Capacitance of a parallel capacitor, V = 2 F Distance between the two plates, d = 0.5 cm = 0.5 x 10-2 m Capacitance of a parallel plate capacitor is given by the relation, C= ε0A/d A =Cd/ε0 Where, ε0 = Permittivity of free space Therefore A= 2x 0.5 x 10⁻²)/(8.85 x 10⁻¹²) Therefore, A = 1130 km² HencRead more
Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = 0.5 x 10-2 m
Capacitance of a parallel plate capacitor is given by the relation,
C= ε0A/d
A =Cd/ε0
Where,
ε0 = Permittivity of free space
Therefore A= 2x 0.5 x 10⁻²)/(8.85 x 10⁻¹²)
Therefore, A = 1130 km²
Hence, the area of the plates is too large.
To avoid this situation, the capacitance is taken in the range of μF.
Total required capacitance, C = 2 pF Potential difference, V = 1, kV = 1000 V Capacitance of each capacitor, C₁ = lμF Each capacitor can withstand a potential difference, V₁= 400 V Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to eacRead more
Total required capacitance, C = 2 pF Potential difference, V = 1, kV = 1000 V
Capacitance of each capacitor, C₁ = lμF
Each capacitor can withstand a potential difference, V₁= 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V.
Hence, number of capacitors in each row is given as 1000/4 = 2.5
Hence, there are three capacitors in each row.
Capacitance of each row = 1/(1+1+1) = 1/3 μF
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
1/3 +1/3 + 1/3 +……..n terms = n/3
However, capacitance of the circuit is given as 2 μF.
Therefore n/3 =2 ,therefore n=6
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 x3 i.e.,18 capacitors are required for the given arrangement.
Ans (a). Zero at both the points Charge - q is located at (0, 0, — a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. ERead more
Ans (a).
Zero at both the points
Charge – q is located at (0, 0, — a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,
V= (1/4π ε0 ) [q/(z-a)] + (1/4π ε0 ) [-q/(z+a)]
= q (z+a -z + a )/ [4π ε0 (z²-a²)]
=2qa/[4π ε0 (z²-a²)]
=p/[4π ε0 (z²-a²)]
Where,
ε0 = Permittivity of free space and p = Dipole moment of the system of two charges = 2qa
Ans (b).
Distance r is much greater than half of the distance between the two charges.
Hence, the potential (V) at a distance r is inversely proportional to square of the distance,
i.e. V ∝ 1/r²
Ans (c).
Zero
The answer does not change if the path of the test is not along the x-axis.
A test charge is moved from point (5, 0, 0) to point (-7, 0, 0) along the x-axis.
Electrostatic potential (V₁) at point (5, 0, 0) is given by,
Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (-7, 0, 0) along the x- axis.
The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.
Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equRead more
Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere.
Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere.
Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EA be the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,
EA/EB = (QA /4π ε0 a2) x (b2 4π ε0 )/QB
EA/EB = (QA /QB) x (b2 / a2) ———————-Eq-1
However , QA /QB =CAV/ CB V
and CA/ CB =a/b
Therefore QA /QB =a/b ———————-Eq-2
Putting the values of Eq-2 in Eq-1 ,we obtain
EA/EB = (a/b ) x (b2 / a2)=b/a
Therefore ,the ratio of electric fields at the surface is b/a.
The distance between electron-proton of a hydrogen atom, d = 0.53 A Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C Ans (a). Potential at infinity is zero. Potential energy of the system, p-e = Potential energy at infinity - Potential energy at distance, d = 0 - (qRead more
The distance between electron-proton of a hydrogen atom, d = 0.53 A
Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C
Ans (a).
Potential at infinity is zero.
Potential energy of the system, p-e
= Potential energy at infinity – Potential energy at distance, d
= 0 – (q₁-q₂)/4πε0 d
Where, ε0 is the permittivity of free space and
1/4πε0 =9x 10⁹ Nm²C⁻²
Therefore Potential Energy = 0- (9x 10⁹) x (1.6 x 10⁻¹⁹)² /(0.53 x 10¹⁰)
= -43.7x 10⁻¹⁹ J , (Since 1.6 x 10⁻¹⁹ J = 1 eV)
Therefore Potential energy =-43.7×10⁻¹⁹ = -(43.7 x 10⁻¹⁹)/(1.6 x 10⁻¹⁹) =-27.2 eV
Therefore, the potential energy of the system is -27.2 eV.
Ans (b).
Kinetic energy is half of the magnitude of potential energy.
Kinetic energy = 1/2 x(-27.2) = 13.6 eV
Total energy = 13.6 – 27.2 = 13.6 eV
Therefore, the minimum work required to free the electron is 13.6 eV.
Ans (c).
When zero of potential energy is taken, d = 1.06 A
Therefore Potential energy of the system = Potential energy at d₁- Potential energy at d
= (q₁q₂)/4πε0 d₁ – 27.2 eV
= (9 x 10⁹) x (1.6 x 10⁻¹⁹)² / ( 1.06 x 10⁻¹⁰) -27.2 eV
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Capacitance of a parallel capacitor, V = 2 F Distance between the two plates, d = 0.5 cm = 0.5 x 10-2 m Capacitance of a parallel plate capacitor is given by the relation, C= ε0A/d A =Cd/ε0 Where, ε0 = Permittivity of free space Therefore A= 2x 0.5 x 10⁻²)/(8.85 x 10⁻¹²) Therefore, A = 1130 km² HencRead more
Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = 0.5 x 10-2 m
Capacitance of a parallel plate capacitor is given by the relation,
C= ε0A/d
A =Cd/ε0
Where,
ε0 = Permittivity of free space
Therefore A= 2x 0.5 x 10⁻²)/(8.85 x 10⁻¹²)
Therefore, A = 1130 km²
Hence, the area of the plates is too large.
To avoid this situation, the capacitance is taken in the range of μF.
See lessAn electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Total required capacitance, C = 2 pF Potential difference, V = 1, kV = 1000 V Capacitance of each capacitor, C₁ = lμF Each capacitor can withstand a potential difference, V₁= 400 V Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to eacRead more
Total required capacitance, C = 2 pF Potential difference, V = 1, kV = 1000 V
Capacitance of each capacitor, C₁ = lμF
Each capacitor can withstand a potential difference, V₁= 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V.
Hence, number of capacitors in each row is given as 1000/4 = 2.5
Hence, there are three capacitors in each row.
Capacitance of each row = 1/(1+1+1) = 1/3 μF
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
1/3 +1/3 + 1/3 +……..n terms = n/3
However, capacitance of the circuit is given as 2 μF.
Therefore n/3 =2 ,therefore n=6
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 x3 i.e.,18 capacitors are required for the given arrangement.
See lessTwo charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively. (a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ? (b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1. (c) How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Ans (a). Zero at both the points Charge - q is located at (0, 0, — a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. ERead more
Ans (a).
Zero at both the points
Charge – q is located at (0, 0, — a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,
V= (1/4π ε0 ) [q/(z-a)] + (1/4π ε0 ) [-q/(z+a)]
= q (z+a -z + a )/ [4π ε0 (z²-a²)]
=2qa/[4π ε0 (z²-a²)]
=p/[4π ε0 (z²-a²)]
Where,
ε0 = Permittivity of free space and p = Dipole moment of the system of two charges = 2qa
Ans (b).
Distance r is much greater than half of the distance between the two charges.
Hence, the potential (V) at a distance r is inversely proportional to square of the distance,
i.e. V ∝ 1/r²
Ans (c).
Zero
The answer does not change if the path of the test is not along the x-axis.
A test charge is moved from point (5, 0, 0) to point (-7, 0, 0) along the x-axis.
Electrostatic potential (V₁) at point (5, 0, 0) is given by,
V₁= -(q /4π ε0) 1/[√(5-0)²+ (-a)²] + (q /4π ε0) 1/[√(5-0)²+ (a)²]
= -(q /4π ε0) 1/[√25+ a²] + (q /4π ε0) 1/[√25+ a²]
= 0
Electrostatic potential ,V₂ at point (-7,0,0) is given by
V₁= -(q /4π ε0) 1/[√(7-0)²+ (-a)²] + (q /4π ε0) 1/[√(7-0)²+ (a)²]
= -(q /4π ε0) 1/[√49+ a²] + (q /4π ε0) 1/[√49+ a²]
= 0
Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (-7, 0, 0) along the x- axis.
The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.
See lessTwo charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equRead more
Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere.
Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere.
Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EA be the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,
EA/EB = (QA /4π ε0 a2 ) x (b2 4π ε0 )/QB
EA/EB = (QA / QB) x (b2 / a2 ) ———————-Eq-1
However , QA / QB =CAV/ CB V
and CA/ CB =a/b
Therefore QA / QB =a/b ———————-Eq-2
Putting the values of Eq-2 in Eq-1 ,we obtain
EA/EB = (a/b ) x (b2 / a2 )=b/a
Therefore ,the ratio of electric fields at the surface is b/a.
See lessIn a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?
The distance between electron-proton of a hydrogen atom, d = 0.53 A Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C Ans (a). Potential at infinity is zero. Potential energy of the system, p-e = Potential energy at infinity - Potential energy at distance, d = 0 - (qRead more
The distance between electron-proton of a hydrogen atom, d = 0.53 A
Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C
Ans (a).
Potential at infinity is zero.
Potential energy of the system, p-e
= Potential energy at infinity – Potential energy at distance, d
= 0 – (q₁-q₂)/4πε0 d
Where, ε0 is the permittivity of free space and
1/4πε0 =9x 10⁹ Nm²C⁻²
Therefore Potential Energy = 0- (9x 10⁹) x (1.6 x 10⁻¹⁹)² /(0.53 x 10¹⁰)
= -43.7x 10⁻¹⁹ J , (Since 1.6 x 10⁻¹⁹ J = 1 eV)
Therefore Potential energy =-43.7×10⁻¹⁹ = -(43.7 x 10⁻¹⁹)/(1.6 x 10⁻¹⁹) =-27.2 eV
Therefore, the potential energy of the system is -27.2 eV.
Ans (b).
Kinetic energy is half of the magnitude of potential energy.
Kinetic energy = 1/2 x(-27.2) = 13.6 eV
Total energy = 13.6 – 27.2 = 13.6 eV
Therefore, the minimum work required to free the electron is 13.6 eV.
Ans (c).
When zero of potential energy is taken, d = 1.06 A
Therefore Potential energy of the system = Potential energy at d₁- Potential energy at d
= (q₁q₂)/4πε0 d₁ – 27.2 eV
= (9 x 10⁹) x (1.6 x 10⁻¹⁹)² / ( 1.06 x 10⁻¹⁰) -27.2 eV
=21.73 x 10⁻¹⁹ J – 27.2 eV =
=13.58 eV – 27.2 eV= – 13.6 eV
See less