Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 77

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Weekly Quiz Competition

Question 1
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Select the correct statements:

A
The electric lines of force are always closed curves.

B
Electric line of forced is parallel to equipotential surface

C
Electric line of force is perpendicular to equipotential surface.

D
Electric line of force is always the path of a positively charged particle.

Solution

Option D is the correct answer because electric line of force is always the path of a positively charged particle.
Question 2
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Three long concentric cylindrical shells have radii R, 2R and $$2\sqrt{2}R$$. Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:

A
$$\dfrac{\dfrac{1}{3}\epsilon_0}{ln 2}$$

B
$$\dfrac{6\pi \epsilon_0}{ln 2}$$

C
$$\dfrac{\pi \epsilon_0}{2 ln 2}$$

D
None
Question 3
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The diagram shows four capacitors with capacitances and break down voltages as mentioned. What should be the maximum value of the external emf source such that no capacitor breaks down?

A
2.5 kV

B
10 / 3kV

C
3 kV

D
1 kV
Question 4
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Three condensers $$C_1, C_2$$ and $$C_3$$ are connected to a 100 volt D.C. source as shown in the figure. If the charges stored on the plates of $$C_1, C_2$$ and $$C_3$$ are $$q_a, q_b$$ and $$q_c, q_d$$ and $$q_e, q_f$$ respectively, then

A
$$q_b+q_d+q_f = \frac{100}{9}$$ coulomb

B
$$q_b+q_d+q_f =0$$

C
$$q_b=q_d=q_f$$

D
$$q_a+q_c+q_e =50$$ coulomb

Solution

As all the three capacitors have same capacitance the charge on all the three will be same.

Charges on plate $$b,d,f$$ also have same magnitude.
Question 5
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In the adjoining circuit, the capacity between the points A and B will be

A
C

B
2C

C
3C

D
4C

Solution
Question 6
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Directions For Questions

While assembling a parallel plate capacitor, the space between
two conducting plates is filled with some insulating material called dielectric
material. If the capacitor is charged, electric field is created inside the
dielectric. Due to this field is an applied in the dietetic, the outer
electrons may get detached from atoms and then the dielectric behaves like a
conductor. This phenomenon is called dielectric breakdown. The minimum field at
which the breakdown occurs is called the electric strength of the material and corresponding
potential is called breakdown potential

There are two capacitor of capacitorance $$C$$ and $$2C$$.
The breakdown potential of each capacitor is $$V_{e}$$.

...view full instructions

If the voltage across the parallel combination is increase, which capacitor will undergo breakdown first

A
$$C$$

B
$$2C$$

C
Both at same moment

D
None of these

Solution
Question 7
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A network of six identical capacitors, each of value $$C$$ is made as shown in the figure. Equivalent capacitance between points $$A$$ and $$B$$ is

A
$$C/4$$

B
$$3C/4$$

C
$$4C/3$$

D
$$3C$$

Solution
Question 8
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A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $$d_{1}$$ and dielectric constant $$k_{1}$$ and the other has thickness $$d_{2}$$ and dielectric constant $$k_{2}$$ as shown in fig. This arrangement can be thought as a dielectric slab of thickness $$d=d_{1}+d_{2}$$ and effective dielectric constant $$k$$. The $$k$$ is then :

A
$$\dfrac{k_{1}d_{1}+k_{2}d_{2}}{d_{1}+d_{2}}$$

B
$$\dfrac{k_{1}d_{1}+k_{2}d_{2}}{k_{1}+k_{2}}$$

C
$$\dfrac{k_{1}k_{2}(d_{1}+d_{2})}{k_{1}d_{2}+k_{2}d_{1}}$$

D
$$\dfrac{2k_{1}k_{2}}{k_{1}+k_{2}}$$

Solution

For the first block

Thickness $$={{d}_{1}}$$

Dielectric constant $$={{k}_{1}}$$

Capacitance $$={{C}_{1}}$$

For second block

Thickness $$={{d}_{2}}$$

Dielectric constant $$={{k}_{2}}$$

Capacitance $$={{C}_{2}}$$

Capacitors C₁ and C₂ are connected in parallel. So equivalent capacitance is :
$$ \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}} $$

$$ {{C}_{eq}}=\dfrac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}........(1) $$

Capacitance in parallel plate capacitor is given as

$${{C}_{1}}=\dfrac{{{k}_{1}}{{\varepsilon }_{0}}A}{{{d}_{1}}}\,\,\And \,\,{{C}_{2}}=\dfrac{{{k}_{2}}{{\varepsilon }_{0}}A}{{{d}_{2}}}$$

So, equation (1) becomes

$${{C}_{eq}}=\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}.............(2)$$

Equivalent capacitance for the combination is

$$C=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}.........(2)$$

From equation (1) and (2)

$$ \dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}} $$

$$ k=\dfrac{{{k}_{1}}{{k}_{2}}({{d}_{1}}+{{d}_{2}})}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}} $$
Question 9
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Three identical charged capacitors each of capacitance$$5 \mu F$$ are connected as shown in figure. Potential difference across capacitor (3), long time after the switches $${ K }_{ 1 }$$ and $${ K }_{ 2 }$$ are closed, is:

A
$$20\ V$$

B
$$10\ V$$

C
$$5\ V$$

D
$$zero$$

Solution
Question 10
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Five identical capacitors, each with capacitance $$C$$ are connected as shown. Find the equivalent capacitance between $$A$$ and $$B$$

A
$$C$$

B
$$5C$$

C
$$C/5$$

D
$$3C$$

Solution

There is no voltage difference across the second capacitor, so it gets removed.

The effective capacitance between the two capacitors is C.