Electrostatic Field MCQ Quiz - Objective Question with Answer for Electrostatic Field - Download Free PDF

Concept

Series connection of capacitors:

When 'n' capacitors are connected in series, the equivalent capacitance is given by:

\({1\over C_{eq}}={1\over C_{1}}+{1\over C_{2}}........{1\over C_{n}}\)

If all capacitors are equal, then:

\(C_{eq}={C\over n}\)

Parallel connection of capacitors:

When 'n' capacitors are connected in parallel, the equivalent capacitance is given by:

\(C_{eq}=C_1+C_2.......C_n\)

If all capacitors are equal, then:

\(C_{eq}=nC\)

Calculation

Given, n = 10

C_eq = 10 μF

\(10={C\over 10}\)

\(C=100\space \mu F\)

When capacitors are connected in parallel.

\(C_{eq}=10\times 100\times 10^{-6}\)

C_eq = 10^-3 F

The energy in the capacitor is given by:

\(E={1\over 2}CV^2\)

\(E={1\over 2}\times 10^{-3}\times (100)^2\)

E = 5 J

The correct answer is - V₀ > VA

Key Points

• Electric potential
  • In a uniform electric field pointing in the positive X direction, the electric potential decreases as we move in the direction of the field.
  • Since point A is at x = +2 cm and the field points in the positive X direction, the potential at A (VA) will be lower than the potential at the origin (V₀).
• Potential difference
  • The potential difference between two points in an electric field is given by V = Ed, where E is the electric field strength and d is the distance.
  • Since A is further from the origin in the direction of the electric field, V₀ > VA.

Additional Information

• Electric field direction
  • The direction of the electric field is defined as the direction in which a positive test charge would move.
  • In this case, the field points in the positive X direction, meaning it pushes positive charges in that direction.
• Potential at point B
  • Point B is on the Y axis at y = +1 cm.
  • Since the electric field is along the X axis, there is no change in potential along the Y axis.
  • Therefore, the potential at B (Vʙ) is the same as the potential at the origin (V₀).
• Electric field strength
  • The electric field strength E is uniform, meaning it has the same magnitude and direction at all points in the region.
  • This uniformity simplifies the calculation of potential differences as it depends only on the distance in the direction of the field.

Explanation:

Electric Potential Difference

Definition: The electric potential difference between two points in an electric field is defined as the work done to move a unit positive charge from one point to another. It is a measure of the potential energy per unit charge at a specific location in the field.

Formula: The electric potential difference (V) between two points is given by:

\( V = \frac{W}{Q} \)

where V is the electric potential difference, W is the work done, and Q is the charge.

Explanation:

When a charge moves in an electric field, work is done by or against the electric field. The amount of work done per unit charge is called the electric potential difference. This concept is analogous to gravitational potential difference, where work is done to move a mass in a gravitational field.

Units: The SI unit of electric potential difference is the volt (V). One volt is equivalent to one joule per coulomb (1 V = 1 J/C).

Correct Option Analysis:

The correct option is:

Option 1: Work done / charge

This option correctly describes the electric potential difference. The work done to move a charge from one point to another divided by the amount of charge gives the electric potential difference. Mathematically, this is represented as:

\( V = \frac{W}{Q} \)

where V is the electric potential difference, W is the work done, and Q is the charge.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Electric charge / time

This option describes the concept of electric current, not electric potential difference. Electric current (I) is defined as the rate of flow of electric charge (Q) through a conductor per unit time (t). Mathematically, it is expressed as:

\( I = \frac{Q}{t} \)

The unit of electric current is the ampere (A).

Option 3: Charge × work done

This option does not represent any standard electrical quantity. Multiplying charge (Q) by work done (W) does not yield electric potential difference or any other commonly used electrical parameter.

Option 4: Work done / time

This option describes the concept of power, not electric potential difference. Power (P) is defined as the rate at which work (W) is done or energy is transferred per unit time (t). Mathematically, it is expressed as:

\( P = \frac{W}{t} \)

The unit of power is the watt (W).

Conclusion:

Understanding the distinction between different electrical concepts is crucial. The electric potential difference specifically refers to the work done to move a unit charge between two points in an electric field, and is accurately represented by the formula \( V = \frac{W}{Q} \). Other options describe different electrical quantities such as electric current and power, but do not pertain to electric potential difference.

Concept:

The work done in moving a charge q between two points in an electric field can be calculated using the formula:

W = q * (V_M - VN)

Where W is the work done, q is the charge, and V_M and V_N are the electric potentials at points M and N, respectively.

Since the points M and N are at the same distance from the charge Q, the potential at both points is the same:

V_M = V_N = 1 / (4πε₀) * Q / r

Thus, the difference in potential between the two points is:

V_M - V_N = 0

The work done in moving the charge between points M and N is given by:

W = (V_M - V_N) * q = 0

∴ The work done in moving the charge is 0 J.

Solution:

Let us first find the number of photons falling on the sphere.

E = n ×  (hC / λ)

n = (E × λ) / (hC)

n = (10^-7 × 200 × 10^-9) / (2 × 10^-25)

n = 10¹¹

Given that one in ten thousand ejects an electron.

Therefore, the total number of electrons ejected = (10¹¹ / 10⁴) = 10⁷

Total charge on the sphere = 10⁷ × 1.6 × 10^-19 = 1.6 × 10^-12 C

Potential = (K × Q) / R

Potential = (9 × 10⁹ × 1.6 × 10^-12) / (4.8 × 10^-2)

Potential = 3 × 10^-1 V

Concept:

Electric Flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity, because it's the dot product of two vector quantities, electric field and the perpendicular differential area.
   ϕ = E.A = EA cosθ 
• The SI unit of the electric flux is N-m2/C.

Electric flux density (D) is a vector quantity because it is simply the product of the vector quantity electric field and the scalar quantity permittivity of the medium, i.e.

\(\overset{\rightharpoonup}{D} = \varepsilon \overset{\rightharpoonup}{E}\)

Its unit is Coulomb per square meter.

Concept:

Coulomb's law: 

It states that the magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance r between them. 

• It is represented mathematically by the equation:

\(F = \frac{1}{4\pi ϵ_0}\frac{q_1 q_2}{r^2}\)

Where ϵ0 is the permittivity of free space (8.854 × 10-12 C2 N-1 m-2).

The value of \(k = \frac{1}{4\pi ϵ_0} = 9 \times 10^9 N m^2 C^{-2}\)

Calculation:

So, initial the force between two charges q1 and q2 is 200 N. 

\(F = k\frac{q_1 q_2}{r^2} = 200N\) -- (1)

If new distance r' = 2 r

New Force is

\(F' = k\frac{q_1 q_2}{r'^2} = k\frac{q_1 q_2}{(2r)^2}\)

\(\implies F' = k\frac{q_1 q_2}{4r^2} = \frac{1}{4}k\frac{q_1 q_2}{(r)^2}\)     ---(2)

From (1) and (2)

\(\implies F' = \frac{F}{4}\)

or

\(\implies F' = \frac{200N}{4} = 50 \ N\)

So, the correct option is 50 N.

Concept:

• The electric potential difference between two points in an electric circuit carrying some current is the work done to move a unit charge from one point to the other.
• The standard metric unit on electric potential difference is the volt, abbreviated V. 
• The potential difference is the work done per unit charge.

Mathematically, it is defined as:

\(V={W\over Q}\)

V = Potential difference

W = Work done

Q = electric charge.

The S.I unit of work is joule and that of the charge is the coulomb.

CONCEPT:

Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

Force (F) ∝ q1 × q2

\(F \propto \frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 109 Nm2/C2 

Calculations:

Consider new charge + 6μC is placed d m apart from old +4 μC charge and (x + 0.6) m apart from -16 μC charge.

Let,
q_A = + 4 μC at point A
q_B = - 16 μC at point B
q_C = + 6 μC at point C

Since, net force on charge q_C will zero.

∴ |F_CA| = |F_CB|

From above concept,

\(\frac{Kq_Aq_C}{d^2}=\frac{Kq_Bq_C}{(0.6+d)^2}\)

\(\frac{24}{d^2}=\frac{96}{(0.6+d)^2}\)

4d² = (0.6 + d)²

4d² = 0.36 + d² + 1.2d

3d² - 1.2d - 0.36 = 0

d₁ = - 0.2 m

d₂ = + 0.6 m

Hence, according to option + 0.6 m distance a third charge of + 6 μC be placed from + 4 μC so that the net force exerted on it will be zero.

Dielectric Materials: The insulating materials which are poor conductor of electric current are called dielectric materials.

When the dielectric is placed in the electric field no current flows through them but exhibits electric dipole i.e. there is the separation of positive and negative electrically charged entities on a molecular or atomic level.

These are used in capacitors, power line and electric insulation, switch bases and light receptacles.

Dielectric strength: The maximum voltage that can be applied to a given material without causing it to break down is called dielectric strength.

It is measured in volts per unit thickness of the material.

EXPLANATION: 

The dielectric strength of an insulating material is measured in volts (V) per unit thickness (m) of the material (V/m). So option 1 is correct.

CONCEPT:

Electric Field Intensity:

The electric field intensity at any point is the strength of the electric field at the point.

It is defined as the force experienced by the unit positive charge placed at that point.

\(\vec E = \frac{{\vec F}}{{{q_o}}}\) Newton/Coulomb

Where F = force and qo = small test charge

The magnitude of the electric field is

\(E = \frac{{kq}}{{{r^2}}}= \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\)

Where K = constant called electrostatic force constant, q = source charge and r = distance

The region around a charged particle in which electrostatic force can be experienced by other charges is called the electric field.

Coulomb’s law:

When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

Force (F) ∝ q1 × q2

\(F \propto \frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 109 Nm2/C2

Calculation:

The given situation can be visualized as shown:

The force on Q because of q₁ will be:

\({F_{Q{q_1}}} = \frac{{kQ{q_1}}}{{{a^2}}}\)

Similarly, the force on Q because of q₂ will be:

\({F_{Q{q_2}}} = \frac{{kQ{q_2}}}{{{{\left( {3a} \right)}^2}}}\)

For the net charge at Q to be zero, we write:

\(\frac{{kQ{q_1}}}{{{a^2}}} + \frac{{kQ{q_2}}}{{9{a^2}}} = 0\)

\(\frac{{{q_1}}}{{{a^2}}} = \frac{{ - {q_2}}}{{9{a^2}}}\)

q₂ = -9q₁

Magnetic circuits are analogous to electric circuits. Some of the quantities are given below.

Mica capacitor:

• Mica capacitors are available in the range between a few pF to thousand pF with voltage rating in between a few hundred volts to a thousand volts
• Mica capacitors are low loss capacitors that are used where the high frequency is required, and their value doesn’t change much over time
• The capacitance of mica capacitors ranges from 10 pF to 5000 pF
• Mica capacitors have low resistive and inductive components associated with it. Hence, they have a high Q factor and because of the high Q factor, their characteristics are mostly frequency independent, which allows this capacitor to work at high frequency.

Applications for Mica Capacitors:

• Ripple filter and Decoupling for general electronic devices
• Resonant circuits
• Coupling circuits
• Time constant circuits
• Harmonic and reactive power compensation
• High-power RF broadcast transmitters
• Defense Electronic Devices
• Power transfer circuits for low-capacitance snubber applications
• Radio or TV transmitters
• Cable TV amplifiers
• High voltage inverter circuits

Additional InformationElectrolytic Capacitor:

• It’s a polarized capacitor whose anode or positive plate is made of metal
• The solid, liquid, or gel electrolyte covers the surface of the oxide layer serving as a cathode or negative plate.

There are three families:

1. Aluminium electrolytic capacitors

2. Tantalum electrolytic capacitors

3. Niobium electrolytic capacitors

• These are generally used when very large capacitance values are required.
• Due to their very thin oxide layer and enlarged anode surface, electrolytic capacitors have a much higher capacitance-voltage product per unit volume than ceramic or film capacitors.

Electrolytic’s are widely used capacitors due to their low cost and small size but there are 3 easy ways to destroy this

1. overvoltage

2. reversed polarity

3. over temperature

The most commonly used dielectric is “Aluminium oxide”.

The main disadvantage is it can’t be used on AC supplies.

Air Capacitor:

Air capacitors are capacitors that use air as their dielectric

Variable air capacitors are used more often because of their simple construction

They are usually made of two sets of semi-circular metal plates separated by air gaps

One set is fixed and the other is attached to a shaft which allows the user to rotate the assembly, therefore changing the capacitance as needed

The larger the overlap between the two sets of plates, the higher the capacitance

The maximum capacitance state is achieved when the overlap between the two sets of plates is highest

The lowest capacitance state is achieved when there is no overlap

Concept:

Electric Current (I):

• It is defined as the net amount of electric charge that flows through a cross-section of a conductor in an electric circuit in a per unit time.
• S.I unit of electric current is Ampere(A).

1 Coulomb (C):

• One Coulomb of charge is equal to the charge transported by an electric current of constant magnitude in one second of time.
• It is defined as the product of a 6.24 × 1018 ​charge of an electron.
• It is the S.I unit of electric charge.

Charge of one electron = 1.6 × 10-19 Coulomb.

Explanation:

We know that e = 1.6 × 10-19 C, i.e charge on the electron.

The total charge required is 1 Coulomb, q = 1C

Therefore we also know that q = n × e, (n = number of electrons on the charge)

Hence, n = q/e....(1)

Substituting all the values in the equation (1)

n = 1/1.6 × 10-19 ​= 625 × 1016

So, if 1A current flows through the conductor then 625 × 1016 electrons flow per second across the cross-section of the conductor.