Electrostatic Potential & Capacitance Important Questions

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Very Short Answer Questions [1 Mark]

Ques. The method wherein a region is freed from any electric field,

1. Electrostatic forcing
2. Electrostatic binding
3. Electrostatic shielding
4. None of the options

Ans: (c) Electrostatic shielding: This phenomenon is witnessed when a Faraday cage is used to block the effects of an electric field.

Ques. The formula for electrostatic potential is 

1. Electrostatic potential = Work done*charge
2. Electrostatic potential = Work done/charge
3. Electrostatic potential = Work done+charge
4. Electrostatic potential = Work done-charge

Ans: (b) Electrostatic potential = Work done/charge

Ques. The amount of work done required to shift a unit positive test charge over a closed path in an electric field,

Ans: It is known that electrostatic forces are conservative in nature because the work done in moving a unit positive test charge in an electric field is zero over a closed path.

Ques. A surface which is known to have the same electrostatic potential at every point on it,

1. Equal-potential surface
2. Same potential surface
3. Equi-magnitude surface
4. Equipotential surface

Ans: (d) Equipotential surface: Equipotential surfaces are known not to intersect one another and are rather closely spaced in the region of strong electric field and vice-versa.

Ques. The capacity of the parallel plate capacitor increases when

1. Area of the plate is decreased
2. Area of the plate is increased
3. Distance between the plates increases
4. None of the option

Ans: Area of the plate is increased: The parallel plate capacitor’s capacity increases when the plate area increases.

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Short Answer Questions [2 Marks]

Ques. Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of the electrostatic force between them exactly given by Q1Q2/4\(??\)0r2, where r is the distance between their centers?

Ans. No, because charge distributions on the spheres will not be uniform.

Ques. If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true?

Ans. No

Ques. A small test charge is released at rest at a point in an electrostatic field configuration. Is it possible for it to pass along the field line moving through that point?

Ans. Not necessarily (the statement, however, is true if only the field line is straight). The field line gives the direction of acceleration and not that of the velocity.

Ques. We’re aware that the electric field is discontinuous across the surface of a charged conductor. Similarly, is electric potential also discontinuous?

Ans. No, the potential is continuous.

Ques. What meaning would you give to the capacitance of a single conductor?

Ans. A single conductor is a capacitor with at least one ‘plate’ at infinity.

Ques. The atmosphere’s top is at about 400 kV with regard to the Earth’s surface, as corresponding to an electric field which decreases with altitude.The field is about 100 Vm^–1 when close to the Earth’s surface. However, what could be the possible reason behind why we get an electric shock just when we step out of our house into the open? (It is presumed that the house to be a steel cage so there is no field inside)

Ans. Typically, our body and the ground can be responsible for creating an equipotential surface. As we step out into the open, the original equipotential surfaces of open-air vary, keeping our head and the ground at the same potential.

(j) A person is seen fixing a two-meter high insulating slab carrying on its top a large aluminum sheet of area 1m2 just outside their house. Is it possible that the person can get an electric shock if they touch the metal sheet the next morning?

Ans. Yes. The steady discharging current in the atmosphere can potentially charge up the aluminum sheet gradually and raise its voltage to the maximum based on the capacitance of the capacitor.

Ques. The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on average over the globe. Determine why doesn’t the atmosphere discharge itself completely and rather become neutral, electrically? Simply, what keeps the atmosphere charged?

Ans. The atmosphere is already charged by the impact of thunderstorms and lightning all over the globe and discharged through regions of ordinary weather. The two opposing currents are usually in equilibrium, if considered on average.

Ques. Highlight the energy types into which the electrical energy of the atmosphere is dissipated during lightning?

Ans. Light energy has been found involved in lightning, while heat and sound energy are seen accompanying thunder.

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Long Answer Questions [3 Marks]

Ques. Considering the separation between the plates is about 0.5 cm, determine the area of plates which have a 2F parallel plate capacitor.

Ans: As per the equation, Capacitance, C = \(\varepsilon\)₀A/d

The Capacitor has a capacitance of, C=2 F

Separation between the plates,

d = 0.5 cm

= 0.5 x 10^-2 m

ε0 = free space permittivity = 8.85 x 10^-12 C²N^-1m^-2

Plate area , A = Cd/\(\varepsilon\)₀

A = [2 x 0.5 x 10^-2]/8.85 x 10^-12

= 1130 x 10⁶ m²

Ques. An electric dipole comprises two opposite charges of 1 μC which are separated by 2 cm. Besides, the dipole is positioned in an external uniform field of 105NC-1 . With the following terms, determine the amount of work done to rotate the dipole through 180° beginning from its initially aligned position.

Ans: As per the equation, q = 1

And, μC = 1 x 10^-6 C

2a = dipole, length = 2 cm = 2 x 10^-2 m.

θ = 0°, θ² = 180°,

E = 10⁵ NC^-1.

Thus, p = 2aq = 2 x 10^-2 x 10^-6 = 2 x 10^-8 cm.

Following the usage of relation, it can be said, W- = pE (cosθ1 – cosθ2)

= 2 x 10^-8 x 10⁵ (cos 0 – cos 180)

= 2 x 10^-3 [1 – (- 1)]

= 2x 10^-3 x 2

= 4 x 10^-3 J

= 0.004 J.

Ques. A charged cylinder is found to have a linear charge density of λ and is encompassed by a hollow coaxial conducting cylinder. Determine the electric field in the space between both cylinders.

Ans: Assuming the length of the charged cylinder and the hollow coaxial conducting cylinder both to be L.

As is mentioned already, the charge density is λ

Now, Let E be the electric field in the space between the two cylinders

With regard to Gauss theorem, the electric flux via Gaussian surface is given as Φ = E (2πd)L

Therefore, d is the distance between the common axis of the cylinders, such that

\(\theta\) = E (2πd)L = q/ε₀

q shows the charge on the inner surface of the outer cylinder

And, ε₀ is the permittivity of the free space

Thus, it can be said, E (2\(\pi \)d)L = λL/ε₀

Henceforth, the electric field λ/2πdε₀ between both cylinders.

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Very Long Answer Questions [5 Marks]

Ques. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected parallel to one another. As per the data, determine,

1. The total capacitance of the combination.
2. Establish the amount of charge on each capacitor, considering the combination is connected to a 100V supply.

Ans: a. As per the given data,

C₁ = 2pF,

C₂ = 3pF

C₃ = 4pF.

Equivalent capacitance for the parallel combination can be shown by, C_eq .

Henceforth,

C_eq = C₁ + C₂ + C₃

= 2 + 3 + 4

= 9pF

2. Supply voltage can be shown by, V = 100V

The three capacitors are known to have the same voltage, V = 100v

As per the formula, q = VC

Where,

q = charge

C = capacitance

V = potential difference

For capacitance, c = 2pF

q = 100 x 2

= 200pC

= 2 x 10^-10C

Now, for capacitance, c = 3pF

q = 100 x 3

= 300pC

= 3 x 10^-10C

Now, for capacitance, c = 4pF

q = 100 x 4

= 400pC

= 4 x 10^-10 C

Ques. A parallel plate capacitor has been developed with a voltage rating of 1kV, by utilizing a material of dielectric constant 3 and dielectric strength about 107Vm−1. For safety, it’s instructed to not surpass a field of 10 dielectric strength. Determine the minimum area of respective plates required to have a capacitance of 50pF.

Ans: As per the given data, Potential rating of a parallel plate capacitor is, V=1 (kV=1000V)

Dielectric constant, \(\varepsilon\)r = 3

Dielectric strength = 107 V/m

The field intensity instructs to not exceed 10% of the dielectric strength.

Which means, the electric field intensity, E=10 of 107= 106 V/m

Capacitance, C = 50 pF = 50 × 10⁻¹² F

Distance between the plates is given by,

d = V/E

d = 1000/106

d = 10⁻³ m

Capacitance can be shown by the relation,

C=\(\varepsilon\)₀\(\varepsilon\)₁A/d

Where,

A= area of each plate

\(\varepsilon\)0= permittivity of free space= 8.85×10⁻¹² N⁻¹m^-2C²

A= CD/∈0∈1

A= 50 × 10⁻¹² × 10⁻³ /8.85×10⁻¹² × 3= 19 cm²

Hence, the area of each plate is about 19 cm².

Ques. A 4 µF capacitor is charged via a 200 V supply. However, after instant disconnection, it is then connected to another uncharged 2 µF capacitor. Determine the amount of electrostatic energy lost from the first capacitor in form of heat and electromagnetic radiation.

Ans: Capacitance, C1= 4 μF

Supply voltage, V₁ = 200 V

Capacitance of the uncharged capacitor, C₂= 2 \(?\)F

Electrostatic energy stored in C1 is given as

E1 = (1/2)C₁V₁₂

= (1/2) x 4 x 10^-6 x (200)²

= 8 x 10^-2 J

When C₁ is disconnected from the power supply and connected to C₂, the voltage retrieved is V₂.

Considering the law of conservation of energy, the initial charge on the capacitor C1 equals the final charge on the capacitors C₁ and C₂.

V₂ (C₁ + C₂) = C₁V₁

V₂ (4+ 2) x 10^-6 = 4 x 10^-6 x 200

V₂ = (\(400\over 3\)) V

Electrostatic energy of the combination is

E₂ = (½) (C₁+C₂) V₁₂

= (½ ) x (2+4) x 10-6 x (\(400\over 3\))²

= 5.33 x 10^-2 J

Which is to say, the lost electrostatic energy by capacitor is, C₁ = E₁ – E₂

= 0.08 – 0.0533 = 0.0267

= 2.67 x 10^-2 

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