Q 1 :

The electrostatic potential due to an electric dipole at a distance 'r' varies as                            [2024]

  • r

     

  • 1r2

     

  • 1r3

     

  • 1r

     

(2)     

          Electric potential due to an electric dipole

           V=14πε0Pcosθr2

           ⇒V∝1r2

 



Q 2 :

The distance between charges +q and −q is 2λ and between +2q and −2q is 4λ. The electrostatic potential at point P at a distance r from center O is -α[qlr2]×109V, where the value of α is _______ . (Use 14πε0=9×109 Nm2C-2)              [2024]



(27)

V=KP→net·r→r3=9×109(6qℓ)r2cos(120°)

=-(27)(qℓr2)×109 Nm2C-2

⇒α=27



Q 3 :

For a short dipole placed at origin O, the dipole moment P is along x-axis, as shown in the figure. If the electric potential and electric field at A are V0 and E0, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y-axis is given by          [2025]

  • V02 and E016

     

  • zero and E08

     

  • zero and E016

     

  • V0 and E04

     

(3)

At point A (axial)

|E0|=2kpr3, V0=kpr2

At point B (equatorial)

|E|=kp(2r)3=E016

V = 0.



Q 4 :

An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σ0. Choose the correct option from the following.         [2025]

  • Torque on dipole is zero and net force is directed away from the sheet.

     

  • Torque on dipole is zero and net force acts towards the sheet.

     

  • Potential energy of dipole is minimum and torque is zero.

     

  • Potential energy and torque both are maximum.

     

(3)

Electric field due to sheet E=σ2ε0 and torque on dipole τ→=P→×E→ here τ→=0 and U=–P→·E→→ should be minimum.



Q 5 :

Two point charges –4 μC and 4 μC, constituting an electric dipole, are placed at (–9, 0, 0) cm and (9, 0, 0) cm in a uniform electric field of strength 104 NC-1. The work done on the dipole in rotating it from the equilibrium through 180° is           [2025]

  • 14.4 mJ

     

  • 18.4 mJ

     

  • 12.4 mJ

     

  • 16.4 mJ

     

(1)

U=–PE cos θ

Wext=∆U=Uf–Ui=–PE cos 180°+PE cos 0°

Wext=2PE=2×(4×10–6)(18)×10–2×104

             =144×10–4=14.4 mJ



Q 6 :

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R)

Assertion (A) : Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field.

Reason (R) : In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions.

In the light of the above statements, choose the most appropriate answer from the options given below:          [2025]

  • (A) is correct but (R) is not correct.

     

  • Both (A) and (R) are correct but (R) is not the correct explanation of (A).

     

  • Both (A) and (R) are correct and (R) is the correct explanation of (A).

     

  • (A) is not correct but (R) is correct.

     

(4)

A: Net dipole moment of polar dielectric substance is zero in absence of electric field as dipoles are randomly oriented.

R: If E→ is absent, polar dielectric remain polar and are randomly oriented.

(A) is not correct but (R) is correct.



Q 7 :

A dipole with two electric charges of 2 μC magnitude each, with separation distance 0.5 μm, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied. Separation between the plates is 0.5 mm. If the dipole is rotated by 30° from the axis, it tends to realign in the direction due to a torque. The value of torque is :          [2025]

  • 5×10–9 Nm

     

  • 5×10–3 Nm

     

  • 2.5×10–12 Nm

     

  • 2.5×10–9 Nm

     

(1)

E=Vd=55×10–4=104 V/m

τ=PE sin θ

Where P=qa=2×10–6×5×10–7=1×10–12 C-m

τ=1×10–12×104×12=5×10–9 N-m



Q 8 :

An electric dipole of dipole moment 6×10–6 cm is placed in uniform electric field of magnitude 106 V/m. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ______ J.          [2025]



(12)

Work done in rotating a dipole =△U

or W=(–PE cos θf)–(–PE cos θi)

            =2PE                                                 (∵ θf=180° and θi=0°)

            =(2×6×10–6×106)J=12 J