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.
A 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.
See less