Charge Q is distributed to two different metallic spheres having radii R and 2R such that both spheres have equal surface charge density, then charge on large sphere is

^4Q⁄₅
^Q⁄₅
^3Q⁄₅
^5Q⁄₅

Solution

An electric dipole is placed at an angle of 30° with an electric field of intensity 2 × 10⁵ NC^-1, It experiences a torque of 4 Nm. Calculate the charge on the dipole if the dipole length is 2cm.

8 mC
4 mC
8 mC
2 mC

Solution

If the dipole of moment 2.57× 10^–17 cm is placed into an electric field of magnitude 3.0 × 10⁴ N/C such that the fields lines are aligned at 30° with the line joining P to the dipole,what torque acts on the dipole?

7.7 × 10^–13 Nm
3.855 × 10^–13 Nm
3.855 × 10^–15 Nm
7.7 × 10^–15 Nm

Solution

An electric dipole has the magnitude of its charge as q and its dipole moment is p.It is placed in uniform electric field E.If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively.

q.E and max.
2 q.E and min.
q.E and min
zero and min.

Solution

When the dipole is in the direction of field then net force is qE + (–qE) =0

and its potential energy is minimum = – p.E.

= –qaE

A hollow insulated conduction sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere if its radius is 2 metres?

Zero
5 μCm^–2
20 μCm^–2
8 μCm^–2

Solution

If electric field in a region is radially outward with magnitude E =Ar, the charge contained in a sphere of radius r centred at the origin is

\(\frac{1}{4\pi \varepsilon _{0}}Ar^{3}\)
4πε₀Ar³
\(\frac{1}{4\pi \varepsilon _{0}}\frac{A}{r^{3}}\)
\(\frac{4\pi \varepsilon _{0}A}{r^{3}}\)

Solution

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is

q E y²
q E² y
q E y
q² E y

Solution

K.E. acquired = work done

= force × distance = q E× y = q E y

There is an electric field E in X-direction. If work done in moving a charge 0.2 C through a distance of 2m along a line making an angle of 60 degree with X-axis is 4.0 joule, what is the value of E?

√3 newton per coulomb
4 netwon per coulomb
5 newton per coulomb
None of these

Solution

Work done, W = F s cos θ= (q E) s cos θ

Two thin infinite parallel sheets have uniform surface densities of charge + σ and – σ. Electric field in the space between the two sheets is

σ/ε₀
σ/2ε₀
2σ/ε₀
zero

Solution

An infinite parallel plane sheet of a metal is charged to charge density σ coulomb per square metre in a medium of dielectric constant K. Intensity of electric field near the metallic surface will be

\(E=\frac{\sigma }{\varepsilon _{0}K}\)
\(E=\frac{\sigma }{2\varepsilon _{0}}\)
\(E=\frac{\sigma }{2\varepsilon _{0}K}\)
\(E=\frac{K\sigma }{2\varepsilon _{0}}\)

Solution

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