JEE Advanced PYQ Physics – Electric Potential Energy and Work Done Questions

Q 1 :

Consider a system of three charges $q / 3$ , $q / 3$ and $- 2 q / 3$ placed at points $A$ , $B$ and $C$ , respectively, as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and angle $C A B = 60 °$ . [2008]

The electric field at point $O$ is $\frac{q}{8 π ε_{0} R^{2}}$ directed along the negative $x$ -axis.
The potential energy of the system is zero.
The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^{2}}{54 π ε_{0} R^{2}}$
The potential at point $O$ is $\frac{q}{12 π ε_{0} R}$

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(3)

(1) The electric field due to charge $\frac{q}{3}$ at A and charge $\frac{q}{3}$ at B at O will get cancelled. The electric field at O due to charge $(\frac{- 2 q}{3})$ is

$E = \frac{1}{4 π ε_{0}} \frac{2 q / 3}{R^{2}} = \frac{q}{6 π ε_{0} R^{2}} .$

In $∆ A B C$ , $\frac{A C}{A B} = \sin 30 °$

$\Rightarrow A C = \frac{A B}{2} = R$

Also, $\frac{B C}{A B} = \sin 60 °$ $\Rightarrow B C = \frac{\sqrt{3} A B}{2} = \sqrt{3} R$

(2) Potential energy of the system $K [\frac{(q / 3) (q / 3)}{2 R}] + K [\frac{(q / 3) (- 2 q / 3)}{R}] + K [\frac{(q / 3) (- 2 q / 3)}{\sqrt{3} R}]$

$= \frac{k q^{2}}{9 R} [\frac{1}{2} - 2 - \frac{2}{\sqrt{3}}] \neq 0$

(3) Magnitude of force between B and C $F = \frac{1}{4 π ε_{0}} \frac{(2 q / 3) (q / 3)}{{(\sqrt{3} R)}^{2}} = \frac{q^{2}}{54 π ε_{0} R^{2}}$

(4) Potential $V_{O} = K [\frac{+ q / 3}{R}] + K [\frac{+ q / 3}{R}] + K [\frac{- 2 q / 3}{R}] = 0$

Q 2 :

Positive and negative point charges of equal magnitude are kept at $(0, 0, \frac{a}{2})$ and $(0, 0, \frac{- a}{2})$ respectively. The work done by the electric field when another positive point charge is moved from $(- a, 0, 0)$ to $(0, a, 0)$ is [2007]

positive
negative
zero
depends on the path connecting the initial and final positions

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(3)

Two charges make an electric dipole. A and B points lie on the equatorial plane of the dipole.

$∴$ Potential at A, $V_{A}$ = potential at B, $V_{B} = 0$

Hence, work done $W = q (V_{A} - V_{B}) = q \times 0 = 0$

Q 3 :

Two equal point charges are fixed at $x = - a$ and $x = + a$ on the $x$ -axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$ , when it is displaced by a small distance $x$ along the $x$ -axis, is approximately proportional to [2002]

$x$
$x^{2}$
$x^{3}$
$1 / x$

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(2)

Initial energy

$U_{i} = \frac{2 Q q}{4 π ε_{0} a} + \frac{1}{4 π ε_{0}} \frac{q^{2}}{2 a}$

Final energy

$U_{f} = \frac{Q q}{4 π ε_{0}} [\frac{1}{a + x} + \frac{1}{a - x}] + \frac{1}{4 π ε_{0}} \frac{q^{2}}{2 a} = \frac{2 Q q a}{4 π ε_{0} (a^{2} - x^{2})}$

$∴$ $U_{i} - U_{f} = \frac{2 Q q}{4 π ε_{0}} [\frac{1}{a} - \frac{a}{(a^{2} - x^{2})}]$

$= \frac{2 Q q}{4 π ε_{0}} [\frac{a^{2} - x^{2} - a^{2}}{a (a^{2} - x^{2})}] = \frac{- 2 Q q x^{2}}{4 π ε_{0} a^{3}}$

When $x ≪ a$ , then $x^{2}$ is neglected in the denominator.

$U_{i} - U_{f} = (\frac{- Q q}{2 π ε_{0} a^{3}}) x^{2}$

$∴$ $U_{i} - U_{f} \propto x^{2}$

Q 4 :

Three charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if $Q$ is equal to [2000]

$\frac{- q}{1 + \sqrt{2}}$
$\frac{- 2 q}{2 + \sqrt{2}}$
$- 2 q$
$+ q$

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(2)

Since net electrostatic energy $Σ u = 0$

$∴$ $\frac{Q q}{a} + \frac{q^{2}}{a} + \frac{Q q}{a \sqrt{2}} = 0$

or $Q \sqrt{2} + q \sqrt{2} + Q = 0$

$∴$ $Q = - \frac{q \sqrt{2}}{\sqrt{2} + 1} = - \frac{2 q}{2 + \sqrt{2}}$

Q 5 :

A point charge $q$ of mass $m$ is suspended vertically by a string of length $l$ . A point dipole of dipole moment $\vec{p}$ is now brought towards $q$ from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is $N \times (m g h)$ , where $g$ is the acceleration due to gravity, then the value of $N$ is ________. (Note that for three coplanar forces keeping a point mass in equilibrium, $\frac{F}{\sin θ}$ is the same for all forces, where $F$ is any one of the forces and $θ$ is the angle between the other two forces.) [2020]

(2)

Initial potential energy, $U_{i} = 0$

Final potential energy

$U_{f} = q V + m g h = \frac{q K p}{d^{2}} + m g h$

or, $U_{f} = \frac{k q P}{{(2 ℓ \sin \frac{α}{2})}^{2}} + m g h$ ...(i)

From $∆ O A B$ ,

$α + 90 ° - θ + 90 ° - θ = 180 °$

$∴$ $α = 2 θ$

From $∆ A B C$ , $h = 2 ℓ \sin (\frac{α}{2}) \sin θ$

$∴$ $h = 2 ℓ \sin (\frac{α}{2}) \sin (\frac{α}{2}) = 2 ℓ \sin^{2} (\frac{α}{2})$

Now charge is in equilibrium at point B, therefore from sine rule,

$\frac{m g}{\sin (90 ° + \frac{α}{2})} = \frac{q E}{\sin (180 ° - 2 θ)}$ $\Rightarrow \frac{m g}{\cos \frac{α}{2}} = \frac{q E}{\sin 2 θ}$

$\Rightarrow \frac{m g}{\cos \frac{α}{2}} = \frac{q E}{\sin α} = \frac{q E}{2 \sin \frac{α}{2} \cos \frac{α}{2}}$

$\Rightarrow q E = m g 2 \sin (\frac{α}{2})$

$\Rightarrow \frac{q^{2} k p}{{(2 ℓ \sin \frac{α}{2})}^{3}} = m g 2 \sin (\frac{α}{2})$ $\Rightarrow \frac{k p q}{{(2 ℓ \sin \frac{α}{2})}^{2}} = m g \sin (\frac{α}{2}) (2 ℓ \sin \frac{α}{2})$

$\Rightarrow \frac{k p q}{{(2 ℓ \sin \frac{α}{2})}^{2}} = m g h$

Substituting this value in equation (i), we get

$U_{f} = 2 m g h$

Work done in bringing the dipole

$W = U_{f} - U_{i} = 2 m g h - 0 = 2 m g h$

$W = Δ U = 2 m g h = N m g h$

$∴$ $N = 2$

Q 6 :

A positive point charge of $10^{- 8} C$ is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2}$ (where $ε_{0}$ is the permittivity of free space), which of the following statements is/are correct? [2025]

Before the grounding, the electrostatic potential of the sphere is 450 V.
Charge flowing from the sphere to the ground because of grounding is $5 \times 10^{- 9} C .$
After the grounding is removed, the charge on the sphere is $- 5 \times 10^{- 9} C .$
The final electrostatic potential of the sphere is 300 V.

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Select one or more options

(1, 2, 3)

Before grounding,

${(V_{sphere})}_{_{net}} = {(V_{c})}_{q} + {(V_{c})}_{_{ind}}$

$V_{sphere} = \frac{k q}{ℓ} + 0 = \frac{9 \times 10^{9} \times 10^{- 8}}{0.2} = \frac{90}{0.2} = \frac{900}{2} = 450 volt$

So, option (1) is correct.

$\frac{k Q}{ℓ} + \frac{k q_{s}}{R} = V'_{sphere} = 0$

$∴$ $q_{s} = - \frac{R}{ℓ} q = - \frac{0.1}{0.2} \times 10^{- 8} = - 5 \times 10^{- 9} C$

$q_{s} = - 5 \times 10^{- 9} C$

Charge flowing from sphere to ground $= 5 \times 10^{- 9} C$

After grounding is removed, the charge on the sphere $= - 5 \times 10^{- 9} C$

So option (2) and (3) are correct.

Final electrostatic potential of the sphere

${(V_{sphere})}_{final} = \frac{k q}{ℓ} + \frac{k q_{s}}{R}$

$= \frac{9 \times 10^{9} \times 10^{2} \times 10^{- 8}}{30} - \frac{9 \times 10^{9} \times 5 \times 10^{- 9} \times 10^{2}}{10}$

$= \frac{9 \times 1000}{30} - 450 = 300 volt - 450 volt = - 150 volt$

So option (4) is incorrect.

Topic Question Set

Q 1 :

Consider a system of three charges $q / 3$ , $q / 3$ and $- 2 q / 3$ placed at points $A$ , $B$ and $C$ , respectively, as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and angle $C A B = 60 °$ . [2008]

Q 2 :

Positive and negative point charges of equal magnitude are kept at $(0, 0, \frac{a}{2})$ and $(0, 0, \frac{- a}{2})$ respectively. The work done by the electric field when another positive point charge is moved from $(- a, 0, 0)$ to $(0, a, 0)$ is [2007]

Q 3 :

Two equal point charges are fixed at $x = - a$ and $x = + a$ on the $x$ -axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$ , when it is displaced by a small distance $x$ along the $x$ -axis, is approximately proportional to [2002]

Q 4 :

Three charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if $Q$ is equal to [2000]

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Topic Question Set

Q 1 : Consider a system of three charges q/3, q/3 and -2q/3 placed at points A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB=60°. [2008]

Q 2 : Positive and negative point charges of equal magnitude are kept at (0,0,a2) and (0,0,-a2) respectively. The work done by the electric field when another positive point charge is moved from (-a,0,0) to (0,a,0) is [2007]

Q 3 : Two equal point charges are fixed at x=-a and x=+a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to [2002]

Q 4 : Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to [2000]

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Q 1 :

Consider a system of three charges $q / 3$ , $q / 3$ and $- 2 q / 3$ placed at points $A$ , $B$ and $C$ , respectively, as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and angle $C A B = 60 °$ . [2008]

Q 2 :

Positive and negative point charges of equal magnitude are kept at $(0, 0, \frac{a}{2})$ and $(0, 0, \frac{- a}{2})$ respectively. The work done by the electric field when another positive point charge is moved from $(- a, 0, 0)$ to $(0, a, 0)$ is [2007]

Q 3 :

Two equal point charges are fixed at $x = - a$ and $x = + a$ on the $x$ -axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$ , when it is displaced by a small distance $x$ along the $x$ -axis, is approximately proportional to [2002]

Q 4 :

Three charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if $Q$ is equal to [2000]