A uniformly charged thin spherical shell of radius R carries uniform surface charge density of sper unit area. It is made of two hemispherical shells, held together by pressing them with force F(see figure). F is proportional to

\(\frac{1}{\varepsilon _{0}}\sigma ^{2}R^{2}\)
\(\frac{1}{\varepsilon _{0}}\sigma ^{2}R\)
\(\frac{1}{\varepsilon _{0}}\frac{\sigma ^{2}}{R}\)
\(\frac{1}{\varepsilon _{0}}\frac{\sigma ^{2}}{R^{2}}\)

Solution

In the given circuit with steady current, the potential drop across the capacitor must be

^2V⁄₃
^V⁄₃
^V⁄₂
V

Solution

The capacitance of the capacitor of plate areas A₁ and A₂(A₁< A₂) at a distance d, as shown in figure is

\(\frac{\epsilon _{0}(A_{1}+A_{2})}{2d}\)
\(\frac{\epsilon _{0}A_{2}}{d}\)
\(\frac{\epsilon _{0}\sqrt{A_{1}A_{2}}}{d}\)
\(\frac{\epsilon _{0}A_{1}}{d}\)

Solution

C =\(\frac{\epsilon _{0}A_{1}}{d}\)

A ⇒ common area, Here A = A₁

Figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the centre of curvature is

\(\frac{Q}{2\pi \varepsilon _{0}R}\)
\(\frac{Q}{4\pi \varepsilon _{0}R}\)
\(\frac{2Q}{\pi \varepsilon _{0}R}\)
\(\frac{Q}{\pi \varepsilon _{0}R}\)

Solution

Two identical particles each of mass m and having charges –q and +q are revolving in a circle of radius r under the influence of electric attraction. Kinetic energy of each particle is\(\left ( K=\frac{1}{4\pi \varepsilon _{0}} \right )\)

kq²/4r
kq²/2r
kq²/8r
kq²/r

Solution

Four point charges q, q, q and –3q are placed at the vertices of a regular tetrahedron of side L. The work done by electric force in taking all the charges to the centre of the tetrahedron is (where K = \(\frac{1}{4\pi \varepsilon _{0}}\))

\(\frac{6Kq^{2}}{L}\)
\(\frac{-6Kq^{2}}{L}\)
\(\frac{12Kq^{2}}{L}\)
zero

Solution

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery.The ratio of the energy stored in the capacitor and the work done by the battery will be

Solution

If a charge – 150 nC is given to a concentric spherical shell and a charge +50 nC is placed at its centre then the charge on inner and outer surface of the shell is

–50 nC, –100 nC
+50 nC, –200 nC
–50 nC, –200 nC
50 nC, 100 nC

Solution

Whenever a charge (+50 nC) is kept inside a hollow metallic spherical shell, it induces an equal and opposite charge on the inner surface and an equal and same type of charges on the outer surface.\Inside, induced charge is – 50 nC and outside, +50 nC – 150 nC already present.

In the circuit given below, the charge in μC, on the capacitor having 5 μF is

Solution

Potential difference across the branch de is 6 V. Net capacitance of de branch is 2.1µF

So, q = CV

⇒ q =2.1 × 6 µC

⇒ q = 12.6 µ C

Potential across 3 µF capacitance is

V = \(\frac{12.6}{3}=4.2\) volt

Potential across 2 and 5 combination in parallel is 6 – 4.2 = 1.8 V

So, q' = (1.8)(5) = 9 µC

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of capacitor so as to fill the space between the plates. If Q, E and W denote respectively,the magnitude of charge on each plate electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab,then which is wrong ?

\(Q=\frac{\varepsilon _{0}AV}{d}\)
\(Q=\frac{\varepsilon _{0}KAV}{d}\)
\(E=\frac{V}{Kd}\)
\(W=\frac{\varepsilon _{0}AV^{2}}{2d}\left ( 1-\frac{1}{K} \right )\)

Solution

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