Electrostatics MCQs – Test 1 6

This entry was posted in Electrostatics

Electrostatics Questions: MCQs on Electrostatics for JEE Main and NEET, Test Number 01

Time limit: 00:05:00
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  1. Answered
  2. Review
  1. 1. Question
    4 points

    An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 105 N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is [NEET 2016]

    • 1.
    • 2.
    • 3.
    • 4.
  2. 2. Question
    4 points

    A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal : [JEE Main 2016]

    • 1.
    • 2.
    • 3.
    • 4.
  3. 3. Question
    4 points

    A, B and C are three points in a uniform electric field. The electric potential is [NEET 2013]

    • 1.
    • 2.
    • 3.
    • 4.
  4. 4. Question
    4 points

    Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is [IIT JEE 2009]

    • 1.
    • 2.
    • 3.
    • 4.
  5. 5. Question
    4 points

    Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is [AIEEE 2010]

    • 1.
    • 2.
    • 3.
    • 4.

Leave a comment

Your email address will not be published. Required fields are marked *

6 thoughts on “Electrostatics MCQs – Test 1”

    • Reply ↓
      Ambuj tiwari

      In q.5 ,Tcos15=mg and Tsin15=F. And when it is immersed in liquid then tan15=F÷K/mg(1-density of liquid÷d.of substance). And tan15 is constant and it is found to divide upper two equation. And when it goes in liquid then Force is changes into F÷K. Where k is dielectric constant which is ask and mg is also changes so easily it is found

    • Reply ↓
      Chirag.K

      Hi AMISHA,
      As mg’ = mg – Upthrust

      = mg – V*sigma*g (since upthrust = v*sigma*g)

      mg’ = mg[1- sigma/rho]

      T cos Theta=mg –(i)

      T sin Theta=F –(ii)

      Therefore from equation (i) and (ii)

      tan Theta = F/mg —-(iii)

      Now for the equilibrium of balls

      tan Theta’ = F’/mg’ = F/Kmg[1-(rho/sigma)]

      = f/(kmg(1-{sigma/rho }) –(iv)

      According to question Theta’=Theta

      from equation (iv) & (iii)

      K=1/(1-(sigma /rho))

      =rho/(rho -sigma ) =1.6/(1.6-0.8)=2

      K=2

      Hope this helps! 🙂

      Chirag K

    • Reply ↓
      king

      Coulombic force –

      F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}

      – wherein

      K – proportionality Constant

      Q1 and Q2 are two Point charge

      T\cos \Theta=mg –(i)

      T\sin \Theta=F_{e} –(ii)

      \therefore from equation (i) & (ii)

      Tan \Theta=\frac{F_{e}}{mg}—(iii)

      if ball is suspended in liquid g’=g[1-\frac{\sigma }{\rho }]

      Tan \Theta’=\frac{F’_{e}}{mg’}

      =\frac{f}{kmg(1-\frac{\sigma}{\rho })} –(iv)

      According to question \Theta’=\Theta

      from equation (iv) & (iii)

      K=\frac{1}{1-\frac{\sigma }{\rho }}

      =\frac{\rho }{(\rho -\sigma )} =\frac{1.6}{(1.6-0.8)}=2
      K=2

    • Reply ↓
      Arghadeep Basu

      As we have studies dielectric in capacitors there’s the concept of dielectric in charges also that dielectric constant (K)= density of material / (density of material – density in medium)