Calculate the area of the plates of a one farad parallel plate capacitor if separation between plates is 1mm and plates are in vacuum

18 × 10⁸ m²
0.3 × 10⁸ m²
1.3 × 10⁸ m²
1.13 × 10⁸ m²

Solution

From a supply of identical capacitors rated 8 mF, 250V, the minimum number of capacitors required to form a composite 16 mF, 1000 V is

Solution

The plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5mm. The capacitor is charged by a 400 volt supply. How much electrostatic energy is stored by the capacitor?

2.55 × 10^–6 J
1.55 × 10^–6 J
8.15 × 10^–6 J
5.5 × 10^–6 J

Solution

A parallel plate capacitor with air between the plates is charged to a potential difference of 500V and then insulated.A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes 75V. The dielectric constant of plastic is

10/3
5
20/3
10

Solution

Two capacitors C₁ and C₂ in a circuit are joined as shown in figure. The potentials of points A and B are V₁ and V₂ respectively; then the potential of point D will be

\(\frac{(V_{1}+V_{2})}{2}\)
\(\frac{C_{2}V_{1}+C_{1}V_{2}}{C_{1}+C_{2}}\)
\(\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}\)
\(\frac{C_{2}V_{1}+C_{1}V_{2}}{C_{1}+C_{2}}\)

Solution

Force between two plates of a capacitor is

\(\frac{Q}{\varepsilon _{0}A}\)
\(\frac{Q^{2}}{2\varepsilon _{0}A}\)
\(\frac{Q^{2}}{\varepsilon _{0}A}\)
None of these

Solution

Two spherical conductors A and B of radii a and b (b>a) are placed concentrically in air. The two are connected by a copper wire as shown in figure. Then the equivalent capacitance of the system is

\(4\pi \mu _{0}\frac{ab}{b-a}\)
4 π ∈ ₀(a + b)
4 π ∈ ₀ b
4 π ∈ ₀ a

Solution

All the charge given to inner sphere will pass on to the outer one. So capacitance that of outer one is 4 π ∈ ₀ b.

Four metallic plates each with a surface area of one side A,are placed at a distance d from each other. The two outer plates are connected to one point A and the two other inner plates to another point B as shown in the figure. Then the capacitance of the system is

\(\frac{\varepsilon _{0}A}{d}\)
\(\frac{2\varepsilon _{0}A}{d}\)
\(\frac{3\varepsilon _{0}A}{d}\)
\(\frac{4\varepsilon _{0}A}{d}\)

Solution

It consists of two capacitors in parallel, therefore, the total capacitance is = \(\frac{2\epsilon _{0}A}{d}\)

(The plates of B, having negative charge do not constitute a capacitor).

The capacitor, whose capacitance is 6, 6 and 3μF respectively are connected in series with 20 voltline. Find the charge on 3μF.

30 μc
60 μF
15 μF
90 μF

Solution

In series \(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\) and charge on each capacitor is same.

The capacity of a parallel plate condenser is 10 μF, when the distance between its plates is 8 cm. If the distance between the plates is reduced to 4 cm, then the capacity of this parallel plate condenser will be

5 μF
10 μF
20 μF
40 μm

Solution

C = 10 μF ; d =8 cm

C' = ? ; d' = 4 cm

C = \(\frac{A\varepsilon _{0}}{d}\Rightarrow C\alpha \frac{1}{d}\)

If d is halved then C will be doubled.

Hence C' = 2C" = 2 × 10 μF =20 μF

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