250+ TOP MCQs on Magnetism and Gauss’s Law | Class12 Physics

Physics Multiple Choice Questions on “Magnetism and Gauss’s Law”.

1. What is the net magnetic flux through a closed surface?
a) Positive
b) Negative
c) Zero
d) Depends on the nature of the surface
Answer: c
Clarification: Gauss’s law indicates that there are no sources or sinks of the magnetic field inside a closed surface. A “closed surface” is a surface that completely encloses a volume with no holes. Therefore, the net magnetic flux through a closed surface is zero.

2. What is the torque exerted by a bar magnet on itself due to its field?
a) Maximum
b) Zero
c) Minimum
d) Depends on the direction of the magnetic field
Answer: b
Clarification: A bar magnet does not exert a force or torque on itself due to its field. But an element of a current-carrying conductor experiences forces due to another element of the conductor. So, the torque exerted by a bar magnet on itself is zero.

3. When does a magnetic dipole possess maximum potential energy inside a magnetic field?
a) Magnetic moment and magnetic field are antiparallel
b) Magnetic moment and magnetic field are parallel
c) The magnetic moment is zero
d) The magnetic field is zero
Answer: a
Clarification: A magnetic dipole possess maximum potential energy when its magnetic moment and the magnetic field are antiparallel. When the magnetic dipole is aligned along the magnetic field, i.e. when θ = 1800, it is in unstable equilibrium having maximum potential energy.

4. Calculate the surface integral of a magnetic field over a surface.
a) Maximum
b) Minimum
c) Zero
d) Equal to its magnetic flux through that surface
Answer: d
Clarification: The Gauss’s law states that the surface integral of a magnetic field over a closed surface is always zero. But the surface integral of a magnetic field over a surface gives magnetic flux through that surface.

5. Which of the following is not a consequence of Gauss’s law?
a) The magnetic poles always exist as unlike pairs of equal strength
b) If several magnetic lines of force enter a closed surface, then an equal number of lines of force must leave that surface
c) There are abundant sources or sinks of the magnetic field inside a closed surface
d) Isolated magnetic poles do not exist
Answer: c
Clarification: Gauss‘s law indicates that there are no sources or sinks of the magnetic field inside a closed surface. In other words, there are no free magnetic charges. Hence, magnetic monopoles do not exist.

6. Isolated magnetic poles exist.
a) True
b) False
Answer: b
Clarification: Gauss’s law indicates that there are no sources or sinks of the magnetic field inside a closed surface. So there is no point at which the field lines start or there is no point at which the field lines terminate. Hence, isolated magnetic poles do not exist.

7. Which among the following is the source of the magnetic field (magnetism)?
a) Mechanical origin
b) Electrical origin
c) Chemical origin
d) Potential origin
Answer: b
Clarification: Magnetism is of electrical origin. The electrons revolving in an atom behave as tiny current loops and these current loops give rise to magnetism. An electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines circling a wire.

8. The line of force in a magnetic field represents the direction at each point that a magnetic needle placed at the point takes up. Do they also represent the direction of the force on a moving charge at each point?
a) Not possible
b) Represent circular motion
c) Represent tangential motion
d) Represents translatory motion
Answer: a
Clarification: No. The force on a charge is perpendicular to the direction of the magnetic field at each point.
F = q (v × B).
It is inappropriate to call magnetic field lines as lines of force.

250+ TOP MCQs on AC Voltage Applied to a Capacitor | Class12 Physics

Physics Multiple Choice Questions on “AC Voltage Applied to a Capacitor”.

1. A 1.5 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance in the circuit.
a) 2120 Ω
b) 21.2 Ω
c) 212 Ω
d) 2.12 Ω
Answer: c
Clarification: Xc = (frac {1}{2πfC})
Xc = (frac {1}{(2 times 3.14 times 50 times 1.50 times 10^{-6})})
Xc = 212 Ω

2. A capacitor of capacitance 10 μF is connected to an oscillator giving an output voltage, E = 10 sin ωt volt. If ω = 10 rad s-1, find the peak current in the circuit.
a) 197 mA
b) 1 mA
c) 179 mA
d) 5 mA
Answer: b
Clarification: I0 = (frac {E_0}{(frac {1}{omega C})})
I0 = ωCE0.
I0 = 10 × 10 × 10-6 × 10
I0 = 10-3A = 1 mA.

3. What is the capacitive reactance of a 5 μF capacitor when it is a part of a circuit whose frequency is 50 Hz?
a) 636.6 Ω
b) 1636.6 Ω
c) 2636.6 Ω
d) 4636.6 Ω
Answer: a
Clarification: Xc = (frac {1}{2πfC})
Xc = (frac {1}{(2 times 3.14 times 50 times 5 times 10^{-6})})
Xc = 636.6 Ω.

4. How will the capacitive reactance be affected if the frequency is doubled?
a) Doubled
b) Insignificant
c) Remains the same
d) Halved
Answer: d
Clarification: If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled. This is because, when capacitive reactance, being a resistance, is lowered, the current can flow more easily through the circuit.

5. Give the SI unit of capacitive reactance.
a) Am
b) Ω
c) Ωm
d) A
Answer: c
Clarification: Capacitive Reactance is the complex impedance of a capacitor whose value changes with respect to the applied frequency. The SI unit of capacitive reactance is the ohm (Ω). Capacitive reactance is denoted by Xc.

6. The capacitive reactance varies directly with the frequency.
a) True
b) False
Answer: b
Clarification: Xc = (frac {1}{omega_C})
Xc = (frac {1}{2πfC}).
Thus the capacitive reactance varies inversely with the frequency. Therefore as frequency increases, the capacitive reactance decreases.

7. Calculate the rms value of current in the circuit wherein an 80 μF capacitor is connected to a 100 V, 80 Hz ac supply.
a) 4 A
b) 2 A
c) 7 A
d) 50 A
Answer: a
Clarification: Xc = (frac {1}{2πfC})
Xc = (frac {1}{(2 times 3.14 times 80 times 80 times 10^{-6})})
Xc = 24.8 Ω.
Irms = (frac {E_{rms}}{X_C})
Irms = (frac {100}{24.8})
Irms = 4.03 A ≈ 4 A.

8. What will be the reactance of a capacitor at 150 Hz, if it has a reactance of 200 Ω at 50 Hz?
a) 67 Ω
b) 40 Ω
c) 150 Ω
d) 60 Ω
Answer: a
Clarification: (frac {X_C^{’}}{X_C} =frac {f}{f^{’}})
Xc = (frac {(50 times 200)}{(150)})
Xc = 66.7 Ω ≈ 67 Ω
Therefore, the reactance at 150 Hz will be 67 Ω.

9. A 1.50 μF capacitor is connected to a 220 V, 50 Hz source. If the frequency is doubled, what happens to the capacitive reactance?
a) Remains the same
b) Doubled
c) Halved
d) Becomes zero
Answer: c
Clarification: When the frequency is doubled, the capacitive reactance will be halved. This is because the capacitive reactance (Xc) is inversely proportional to the frequency (f). This is the effect of capacitive reactance on frequency.

10. Which among the following is the correct expression for finding capacitive reactance for an ac circuit containing capacitor only?
a) Xc = 2πf
b) Xc = (frac {1}{2πfC})
c) Xc = 2πfC
d) Xc = (frac {2πf}{C})
Answer: b
Clarification: For an ac circuit containing capacitor only,
Xc = (frac {1}{omega_C})
Xc = (frac {1}{2πfC}).

250+ TOP MCQs on Coherent and Incoherent Addition of Waves | Class12 Physics

Physics Multiple Choice Questions on “Coherent and Incoherent Addition of Waves”.

1. Which of the following is a form of light whose photons share the same frequency and whose wavelengths are in phase with one another?
a) Coherent sources
b) Incoherent sources
c) Electromagnetic waves
d) Sunlight
Answer: a
Clarification: Coherent light is a form of light whose photons share the same frequency and whose wavelengths are in phase with one another. The phase difference between the waves should be constant in case of coherent sources.

2. Which among the following is an example of coherent sources?
a) Fluorescent tubes
b) LED light
c) LASER
d) Tungsten filament lamps
Answer: c
Clarification: LASER is the short form for Laser Amplification by Stimulated Emission Radiation. The amplified light beam coming out of a LASER is essentially due to the emission of electrons stimulated by incident radiation consisting of photons. As a result, this causes the coherent behavior of the LASER beam.

3. Pick the odd one out.
a) LASER
b) LED
c) Sound waves
d) Radio transmitters
Answer: b
Clarification: LED is the odd one out. LED is short for light-emitting diode. LED is not a coherent source, whereas, others are examples of coherent sources. The light emitted from an LED is neither spectrally coherent nor even highly monochromatic.

4. Scattering of waves can be coherent and incoherent.
a) True
b) False
Answer: a
Clarification: Yes, the scattering of waves can be coherent and incoherent. The scattering of a wave is coherent and constructive if the phase delay is the same for all the scattered waves. If it varies randomly, then it is considered to be incoherent.

5. Identify the factor is not the same for coherent waves.
a) Frequency
b) Phase difference constant
c) Amplitude
d) Wavelength in phase with each other
Answer: c
Clarification: Coherent waves are the waves with the same frequency and the wavelength of the waves are in phase as well. Therefore, the phase difference is constant. But the coherent waves do not have the same amplitude. Since the amplitude is different, there will be no complete constructive interference where they meet, so they will contribute poorly to an interference pattern.

6. Which of the following is the formula for calculating coherence time?
a) Τc = (frac {lambda^3}{(cDelta lambda)})
b) Τc = (frac {lambda}{(cDelta lambda)})
c) Τc = (frac {lambda^2}{(cDelta lambda)})
d) Τc = (frac {lambda^2}{(cDelta lambda)})
Answer: d
Clarification: The formula for calculating coherence time is given as:
Τc = (frac {lambda^2}{(cDelta lambda)})
Where Τc is the coherence time, λ is the wavelength, ∆λ is the spectral width of the source, and c is the speed of light in a vacuum (i.e. 3 × 108 m/s).

7. When is the wave interference strong?
a) When the paths taken by all of the interfering waves are greater than the coherence length
b) When the paths taken by all of the interfering waves are lesser than the coherence length
c) When the paths taken by all of the interfering waves are equal than the coherence length
d) When the paths taken by all of the interfering waves are independent of the coherence length
Answer: b
Clarification: Coherence length is defined as the propagation distance over which a coherent wave maintains a specified degree of coherence. Wave interference is strong when the paths taken by all of the interfering waves are lesser than the coherence length.

250+ TOP MCQs on Size of the Nucleus | Class12 Physics

Physics Online Quiz for IIT JEE Exam on “Size of the Nucleus”.

1. Identify the expression for the nuclear radius from the following.
a) R = R0 ∛A
b) R = R0 √A
c) R = R0 A3
d) R = R0 A2
Answer: a
Clarification: Most of the nuclei of atoms are spherical in structure. The expression for the nuclear radius is given by:
R = R0 ∛A
Where R0 is a constant and A is the mass number. The nuclear radius is measured in Fermi meter.
1 fm = 10-15 m.

2. Who measured the size of the nucleus first?
a) Bohr
b) Einstein
c) Rutherford
d) Geiger and Marsden
Answer: c
Clarification: Rutherford was the form to measure the size of the nuclei of an atom by 1911. He discovered that these were about 104 times smaller than the atoms that constituted them and this was due to the fascinating way that the alpha particles scattered from metal foils.

3. Which of the following is a stable nucleus?
a) The nucleus with even protons and odd electrons
b) The nucleus with even number of protons and neutrons
c) The nucleus with even neutrons and odd protons
d) The nucleus with odd protons and neutrons
Answer: b
Clarification: The nuclei of atoms having even numbers of both protons and neutrons are the most stable ones and this also means that they are less radioactive than nuclides containing even numbers of protons and odd numbers of neutrons.

4. If the internal energy of a nucleus is high, then it is radioactive.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. An atom is said to be unstable, or in other words, radioactive when the forces that it is subjected to it are unbalanced. Therefore, when there is an excess of internal energy, instability of an atom’s nucleus occurs and hence they can become radioactive. This may also result from an excess of either neutrons or protons in the nucleus.

5. A nucleus at rest splits into two nuclear parts having radii in the ratio of 1:3. Find the ratio of their velocities.
a) 1:9
b) 3:1
c) 1:27
d) 27:1
Answer: d
Clarification: Given: R1 = R0 (A1)1/3 and R2 = R0 (A2)1/3
(frac {R1}{R2} = (frac {A1}{A2} ) ^{frac {1}{3}} ) or ( frac {A1}{A2} = (frac {R1}{R2} )^3 = (frac {1}{3} )^3 )
( frac {A1}{A2} = frac { 1}{27})
So, the ratio of their masses is given as:
( frac {m1}{m2} = frac { 1}{27})
According to the principle of conservation of momentum, the magnitude of p1 = magnitude of p2
m1v1 = m2v2
( frac {v1}{v2} = frac {m2}{m1} = frac {27}{1}) = 27:1.

Physics Online Quiz for IIT JEE Exam,

250+ TOP MCQs on Electric Field | Class12 Physics

Physics Multiple Choice Questions on “Electric Field”.

1. The amount of force exerted on a unit positive charge in an electric field is known as _____
a) Electric field intensity
b) Electric flux
c) Electric potential
d) Electric lines of force

Answer: a
Clarification: The zone near a charge where its attraction or repulsion force works, is known as the electric field of that charge. Theoretically, it is up to infinite but practically it has limitations. If a unit positive charge is kept in that field, it will undergo some force which is known as electric field intensity at that point.

2. The direction of electric field created by a negative charge is ___________
a) Directed outwards
b) Directed towards the charge
c) Maybe outwards or towards the charge
d) Circular in shape

Answer: b
Clarification: If a unit positive charge is kept near a negative charge, the unit positive charge will be attracted towards the negative charge. That means the electric field is towards the negative charge. But in case of positive charge, the field is directed away from the charge.

3. Electric field inside a hollow conducting sphere ______
a) Increases with distance from the center of the sphere
b) Decreases with distance from the center of the sphere
c) Is zero
d) May increase or decrease with distance from the center

Answer: c
Clarification: According to Gauss’s law, if there is no charge inside a closed surface, the field inside the closed surface will always be zero. We know the charge is distributed on the outer surface of a conducting hollow sphere because the charges want to maintain maximum distance among them due to repulsion. So there is no charge inside the sphere and hence no electric field.

4. Electric field due to a uniformly charged hollow sphere at a distance of r (where r is greater than the radius of the sphere) is __________
a) Proportional to r
b) Inversely proportional to r
c) Proportional to r2
d) Inversely proportional to r2

Answer: d
Clarification: If the total charge of the sphere is q then the electric field at a distance of r is equal to (frac {q}{4pivarepsilon_o r^2}).Therefore the electric field is proportional is (frac {1}{r^2}) (if r > radius of the sphere). But if r < radius of the sphere the electric field will be zero i.e. electric field inside a hollow sphere is always zero.

5. Two point charges q1 and q2 are situated at a distance d. There is no such point in between them where the electric field is zero. What can we deduce?
a) There is no such point
b) The charges are of the same polarity
c) The charges are of opposite polarity
d) The charges must be unequal

Answer: c
Clarification: If both the charges are of the same polarity (maybe of unequal magnitude), there must be a point in between them where the electric field intensities of the charges are of equal magnitude and in opposite direction. Hence they balance each other and the net field intensity must be zero. But if the charges are of opposite polarities their field intensities aid each other and net field intensity can never be zero.

6. A uniformly charged sphere of radius R has charge +Q. A point charge –q is placed at a distance of 2R from the center of the sphere. The point charge will execute the simple harmonic motion. The statement is _____
a) False
b) True

Answer: a
Clarification: In the case of SHM, the force on a body is inversely proportional to the distance of the body from the mean position. But in this case, we know the force acting on the body is inversely proportional to the square of the distance [F =(frac {qQ}{4pivarepsilon_o(2R)^2})]. So the motion of the body will be oscillatory but not SHM.

7. Two point charges of the same polarities are hung with the help of two threads and kept close. The angle between the threads will be _________ if the system is taken to space.
a) 180 degree
b) 90 degree
c) 45 degree
d) 60 degree

Answer: a
Clarification: There is gravitational field on earth, so if we hang the two same charges there will be an interaction of vertical gravitational field and horizontal electric field. The system will achieve equilibrium by creating a certain angle between the threads and hence the vertical and horizontal components of forces will balance. But in space, there is no gravity. So the charges will be at 180-degree separation.

8. An electron of mass m is kept in a vertical electric field of magnitude E. What must be the value of E so that the electron doesn’t fall due to gravity?
a) m*g*e
b) (frac {e}{(m*g)})
c) (frac {(m*g)}{e})
d) (frac {1}{(m*g*e)})

Answer: c
Clarification: Gravitational force on the electron is m*g (weight of the electron). Electrical force on the body is e*E. If the electron doesn’t fall then these two forces balance each other, so m*g=E*e. Therefore E= (frac {(m*g)}{e}).

9. Electric field is a _______
a) Scalar quantity
b) Vector quantity
c) Tensor quantity
d) Quantity that has properties of both scalar and vector

Answer: b
Clarification: A scalar quantity is a quantity with magnitude only but no direction. But a vector quantity possesses both magnitude and direction. An electric field has a very specific direction (away from a positive charge or towards a negative charge). Hence electric field is a vector quantity. Moreover, we have to use a vector addition for adding two electric fields.

10. Two point charges +4q and +q are kept at a distance of 30 cm from each other. At which point between them, the field intensity will be equal to zero?
a) 15cm away from the +4q charge
b) 20cm away from the +4q charge
c) 7.5cm away from the +q charge
d) 5cm away from the +q charge

Answer: b
Clarification: The electric field at a distance of r from a charge q is equal to (frac {q}{4pivarepsilon_or^2}). Let the electric field intensity will be zero at a distance of x cm from +4q charge, so the fields due to the two charges will balance each other at that point. Therefore (frac {4q}{4pivarepsilon_ox^2}=frac {q}{4pivarepsilon_o(30-x)^2}). Solving this we get x=20cm. Therefore the point will be 20cm away from the +4q charge.

11. What is the dimension of electric field intensity?
a) [M L T-2 I-1]
b) [M L T-3 I-1]
c) [M L T-2 I-2]
d) [M L T-3 I]

Answer: b
Clarification: Electric field intensity is defined as the force on a unit positive charge kept in an electric field. Hence we can simply consider its dimension as(frac {the , dimension , of , force}{the , dimension , of , charge}). The dimension of force is [MLT-2] and the dimension of charge is [IT]. Therefore the dimension of field intensity is [M L T-3 I-1].

12. V/m is the unit of ______
a) Electric field intensity
b) Electric flux
c) Electric potential
d) Charge

Answer: a
Clarification: E=-(frac {dV}{dx}) where E is the field intensity, V is potential and x is distance. Therefore unit of electric field intensity will be (frac {unit , of , potential}{unit , of , distance} = frac {V}{m}). Electric flux has unit V*m, V is the unit of electric potential whereas charge has a unit of Coulomb or esu.

13. Electric field intensity at the center of a square is _____ if +20 esu charges are placed at each corner of the square having side-length as 10 cm.
a) 0
b) 0.4 dyne/esu
c) 2 dyne/esu
d) 1.6 dyne/esu

Answer: a
Clarification: Distance of center from each corner point of the square is = (frac {10sqrt2}{2}) = 5√2. Therefore field intensity at the center due to a single charge is = (frac {20}{(5sqrt2)^2}) dyne/esu. But the fields due to the four charges are equal and are at perpendicular to each other. So the fields balance each other and the net electric field at the center will be equal to zero.

14. A ball of 80mg mass and a 2*10-8 charge is hung with a thread in a uniform horizontal electric field of 2*104V/m. What is the angle made by the thread with vertical?
a) 27 degree
b) 30 degree
c) 45 degree
d) 0 degree

Answer: a

15. Find the electric field intensity at 10cm away from a point charge of 100 esu.
a) 1 dyne/esu
b) 10 dyne/esu
c) 100 dyne/esu
d) 9*109 dyne/esu

Answer: a
Clarification: We know that electric field intensity at r distance from a point charge q is defined as (frac {q}{r^2}) in the CGS system. Here q=100 esu, r=10cm. Substituting the value we get E=1dyne/esu. We must remember that factor (frac {1}{4pivarepsilon}) is considered only in the SI system and its value is 9*109, but in the CGS system, we simply take the value of this constant as unity.

250+ TOP MCQs on Parallel Plate Capacitor | Class12 Physics

Physics Multiple Choice Questions on “Parallel Plate Capacitor”.

1. Identify the simplest and the most widely used capacitor among the following.
a) Electrolytic capacitor
b) Spherical Capacitor
c) Parallel plate capacitor
d) Cylindrical capacitor

Answer: c
Clarification: The simplest and the most widely used capacitor is the parallel plate capacitor. It consists of two large plane parallel conducting plates, separated by a small distance.

2. How is the electric field between the two plates of a parallel plate capacitor?
a) Zero
b) Uniform
c) Maximum
d) Minimum

Answer: b
Clarification: The direction of the electric field is from the positive to the negative plate. In the inner region, between the two capacitor plates, the electric fields due to the two charged plates add up. Hence, the field is uniform throughout.

3. Identify the factor on which the capacitance of a parallel plate capacitor does not depend.
a) Permeability of the medium between the plates
b) Area of the plates
c) Distance between the plates
d) The permittivity of the medium between the plates

Answer: a
Clarification: The capacitance of a parallel plate capacitor is directly proportional to the area of the plates and permittivity of the medium between the plates. It is indirectly proportional to the distance between the plates.

4. Calculate the capacitance of the capacitor, if 1012 electrons are transferred from one conductor to another of a capacitor and a potential difference of 10 V develops between the two conductors.
a) 1.6 × 10-7 F
b) 160 × 10-8 F
c) 16 × 10-8 F
d) 1.6 × 10-8 F

Answer: d
Clarification: q = ne = 1012 × 1.6 × 10-19 = 1.6 × 10-7 C.
V = 10 V.
C = (frac {q}{V}) = 1.6 × (frac {1.6}{10^{-7}})

C = 1.6 × 10-8 F.

5. What is the net electric field in the outer regions above the upper plate and below the lower plate in a parallel plate capacitor?
a) Maximum
b) Uniform
c) Zero
d) Minimum

Answer: c
Clarification: In the outer regions above the upper plate and below the lower plate, the electric fields due to the two charged plates cancel out. Hence, the net electric field in the outer regions above the plate and below the lower plate is zero.

6. In the inner region between the two capacitor plates, the electric fields due to the two charged plates are zero.
a) True
b) False

Answer: b
Clarification: In the inner region between the two capacitor plates, the electric fields due to the two charged plates add up. The net field is given by:
(frac {sigma}{2varepsilon_0}+frac {sigma}{2varepsilon_0}=frac {sigma}{varepsilon_0} ).

7. ‘X’ is a widely used capacitor which consists of two large plane parallel conducting plates separated by a small distance. Identify X.
a) Spherical capacitor
b) Parallel plate capacitor
c) Cylindrical capacitor
d) Electrolytic capacitor

Answer: b
Clarification: The simplest and the most widely used capacitor is the parallel plate capacitor. It consists of two large plane parallel conducting plates, separated by a small distance.
Capacitance of a parallel plate capacitor = (frac {Q}{V} = [ frac {sigma A}{(frac {sigma d}{varepsilon_0})} ] = frac {Avarepsilon_0}{d}).

8. Find out the correct expression of the capacitance of a parallel plate capacitor where ‘A’ is the area of the plates, ‘d’ is the distance between the plates and ‘ε0’is the permittivity of the medium.
a) (frac {Avarepsilon_0}{d})
b) (frac {Ad}{varepsilon_0})
c) (frac {dvarepsilon_0}{A})
d) Aε0d

Answer: a
Clarification: The capacitance of a parallel plate capacitor is directly proportional to the area of the plates and permittivity of the medium between the plates. It is indirectly proportional to the distance between the plates.
C = (frac {Q}{V} = [ frac {sigma A}{(frac {sigma d}{varepsilon_0})} ] = frac {Avarepsilon_0}{d}).

9. A parallel plate capacitor has a plate area of 100 cm2 and is separated by a distance of 20 mm. Find its capacitance.
a) 6.425 × 10-12 F
b) 5.425 × 10-12 F
c) 4.425 × 10-12 F
d) 3.425 × 10-12 F

Answer: c
Clarification: C = (frac {Q}{V} = [ frac {sigma A}{(frac {sigma d}{varepsilon_0})} ] = frac {Avarepsilon_0}{d}).
C = (frac {100 times 10^{-4} times 8.85 times 10^{-12})}{(20 times 10^{-3})})
C = 4.425 × 10-12 F.