This energy possessed by a system of charges by virtue

Case study question of Physics – Electrostatic potential and capacitance

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:

Q 1. Calculate work done in separating two electrons from a distance of 1 m to 2 m in air, where e is electric charge and K is electrostatic force constant.

(a) Ke² (b) e²/2

Q 2. Two points A and B are located in diametrically opposite direction of a point charge +2 μC at distance 2 m and 1 m respectively from it. The potential difference between A and B is :

(a) $3\times 10^3$ V

(b) $6 \times 10^4$ V

(d) $-3 \times 10^3$ V

Q 3. Two points charges A = + 3 nC and B = +1 nC are placed 5 cm apart in air. The work done to move charge B towords A by 1 cm is :

(a) $2.0 \times 10^{-7}$ J

(b) $1.35 \times 10^{-7}$ J

(d) $12.1 \times 10^{-7}$ J

Q 4. A charge Q is placed at the origin. the electric potential due to this charge at a given point in space is V. The work done by an external force in bringing another charge q from infinity up to the point is :

(a) V/q (b) Vq

Answer : 1. (c) -Ke²/2

$W = (PE)_{final} - (PE)_{initial}$

$=\dfrac{Ke^2}{2} - \dfrac{Ke^2}{1} = -\dfrac{Ke^2}{2}$

2. (c) $-9 \times 10^3$ V

Here, q = 2 μC = $2 \times 10^{-6}C, r_A = 2m, r_B = 1m$

$\therefore V_A - V_B = \dfrac{q}{4\pi \epsilon _0}[\dfrac{1}{r_A} - \dfrac{1}{r_B}]$

$= 2\times 10^{-6} \times 9 \times 10^9[\dfrac{1}{2} - \dfrac{1}{1}]$ V

$= -9\times 10^3$ V

3. (b) $1.35 \times 10^{-7}$ J

Given that,

$A = +3 nC = 3\times 10^{-9}$ C

$A = +1 nC = 1\times 10^{-9}$ C

Distance $r_1 = 5 cm = 5\times 10^{-2}$ m

and $r_1 = r_2 = 1$

= 5 – 1 = 4 cm

$= 4\times 10^{-2}$ m

Required work done = change in potential energy of the system

$W = U_f - U_i = K\dfrac{q_1q_2}{r_f} - K\dfrac{q_1q_2}{r_i}$

$= Kq_1 q_2[\dfrac{1}{r_f}-\dfrac{1}{r_i}]$

$W = (9\times 10^9)(3 \times 10^{-9}\times 1 \times 10^{-9})\times [\dfrac{1}{4\times 10^{-2}} - \dfrac{1}{5\times 10^{-2}}]$

$= 27\times 10^{-9}\times \dfrac{1}{20\times 10^{-2}} = 1.35 \times 10^{-7}$ J

4 (b) Vq

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.