Two capacitors when connected in series have a capacitance of 3 μF,and when connected in parallel have a capacitance of 16 μF.Their individual capacities are

1 μF,2 μF
6 μF,2 μF
12 μF,4 μF
3 μF,16 μF

Solution

A capacitor is charged to store an energy U. The charging battery is disconnected. An identical capacitor is now connected to the first capacitor in parallel. The energy in each of the capacitors is

3 U/2
U
U/4
U/2

Solution

As battery is disconnected, total charge Q is shared equally by two capacitors.

Energy of each capacitor = \(\frac{(Q/2)^{2}}{2C}=\frac{1}{4}\frac{Q^{2}}{2C}=\frac{1}{4}U\)

An air capacitor of capacity C= 10 μF is connected to a constant voltage battery of 12 volt. Now the space between the plates is filled with a liquid of dielectric constant 5. The(additional) charge that flows now from battery to the capacitor is

120 μ C
600 μ C
480 μ C
24 μ C

Solution

q₁ = C₁V = 5 × 12 = 120 μ C

q₂ = C₂V = KC₁ × V = 5 × 10 × 12 = 120 μ C

Additional charge that flows

= q₂- q₁ = 600 - 120 = 480 μ C.

The four capacitors, each of 25 μ F are connected as shown in fig. The dc voltmeter reads 200 V. The charge on each plate of capacitor is

± 2 × 10^-3 C
± 5 × 10^-3 C
± 2 × 10^-2 C
± 5 × 10^-2 C

Solution

Charge on each plate of each capacitor

Q = ± CV = ± 25 × 10⁶ × 200 = ± 5 × 10^-3 C

Three point charges +q , + 2q and – 4q where q = 0.1 mC, are placed at the vertices of an equilateral triangle of side 10 cm as shown in figure. The potential energy of the system is

3 × 10^–3 J
–3 × 10^–3 J
9 × 10^–3 J
–9 × 10^–3 J

Solution

Two concentric, thin metallic spheres of radii R₁ and R₂(R₁> R₂) bear charges Q₁ and Q₂ respectively. Then the potential at distance r between R1 and R₂ will be \(\left ( k=\frac{1}{4\pi \epsilon _{0}} \right )\)

\(k\left ( \frac{Q_{1}-Q_{2}}{r} \right )\)
\(k\left ( \frac{Q_{1}}{r}+ \frac{Q_{2}}{R_{2}}\right )\)
\(k\left ( \frac{Q_{2}}{r}+ \frac{Q_{1}}{R_{1}}\right )\)
\(k\left ( \frac{Q_{2}}{R_{1}}+ \frac{Q_{2}}{R_{2}}\right )\)

Solution

A battery of e.m.f.V volt,resistors R₁ and R₂, a condenser C and switches S₁ and S₂ are connected in a circuit shown.The condenser will get fully charged to V volt when

S₁ and S₂ are both closed
S₁ and S₂ are both open
S₁ is open and S₂ is closed
S₁ is closed and S₂ is open

Solution

When S₁ is closed and S₂ is opened, the capacitor will get charged to a potential difference of V volts.

Find the dipole moment of a system where the potential 2.0 × 10^–5V at a point P, 0.1m from the dipole is 3.0 × 10⁴.(Use θ= 30°).

2.57 × 10^–17 Cm
1.285 × 10^–15 Cm
1.285 × 10^–17 Cm
2.57 × 10^–15 Cm

Solution

A positive point charge q is carried from a point B to a point A in the electric field of a point charge + Q at O. If the permitivity of free space is e0, the work done in the process is given by

\(\frac{q\: Q}{4\pi \varepsilon _{0}}\left ( \frac{1}{a}+\frac{1}{b} \right )\)
\(\frac{q\: Q}{4\pi \varepsilon _{0}}\left ( \frac{1}{a}-\frac{1}{b} \right )\)
\(\frac{q\: Q}{4\pi \varepsilon _{0}}\left ( \frac{1}{a^{2}}-\frac{1}{b^{2}} \right )\)
\(\frac{q\: Q}{4\pi \varepsilon _{0}}\left ( \frac{1}{a^{2}}+\frac{1}{b^{2}} \right )\)

Solution

Three charges 2 q, – q and – q are located at the vertices of an equilateral triangle. At the centre of the triangle

the field is zero but potential is non-zero
the field is non-zero, but potential is zero
both field and potential are zero
both field and potential are non-zero

Solution

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