Surface charge density is defined as charge per unit

Case study question of Physics – Electric charges and field

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.

Surface charge density is defined as charge per unit

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

Q 1:- What is the value of E in the outer region of the first plate ?

Q 2:- What is the value of E in the outer region of the second plate ?

Q 3:- What will be the value of E between the plates ?

Q 4:- What is the ratio of E from right side of b at distance of 2 cm and 4 cm, respectively ?

Q 5:- In order to estimate the electric field due to a thin finite plane metal plate, what is the shape of the Gaussian surface ?

Answer :- 1. There are two plates A and b having surface charge densities, $\sigm _A = 17.0\times 10^{-22}$ C/m² on B. respectively.

Surface charge density is defined as charge per unit

According to the Gauss’s theorem, if the plates have same surface charge density but having opposite sign, then the electric field in region is zero.

$E = E_A + E_B = \dfrac{\sigma}{2\epsilon _0}+(-\dfrac{\sigma}{2\epsilon _0}) = 0$

2. The electric field in region III is also zero.

$E_{III} = E_A + E_B = \dfrac{\sigma}{2\epsilon _0}+(-\dfrac{\sigma}{2\epsilon _0}) = 0$

3. In region II or between the plates, the electric field

$E = E_A - E_B = \dfrac{\sigma}{2\epsilon _0}+\dfrac{\sigma}{2\epsilon _0}$

$= \dfrac{\sigma(\sigma _A \text{ or } \sigma_B)}{\epsilon _0} = \dfrac{17.0\times 10^{-22}}{8.85\times 10^{-12}}$

$E = 1.9\times 10^{-10}$ N/C

4. Electric field due to an infinite plane sheet of charge does not depend on the distance of observation point from the plane sheet of charge.

so, ratio of E will be 1:1.

5. We take a cylindrical cross-section area A and length 2r as Gaussian surface.

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

Answer :- See all answer

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

Answer :- See the 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 4:- When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net forceon electric dipole in uniform electric field is zero. However, these forces are not collinear, so they give rise to some torque on the dipole. Since, net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However, some work is done in rotating the dipole against the torque acting on it.

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

Q 1:- The dipole moment of a dipole in a uniform external field $\vec{E}$ is $\vec{P}$ . What will be the torque $\vec{\Tau}$ acting on the dipole ?

Solution :- See full solution

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