The magnitude of the electric field E in the annular region of a charged cylindrical capacitor

is same throughout
is higher near the outer cylinder than near the inner cylinder
varies as ¹⁄_r, where r is the distance from the axis
varies as ¹⁄_r², where r is the distance from the axis

Solution

\(E\alpha \frac{\lambda }{2\pi \varepsilon _{0}r}\: hence\: E\alpha \frac{1}{r}\)

A conductor carries a certain charge.When it is connected to another uncharged conductor of finite capacity, then the energy of the combined system is

more than that of the first conductor
less than that of the first conductor
equal to that of the first conductor
uncertain

Solution

Energy will be lost during transfer of charge (heating effect).

When air is replaced by a dielectric medium of force constant K, the maximum force of attraction between two charges, separated by a distance

decreases K-times
increases K-times
remains unchanged
becomes ¹⁄_K² times

Solution

If the potential of a capacitor having capacity 6 μF is increased from 10 V to 20V, then increase in its energy will be

4 × 10^–4 J
4 × 10^–4 J
9 × 10^–4 J
12 × 10^–6 J

Solution

Capacitance (in F) of a spherical conductor with radius 1 m is

1.1 × 10^–10
10⁶
9 × 10^–9
10^–3

Solution

A one microfarad capacitor of a TV is subjected to 4000 V potential difference. The energy stored in capacitor is

8 J
16 J
4 × 10^–3 J
2 × 10^–3 J

Solution

E = ¹⁄₂ CV² = ¹⁄₂ × 1 × 10^-6 × (4000)² = 8J.

Two capacitors of capacitances C₁ and C₂ are connected in parallel across a battery. If Q₁ and Q₂ respectively be the charges on the capacitors, then ^Q₁⁄_Q₂ will be equal to

^C₂⁄_C₁
^C₁⁄_C₂
\(\frac{C_{1}^{2}}{C_{2}^{2}}\)
\(\frac{C_{2}^{2}}{C_{1}^{2}}\)

Solution

In parallel, potential is same, say V

\(\frac{Q_{1}}{Q_{2}}=\frac{C_{1}V}{C_{2}V}=\frac{C_{1}}{C_{2}}\)

An alpha particle is accelerated through a potential difference of 10⁶ volt. Its kinetic energy will be

1 MeV
2 MeV
4 MeV
8 MeV

Solution

Charge on α particle, q = 2e.

K.E. = work done = q × V = 2e × 10⁶V = 2 MeV.

A ball of mass 1 g carrying a charge 10^–8 C moves from a point A at potential 600 V to a point B at zero potential. The change in its K.E. is

–6 × 10^–6erg
–6 × 10^–6 J
6 × 10^–6 J
6× 10^–6erg

Solution

As work is done by the field, K.E. of the body increases by

K.E. = W = q (V_AV - V_B)

=10^-8(600 - 0) = 6 × 10⁶J

The electric potential at the surface of anatomic nucleus(Z = 50) of radius of 9 × 10^–15 m is

80 V
8 × 10⁶V
9 V
9 × 10⁵ V

Solution

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete