A dielectric slab is a substance which does not allow

Case study question of Physics – Electrostatic potential and capacitance

Case study 3:- A dielectric slab is a substance which does not allow the flow of charges through it but permits them to exert electrostatic forces on one another. When a dielectric slab is placed between the plates, the field $E_0$ polarises the dielectric. This induce charge $-Q_p$ on the upper surface are $+Q_p$ on the lower surface of the dielectric. These induced charges set up a field $E_p$ inside the dielectric in the opposite direction of $\vec{E_0}$ as shown.

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

Q 1. In a parallel plate capacitor, thee capacitance iincreases from 4 μF to 80 μF, on introducing a dielectric medium between the plates. whaat is the dielectric constant of the medium ?

(a) 10 (b) 20

Q 2. A paarallel plate capacitor with air between the plates has a capacitance of 8 pF. the separation between the plates is now reduced half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in second case.

(a) 8 pF (b) 10 pF

Q 3. A dielectric introduced between the plates of a parallel plate condenser:

(a) Decreases the electric field between the plates

(b) Increases the capacity of the condenser

(d) Increases the capacity of the condenser

Q 4. A parallel plate capacitor of capacitance 1 pF has separation between the platess d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it, the capacitance becomes 2 pF. what is the value of x ?

(a) 2 (b) 4

Q 5. A parallel plate capacitor having area A and separated by distance d filled by copper plate of thickness b. The new capacity is:

(a) $\dfrac{\epsilon_0 A}{d + \frac{b}{2}}$

(b) $\dfrac{\epsilon_0 A}{2d}$

(d) $\dfrac{2\epsilon A}{d + \frac{b}{2}}$

Solution:- 1. (b) 20

Dielectric constant

$E = \dfrac{\text{Capacitance with dielectric}}{\text{Capacitance without dielectric}}$

$= \dfrac{80}{4} = 20$

2. (c) 80 pF

Capacitance of the capacitor with air between plates

$C' = \dfrac{\epsilon_0 A}{d} = 8pF$

With the capacitor filled with dielectric (k = 5) between its plates and the distance between the plates is reduced by half, capacitance become

$C = \dfrac{\epsilon_0 KA}{d/2} = \dfrac{\epsilon_0 \times 5\times A}{d/2}$

$= 10C' = 10\times 8 = 80pF$

3 (d) Increases the capacity of the condenser

If a dielectric medium of dielectric constant K is filled completely between the plates, then capacitance increases by K times.

4. (b) 4

$C = \dfrac{\epsilon_0 A}{d} = 1pF$ . . .(i)

$C' = \dfrac{X\epsilon_0 A}{(2d)} = 2pF$ . . . (ii)

Dividing eq (ii) by eq (i)

$\dfrac{x}{2} = \dfrac{2}{1}$

⇒ x = 4

5. (c) $\dfrac{\epsilon_0 A}{d - b}$

As capacitance, $C_0 = \dfrac{\epsilon0 A}{d}$

$\therefore After inserting copper plate, [latex]C_0 = \dfrac{\epsilon0 A}{d-b}$

Case study 4:- Potential difference (ΔV) between two points A and B separated by a distance x in a uniform electric field E is given by ΔV = -Ex, where, x is measured parallel to the field lines. If a charge $q_0$ moves from P to Q, the changes in potential energy (ΔU) is given as ΔU = $q_0\delta V$ . A proton is released from rest in uniform electric field of magnitude $4.0\times 10^8 Vm^{-1}$ directly along the positive X-axis. The proton undergoes a displacement of 0.25 m in the direction of E.

Mass of a proton $= 1.66 \times 10^{-27}$ kg and charge of proton $=1.6 \times 10^{-19}$ C

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

Q 1:- What will be the change in electric potential of the proton between the points A and B ?

Solution:- See full solution

Case study 1:- The potential at any observation point P of a static electric field is defined as the work done by the external agent (or negative of work done by electrostatic field) in slowly bringing a unit positive point charge from infinity to the observation point. Figure shows the potential variation along the line of charges. Two point charges $Q_1$ and $Q_2$ lie along a line at a distance from each other.

The potential at any observation point P of a static electric

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

Q 1. At which of the points 1, 2 and 3 is the electric field zero ?

(a) 1 (b) 2

Solution:- See full solution

Case study 2:- This energy possessed by a system of charges by virtue of their positions. When two like charges lie infinite distance apart, their potential energy is zero because no work has to be done in moving one charge at infinite distance from the other. In carrying a charge q from point A to point B, work done $W = q(V_A - V_B)$ . this may appear as change in KE/PE of the charge. The potential energy of two charges $q_1$ and $q_2$ at a distance r in air is $\dfrac{q_1q_2}{4\pi \epsilon _0 r}$ . It is measured in joule. It may be positive, negative or zero depending on the signs of $q_1$ and $q_2$ .

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

Solution:- See full solution

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