Potential difference (ΔV) between two points A and B separated

Case study question of Physics – Electrostatic potential and capacitance

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 ?

Q 2:- What will the change in electric potential energy of the proton for displacement from A to B ?

Q 3:- Calculate the mutual electrostatic potential energy between two protons which are at a distance of $9\times 10^{-15}$ m, in ${}_{92}U^{235}$ nucleus.

Q 4:- If a system consists of two charges 4 μC and -3 μC with no external field placed at (-5 cm, 0, 0) and (5 cm, 0, 0) respectively. find the amount of work required to separate the two charges infinitely away from each other.

Q 5:- As the proton moves from P to Q, then what will happen ?

Answer :- 1. ΔV = -EΔX

$= -(4\times 10^8)\times 0.25 = -10^8$ V

2. As ΔU = qΔV

$= 1.6\times 10^{-19} \times(-1.0 \times 10^8)$

$= -1.6 \times 10^{-11}$ J

3. Potential energy. $U = \dfrac{1}{4\pi \epsilon _0}\dfrac{q_1q_2}{r}$

$=\dfrac{9\times 10^9 \times 1.6 \times 10^{-19}\times 1.6\times 10^{-19}}{9\times 10^{-15}}$

$= 2.56 \times 10^{-14}$ J

4. $U = \dfrac{1}{4\pi \epsilon _0}\dfrac{q_1q_2}{r}$

$= \dfrac{9\times 10^9 \times 4 \times 10^{-6} \times (-3) \times 10^{-6}}{0.1}$

$= -11$ J

5. As proton moves in the direction of the electric field or from P to Q then its potential energy decreases

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

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