Two identical thin rings, each of radius a meter,are coaxially placed at a distance R meter apart. If Q₁ coulomb and Q₂ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q coulomb from the centre of one ring to that of the other is

Zero
\(\frac{q(Q_{1}-Q_{2})(\sqrt{2}-1)}{4\sqrt{2}\pi \varepsilon _{0}a}\)
\(\frac{q\sqrt{2}(Q_{1}-Q_{2})}{4\pi \varepsilon _{0}a}\)
\(\frac{q(Q_{1}+Q_{2})(\sqrt{2}+1)}{4\sqrt{2}\pi \varepsilon _{0}a}\)

A simple pendulum has a length l, mass of bob m. The bob is given a charge q. The pendulum is suspended between the vertical plates of the charged parallel plate capacitor. If E is the field strength between the plates, then time period T equal to

\(2\pi \sqrt{\frac{l}{g}}\)
\(2\pi \sqrt{\frac{l}{g+\frac{qE}{m}}}\)
\(2\pi \sqrt{\frac{l}{g-\frac{qE}{m}}}\)
\(2\pi \sqrt{\frac{l}{g^{2}+(\frac{qE}{m})^{2}}}\)

An electric dipole is placed along the x-axis at the origin O.A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the x-axis. If the electric field at P makes an angle q with the x-axis, the value of θ would be

^π⁄₃
\(\frac{\pi }{3}+tan^{-1}\left ( \frac{\sqrt{3}}{2} \right )\)
^2π⁄₃
\(tan^{-1}\left ( \frac{\sqrt{3}}{2} \right )\)

Solution

ABC is an equilateral triangle. Charges +q are placed at each corner as shown as fig. The electric intensity at centre O will be

\(\frac{1}{4\pi \epsilon _{0}}\frac{q}{r}\)
\(\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}}\)
\(\frac{1}{4\pi \epsilon _{0}}\frac{3q}{r^{2}}\)
zero

Solution

Unit positive charge at O will be repelled equally by three charges at the three corners of triangle.By symmetry,resultant \(\overrightarrow{E}\) at O would be zero.

A drop of oil of density rand radius r carries a charge q when placed in an electric field E, it moves upwards with a velocity v. If r₀ is the density of air,h be the viscosity of the air,then which of the following force is directed upwards?

q E
6 π η r v
⁴⁄₃ π r³(ρ - ρ₀)g
None of these

Solution

Force due to electric field (F = qE) acts upwards.

The spatial distribution of electric field due to charges(A, B) is shown in figure. Which one of the following statements is correct ?

A is +ve and B –ve, |A| > |B|
A is–ve and B+ve, |A| = |B|
Both are +ve but A > B
Both are –ve but A > B

Solution

Since lines of force starts from A and ends at B, so A is +ve and B is –ve. Lines of forces are more crowded near A, so A > B.

Two particle of equal mass m and charge q are placed at a distance of 16 cm. They do not experience any force. The value of ^q⁄_m is

1
\(\sqrt{\frac{\pi \varepsilon _{0}}{G}}\)
\(\sqrt{\frac{G}{4\pi \varepsilon _{0}}}\)
\(\sqrt{4\pi \varepsilon _{0}G}\)

Solution

Among two discs A and B, first have radius 10 cm and charge 10^–6 µC and second have radius 30 cm and charge 10^–5C. When they are touched, charge on both q_A and q_B respectively will, be

q_A = 2.75μC,q_B = 3.15μC
q_A = 1.09μC,q_B = 1.53μC
q_A = q_B =5.5μC
None of these

Solution

The charge on disc A is 10^–6 μC. The charge on disc B is 10 × 10^–6 μC. The total charge on both =11 μC. When touched, this charge will be distributed equally i.e.5.5 μC on each disc.

Three charge q, Q and 4q are placed in a straight line of length lat points distant 0,¹⁄₂ and l respectively from one end. In order to make the net froce on q zero, the charge Q must be equal to

–q
–2q
^-q⁄₂
q

Solution

Two point charges q₁ = 4μC and q₂= 9μC are placed 20 cmapart. The electric field due to them will be zero on the line joining them at a distance of

8 cm from q₁
8 cm form q₂
180 cm from q₁₃
280 cm from q

Solution

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete