Coulamb's law states that the electrostatic force of attraction

Case study question of Physics – Electric charges and field

Case study 2:- Coulamb’s law states that the electrostatic force of attraction or repulsion acting between two stationary points charges is given by

$F = \dfrac{1}{4\pi \epsilon _0}\dfrac{q_1q_2}{r^2}$

Where F denotes the force between two charges $q_1$ and $q_2$ separated by a distance r in free space, $\epsilon _0$ is a constant known as permittivity of free space. Free space is vacuum and may be taken to be air practically.

If free space is replaced by a medium, then $\epsilon _0$ is repaced by $(\epsilon _0 k)$ or $\epsilon _0 \epsilon$ , where k is known as dielectric constant or relative permittivity.

Read the given passage carefully and give the answer of the following questions:

Q 1:- In coulamb’s law, $F = k\dfrac{q_1 q_2}{r^2}$ , then on which of the following factors does the proportionality constant k depends ?

(a) Electrostatic force acting between the two charge

(b) Nature of the medium between the two charges

(d) Distance between the two charges

Q 2:- Dimensional formula for the permittivity constant $\epsilon _0$ of free space is :

(a) $[ML^{-3}T^4A^2]$ (b) $[M^{-1}L^{3}T^2A^2]$

Q 3:- The force of repulsion between two charges of 1 c each, kept 1 m apart in vacuum is:

(a) $\dfrac{1}{9\times 10^9}$ N

(b) $9\times 10^9$ N

(d) $\dfrac{1}{9\times 10^{12}}$ N

Q 4:- Two identical charges repel each other with a force equal to 10 mg wt when they are 0.6 m apart in air. (g = 10 m/s²). The value of each charge is :

(a) 2 mC (b) $2\times 10^{-7}$ mC

Q 5:- Coulamb’s law for the force betweenelectric charges most closely resembles with:

(a) law of conservation of energy

(b) Newton’s law of gravitation

(d) Law of conservation of charge

Answer :- 1. (b) Nature of the medium between the two charges.

2.(c) $[M^{-1}L^{-3}T^4A^2]$

As, we know that.

$[\epsilon _0] = \dfrac{1}{4\pi F}\dfrac{q_1 q_2}{r62}$

⇒ $[\epsilon _0] = \dfrac{[AT]^2}{[MLT^{-2}][L^2]}$

⇒ $[\epsilon _0] = [M^{-1}L^{-3}T^4A^2]$

3. (b) $9\times 10^9$ N

4 (d) 2 μC

Given that, $F = 10 \text{ mg wt} = 10 \times 10^{-3} \times 10$ N

⇒ $F = \dfrac{1}{\4\pi \epsilon _0}\dfrac{q_1 q_2}{d^2}$

∴ $(10\times 10^{-3})\times 10 = \dfrac{(9\times 10^9)\times q^2}{(0.6)^2}$

or $q^2 = \dfrac{10^{-1}\times 0.36}{9\times 10^9} = 4\times 10^{-12}$

Or $q = 2\times 10^{-6}$ C = 2 μC

5. (b) Newton’s law of gravitation.

Case study 1:- A charge is a property associated with the matter due to which it experiences and produces an electric and magnetic fields. Charges are scalar in nature and they add up like real numbers. Also, the total charge of an isolated system is always conserved. When the objects rub against each other, charges acquired by them must be equal and opposite.

Read the given passage carefully and give the answer of the following questions:

Q 1:- The cause of charging is:

(a) The actual transfer of protons

(b) The actual transfer of electrons

(d) None of the above

Q 2:- When glass rod is rubbed with silk, then:

(a) Negative charge is produced on silk but no charge on glass rod

(b) Equal but opposite charge are produced on both

(d) Positive charge is produced on glass rod but no charge on silk

Q 3:- If an object is positively charged theoretically, the mass of the object:

(a) Remains the same

(b) Increases slightly by a factor of $9.11 \times 10^{-31}$ kg

(d) Decreases slightly by a factor of $9.11 \times 10^{-31}$ kg.

Q 4:- We have two bodies with charges $q_1$ and $q_2$ on them, then $q_1 + q_2 = 0$ signify:

(a) $q_1$ and $q_2$ are equal charges with opposite signs

(b) $q_1$ and $q_2$ are equal charges with same signs

(d) $q_1$ and $q_2$ are equal charges.

Q 5:- The cause of quantisation of electric charge is:

(a) Transfer of an integral number of neutrons

(b) Transfer of an integral number of protons

(d) None of the above.

Answer :- See all answer

Case study 3:- Animals emit low frequency electric fields due to a process known as osmoregulation. This process allows the concentration of ions (charged atoms or molecules) to flow between the inside of our bodies and the outside. In order for our cells to stay intact, the flow of ions needs to be balanced. But balanced doesn’t necessarily mean equal. The concentration of ions within a shrimp’s body is much lower than that of the sea water it swims in. Their voltage or potential sifference generated between the two concentrations across ‘leaky’ surfaces, can then be measured.

Read the given passage carefully and give the answer of the following questions:

Solution:- See full solution

Case study 5:- Surface charge density is defined as charge per unit surface area of surface charge distribution i.e., σ = dq/ds. Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sign having magnitude of $17.0\times 10^{-22}$ C/m² as shown. The intensity of electric field at a point is $E =\dfrac{\sigma}{\epsilon_0}$ , where $\epsilon _0=$ permittivity of free space.