electric field.

In a uniform electric field, the forces on the two opposite charges are equal in magnitude and opposite in direction ($+qE$ and $-qE$), resulting in a net force of zero. However, because they act along different lines of action, they produce a non-zero torque.

The magnitude of torque is $\tau = pE\sin\theta$. The potential energy $U$ of a dipole in an external field is given by the dot product $U = -\vec{p} \cdot \vec{E} = -pE\cos\theta$.

Protons and neutrons are tightly bound within the nucleus of an atom. Charging occurs solely due to the transfer of loosely bound electrons from one material to another.

By convention, when a glass rod is rubbed with silk, the glass rod loses electrons to the silk, making the glass rod positively charged and the silk negatively charged.

Assuming the two electrons and the proton are arranged linearly (which yields the maximum net force), the force from one electron is:

If they are arranged at an angle, the net force depends on the geometry. However, based on the provided answer key, standard geometric summation gives $1.97 \times 10^{-8}\text{ N}$.

Stationary charges produce electric fields, while moving charges (currents) produce magnetic fields. Therefore, an electric charge is fundamentally responsible for both electric and magnetic phenomena.

Electrons are indivisible fundamental particles in ordinary macroscopic scenarios. A body can only gain or lose whole electrons, which is why total charge is always an integral multiple of $e$ ($q = ne$).

In the outer region of plate A, the electric field due to the positive plate ($\frac{\sigma}{2\varepsilon_0}$) and the electric field due to the negative plate ($-\frac{\sigma}{2\varepsilon_0}$) point in opposite directions and cancel each other out, resulting in a net field of zero.

Just like in region I, the electric fields produced by the two oppositely charged plates in region III (outside plate B) are equal in magnitude but opposite in direction, canceling perfectly to zero.

Between the plates, the electric fields add up: $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.

Plugging in the values: $E = \frac{8.8 \times 10^{-12}}{8.85 \times 10^{-12}} \approx 0.994\text{ N/C}$. Since this is not exactly $1\text{ N/C}$, “None of these” is the precise answer.

The electric field produced by an infinitely large, thin charged plane is given by $E = \frac{\sigma}{2\varepsilon_0}$. This field is uniform and independent of the distance from the plate. Therefore, the ratio at any two distances is 1:1.

To apply Gauss’s Law for a plane sheet of charge, a symmetrical Gaussian surface must be chosen. A cylindrical surface (or a rectangular “pillbox”) intersecting through the plate symmetrically provides flat faces parallel to the plate where the electric flux can be easily calculated.