Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electrostatics question

2023 · 13 Apr · Shift 1 · Q53
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Electrostatics
  5. /2023 · 13 Apr · Shift 1 · Q53

Electrostatics question

2023 · 13 Apr · Shift 1 · Q53

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two charges each of magnitude 0.01 C0.01 ~\mathrm{C}0.01 C and separated by a distance of 0.4 mm0.4 \mathrm{~mm}0.4 mm constitute an electric dipole. If the dipole is placed in an uniform electric field 'E⃗\vec{E}E' of 10 dyne/C making 30∘30^{\circ}30∘ angle with E⃗\vec{E}E, the magnitude of torque acting on dipole is:
  1. A
    4⋅0×10−10 Nm4 \cdot 0 \times 10^{-10} ~\mathrm{Nm}4⋅0×10−10 Nm
  2. B
    1.5×10−9 Nm1.5 \times 10^{-9} ~\mathrm{Nm}1.5×10−9 Nm
  3. C
    1.0×10−8 Nm1.0 \times 10^{-8} ~\mathrm{Nm}1.0×10−8 Nm
  4. D
    2.0×10−10 Nm2.0 \times 10^{-10} ~\mathrm{Nm}2.0×10−10 Nm
View written solutionFree

Correct answer: D

  1. Use the torque formula for an electric dipole

For a dipole in a uniform electric field,

τ=pEsin⁡θ\tau = pE\sin\thetaτ=pEsinθ

where

  • p=q dp = q\,dp=qd is the dipole moment,
  • q=0.01 Cq = 0.01\,\text{C}q=0.01C,
  • d=0.4 mm=0.4×10−3 m=4×10−4 md = 0.4\,\text{mm} = 0.4 \times 10^{-3}\,\text{m} = 4 \times 10^{-4}\,\text{m}d=0.4mm=0.4×10−3m=4×10−4m,
  • θ=30∘\theta = 30^\circθ=30∘.
  1. Calculate dipole moment

p=qd=0.01×4×10−4=4×10−6 C mp = qd = 0.01 \times 4 \times 10^{-4} = 4 \times 10^{-6}\,\text{C m}p=qd=0.01×4×10−4=4×10−6C m

  1. Convert electric field into SI units

Given:

E=10 dyne/CE = 10\,\text{dyne/C}E=10dyne/C

In CGS electrostatic questions of this type, this is intended as force per unit charge equivalent to

1 dyne=10−5 N1\,\text{dyne} = 10^{-5}\,\text{N}1dyne=10−5N

So,

E=10×10−5=10−4 N/CE = 10 \times 10^{-5} = 10^{-4}\,\text{N/C}E=10×10−5=10−4N/C

  1. Now compute torque

τ=pEsin⁡30∘\tau = pE\sin 30^\circτ=pEsin30∘

τ=(4×10−6)(10−4)(12)\tau = (4 \times 10^{-6})(10^{-4})\left(\frac{1}{2}\right)τ=(4×10−6)(10−4)(21​)

τ=2×10−10 N m\tau = 2 \times 10^{-10}\,\text{N m}τ=2×10−10N m

  1. Match with options

Thus the correct option is

D  :  2.0×10−10 N m\boxed{D\;:\;2.0 \times 10^{-10}\,\text{N m}}D:2.0×10−10N m​

PreviousNext

More from Electrostatics

  • A thin infinite sheet charge and an infinite line charge of respective charge densities +σ and +λ are placed parallel at 5 m distance from each other. Points 'P' and 'Q' are at π3​ m and π4​…2023 · Numerical
  • A 10 μC charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are:2023 · MCQ
  • Three point charges q,−2q and 2q are placed on x-axis at a distance x=0,x=43​R and x=R respectively from origin as shown. If q=2×10−6C and R=2 cm… Includes diagram2023 · Numerical
  • The electric field due to a short electric dipole at a large distance (r) from center of dipole on the equatorial plane varies with distance as :2023 · MCQ
  • If two charges q 1​ and q 2​ are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?2023 · MCQ
  • A stream of a positively charged particles having mq​=2×1011kgC​ and velocity v0​=3×107im/s is deflected by an electric field 1.8j​ kV/m. The…2023 · Numerical
  • The electric potential at the centre of two concentric half rings of radii R 1​ and R 2​, having same linear charge density λ is : Includes diagram2023 · MCQ
  • A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The… Includes diagram2023 · Numerical