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
Charge density of the long charged cylinder of length L and radius r is λ .Another cylinder of same length surrounds the previous cylinder .The radius of the cylinder is R. Let E be the electric field produced in the space between the two cylinders. Electric flux through the Gaussian surface is givRead more
Charge density of the long charged cylinder of length L and radius r is λ .Another cylinder of same length surrounds the previous cylinder .The radius of the cylinder is R.
Let E be the electric field produced in the space between the two cylinders.
Electric flux through the Gaussian surface is given by Gauss’s theorem as ,
φ=E (2πd ) L
Where ,d = D= distance of a point from the common axis of the cylinder .
Let q be the total charge on the cylinder.
It can be written as
φ=E (2πdL ) = q/ε0
Where ,q = Charge on the inner sphere of the outer cylinder
ε0 = Permittivity of free space
E (2πdL) =λ L/ε0
E= λ/2πε0d
Therefore ,the electric field in the space between the two cylinders is λ/2πε0d
Electric field on one side of a charged body is E₁ and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by, Ē1= (σ /2 ε0 )ñ --------------------Eq-1 Where , ñ = Unit veRead more
Electric field on one side of a charged body is E₁ and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
Ē1= (σ /2 ε0 )ñ ——————–Eq-1
Where ,
ñ = Unit vector normal to the surface at a point
σ= Surface charge density at that point
Electric field due to the other surface of the charge body ,
Ē₂= – (σ /2 ε0 )ñ ——————–Eq-2
Electric field at any point due to the two surfaces,
Ē₂ – Ē1= (σ /2 ε0) ñ + (σ /2 ε0 )ñ = (σ /ε0 )ñ
(Ē₂ – Ē1) ñ = σ /ε0 ———————-Eq-3
Since inside a closed conductor ,Ē1=0.
Therefore Ē=Ē₂= (σ /2 ε0)ñ
,the electric field just outside the conductor is σ /ε0 ñ
Ans (b).
When a charged particle is moved from one point to the other on a closed loop ,the work done by the electrostatic field is zero. Hence ,the tangential component of electrostatic field is continuous from one side of a charged surface to the other.
Ans (a). Charge placed at the centre of a shell is +q. Hence, a charge of magnitude -q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is -q. Surface charge density at the inner surface of the shell is given by the relation, σ₁ =Total CharRead more
Ans (a).
Charge placed at the centre of a shell is +q. Hence, a charge of magnitude -q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is -q.
Surface charge density at the inner surface of the shell is given by the relation,
σ₁ =Total Charge/Inner Surface Area = -q/4π r₁²——————–Eq-1
A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,
σ₂ =Total Charge/Inner Surface Area = (Q+q)/4π r2² ——————Eq-2
(b) Yes
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.
Capacitance of the capacitor, C = 600 pF Potential difference, V = 200 V Electrostatic energy stored in the capacitor is given by, E₁=1/2 x CV2 = 1/2 x (600 x 10⁻12) x (200)2 J = 1.2 x 10⁻5 J If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connectedRead more
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
E₁=1/2 x CV2
= 1/2 x (600 x 10⁻12) x (200)2 J = 1.2 x 10⁻5 J
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (Ceq) of the combination is given by,
1/Ceq=1/C + 1/C
=> 1/Ceq =1/600 + 1/600=2/600 =1/300
=> Ceq =300 pF
New electrostatic energy can be calculated as
E2=1/2 Ceq V2
=1/2 x 300 x (200)2 J = 0.6 x 10⁻5 J
Loss in electrostatic energy = E₁ – E2
= 1.2 x 10⁻5 – 0.6 x 10⁻5 J = 0.6 x 10⁻5J = 6x 10⁻6J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J .
Capacitor of the capacitance, C = 12 pF = 12 x 10-12 F Potential difference, V = 50 V Electrostatic energy stored in the capacitor is given by the relation, E=1/2 x CV2=1/2 x 12 x 10-12 x (50)2 J =1.5 x 10⁻⁸J Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10-8 J.
Capacitor of the capacitance, C = 12 pF = 12 x 10-12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
E=1/2 x CV2=1/2 x 12 x 10-12 x (50)2 J =1.5 x 10⁻⁸J
Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10-8 J.
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 lessA long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Charge density of the long charged cylinder of length L and radius r is λ .Another cylinder of same length surrounds the previous cylinder .The radius of the cylinder is R. Let E be the electric field produced in the space between the two cylinders. Electric flux through the Gaussian surface is givRead more
Charge density of the long charged cylinder of length L and radius r is λ .Another cylinder of same length surrounds the previous cylinder .The radius of the cylinder is R.
Let E be the electric field produced in the space between the two cylinders.
Electric flux through the Gaussian surface is given by Gauss’s theorem as ,
φ=E (2πd ) L
Where ,d = D= distance of a point from the common axis of the cylinder .
Let q be the total charge on the cylinder.
It can be written as
φ=E (2πdL ) = q/ε0
Where ,q = Charge on the inner sphere of the outer cylinder
ε0 = Permittivity of free space
E (2πdL) =λ L/ε0
E= λ/2πε0d
Therefore ,the electric field in the space between the two cylinders is λ/2πε0d
See less(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by ( E₂ -E₁ ). ñ =σ/ ε0 where ñ is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of ñ is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ nˆ /ε0. (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
Electric field on one side of a charged body is E₁ and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by, Ē1= (σ /2 ε0 )ñ --------------------Eq-1 Where , ñ = Unit veRead more
Electric field on one side of a charged body is E₁ and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
Ē1= (σ /2 ε0 )ñ ——————–Eq-1
Where ,
ñ = Unit vector normal to the surface at a point
σ= Surface charge density at that point
Electric field due to the other surface of the charge body ,
Ē₂= – (σ /2 ε0 )ñ ——————–Eq-2
Electric field at any point due to the two surfaces,
Ē₂ – Ē1= (σ /2 ε0) ñ + (σ /2 ε0 )ñ = (σ /ε0 )ñ
(Ē₂ – Ē1) ñ = σ /ε0 ———————-Eq-3
Since inside a closed conductor ,Ē1=0.
Therefore Ē=Ē₂= (σ /2 ε0)ñ
,the electric field just outside the conductor is σ /ε0 ñ
Ans (b).
When a charged particle is moved from one point to the other on a closed loop ,the work done by the electrostatic field is zero. Hence ,the tangential component of electrostatic field is continuous from one side of a charged surface to the other.
See lessA spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Ans (a). Charge placed at the centre of a shell is +q. Hence, a charge of magnitude -q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is -q. Surface charge density at the inner surface of the shell is given by the relation, σ₁ =Total CharRead more
Ans (a).
Charge placed at the centre of a shell is +q. Hence, a charge of magnitude -q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is -q.
Surface charge density at the inner surface of the shell is given by the relation,
σ₁ =Total Charge/Inner Surface Area = -q/4π r₁²——————–Eq-1
A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,
σ₂ =Total Charge/Inner Surface Area = (Q+q)/4π r2² ——————Eq-2
(b) Yes
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.
See lessA 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Capacitance of the capacitor, C = 600 pF Potential difference, V = 200 V Electrostatic energy stored in the capacitor is given by, E₁=1/2 x CV2 = 1/2 x (600 x 10⁻12) x (200)2 J = 1.2 x 10⁻5 J If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connectedRead more
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
E₁=1/2 x CV2
= 1/2 x (600 x 10⁻12) x (200)2 J = 1.2 x 10⁻5 J
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (Ceq) of the combination is given by,
1/Ceq=1/C + 1/C
=> 1/Ceq =1/600 + 1/600=2/600 =1/300
=> Ceq =300 pF
New electrostatic energy can be calculated as
E2=1/2 Ceq V2
=1/2 x 300 x (200)2 J = 0.6 x 10⁻5 J
Loss in electrostatic energy = E₁ – E2
= 1.2 x 10⁻5 – 0.6 x 10⁻5 J = 0.6 x 10⁻5J = 6x 10⁻6J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J .
See lessA 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Capacitor of the capacitance, C = 12 pF = 12 x 10-12 F Potential difference, V = 50 V Electrostatic energy stored in the capacitor is given by the relation, E=1/2 x CV2=1/2 x 12 x 10-12 x (50)2 J =1.5 x 10⁻⁸J Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10-8 J.
Capacitor of the capacitance, C = 12 pF = 12 x 10-12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
E=1/2 x CV2=1/2 x 12 x 10-12 x (50)2 J =1.5 x 10⁻⁸J
Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10-8 J.
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