Electrostatics Questions: MCQs on Electrostatics for JEE Main and NEET, Test Number 01
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Question 1 of 5
1. Question
4 pointsAn electric dipole is placed at an angle of 30° with an electric field intensity 2 × 10^{5} 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]
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Question 2 of 5
2. Question
4 pointsA 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]
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Question 3 of 5
3. Question
4 pointsA, B and C are three points in a uniform electric field. The electric potential is [NEET 2013]
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Question 4 of 5
4. Question
4 pointsThree concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q_{1}, Q_{2}, Q_{3}, 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, Q_{1} : Q_{2} : Q_{3}, is [IIT JEE 2009]
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Question 5 of 5
5. Question
4 pointsTwo 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]
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It was a very useful experience .I learnt a lot.
Thankyou very much!!!
Give the explaination of question 5
In q.5 ,Tcos15=mg and Tsin15=F. And when it is immersed in liquid then tan15=F÷K/mg(1density 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
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/(1.60.8)=2
K=2
Hope this helps! 🙂
Chirag K
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}{(1.60.8)}=2
K=2
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)