Ans (a). Dielectric constant of the mica sheet, k= 6 If voltage supply remained connected, voltage between two plates will be constant. Supply voltage, V = 100 V Initial capacitance, C= 1.771 x 10-11 F New capacitance, C₁ = kC = 6 x 1.771 x 10-11 F = 106 pF New charge, q₁ = C₁V = 106 x 100 pC = 1.06Read more
Ans (a).
Dielectric constant of the mica sheet, k= 6
If voltage supply remained connected, voltage between two plates will be constant.
Supply voltage, V = 100 V
Initial capacitance, C= 1.771 x 10-11 F
New capacitance, C₁ = kC = 6 x 1.771 x 10-11 F = 106 pF
New charge, q₁ = C₁V = 106 x 100 pC = 1.06 x 10–8 C
Potential across the plates remains 100 V.
Ans (b).
Dielectric constant, k= 6
Initial capacitance, C = 1.771 x 10-11 F
New capacitance, C₁= kC = 6 x 1.771 x 10-11 F = 106 pF
If supply voltage is removed, then there will be constant amount of charge in the plates.
Area of each plate of the parallel plate capacitor, A = 6 x 10-3 m2 Distance between the plates, d = 3 mm = 3 x 10-3 m Supply voltage, V = 100 V Capacitance C of a parallel plate capacitor is given by, C = ε0 A/d Where, ε0 = Permittivity of free space = 8.854 x 10-12 N-1m-2 C-2 Therefore C = (8.85Read more
Area of each plate of the parallel plate capacitor, A = 6 x 10-3 m2
Distance between the plates, d = 3 mm = 3 x 10–3 m
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by, C = ε0 A/d
Where,
ε0 = Permittivity of free space = 8.854 x 10-12 N-1m-2 C-2
Therefore C = (8.854 x 10-12 x 6 x 10–3)/3 x 10–3
= 17.71 x 10-12 F = 17.71 pF
So, charge on each plate of the capacitor
q = VC = 100 x 17.71 x 10-12 C = 1.771 x 10–9 C
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 x 10–9 C.
Ans (a). Capacitances of the given capacitors: C₁ = 2 pF, C2 = 3 pF and C3 = 4 pF For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum, Therefore Ceq = C₁+C2 + C3 = 2 + 3+ 4 = 9pF Therefore, total capacitance of the combination is 9 pF. Ans (b). SuppRead more
Ans (a).
Capacitances of the given capacitors: C₁ = 2 pF, C2 = 3 pF and C3 = 4 pF
For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,
Therefore Ceq = C₁+C2 + C3 = 2 + 3+ 4 = 9pF
Therefore, total capacitance of the combination is 9 pF.
Ans (b).
Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation,
q=VC… (i)
For C = 2 pF, charge = VC = 100 x 2 = 200 pC = 2 x lO⁻10 C
For C = 3 pF, charge = VC = 100 x 3 = 300 pC = 3 x lO⁻10 C
For C = 4 pF, charge = VC = 100 x 4 = 400 pC = 4 x 10⁻10 C
(a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (Ceq) of the combination of the capacitors is given by the relation, 1/Ceq =1/C + 1/C + 1/C = 3/C = 3/9 = 1/3 => 1/Ceq = 1/3 => Ceq = 3 pF Therefore, total capacitance of the combination is 3 pF. (b) Supply voltaRead more
(a) Capacitance of each of the three capacitors, C = 9 pF
Equivalent capacitance (Ceq) of the combination of the capacitors is given by the relation,
1/Ceq =1/C + 1/C + 1/C = 3/C = 3/9 = 1/3
=> 1/Ceq = 1/3 => Ceq = 3 pF
Therefore, total capacitance of the combination is 3 pF.
(b) Supply voltage, V = 100 V
Potential difference (V₁) across each capacitor is equal to one-third of the supply voltage.
Therefore V1=V/3=120/3=40V
Therefore, the potential difference across each capacitor is 40 V.
Capacitance between the parallel plates of the capacitor, C = 8 pF Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1 Capacitance, C, is given by the formula, C = (kε0 A)/d = ε0 A/d -----------------------Eq-1 Where, A = Area of each pRead more
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air.
Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
C = (kε0 A)/d = ε0 A/d ———————–Eq-1
Where,
A = Area of each plate
ε0 = Permittivity of free space
If distance between the plates is reduced to half, then new distance, di = d/2 Dielectric constant of the substance filled in between the plates, k₁= 6 Hence, capacitance of the capacitor becomes
Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.
Ans (a). Dielectric constant of the mica sheet, k= 6 If voltage supply remained connected, voltage between two plates will be constant. Supply voltage, V = 100 V Initial capacitance, C= 1.771 x 10-11 F New capacitance, C₁ = kC = 6 x 1.771 x 10-11 F = 106 pF New charge, q₁ = C₁V = 106 x 100 pC = 1.06Read more
Ans (a).
Dielectric constant of the mica sheet, k= 6
If voltage supply remained connected, voltage between two plates will be constant.
Supply voltage, V = 100 V
Initial capacitance, C= 1.771 x 10-11 F
New capacitance, C₁ = kC = 6 x 1.771 x 10-11 F = 106 pF
New charge, q₁ = C₁V = 106 x 100 pC = 1.06 x 10–8 C
Potential across the plates remains 100 V.
Ans (b).
Dielectric constant, k= 6
Initial capacitance, C = 1.771 x 10-11 F
New capacitance, C₁= kC = 6 x 1.771 x 10-11 F = 106 pF
If supply voltage is removed, then there will be constant amount of charge in the plates.
Charge = 1.771 x 10–9 C
Potential across the plates is given by,
V₁q/C₁= (1.771 x 10–9)/(106 x 10⁻¹²)
=16.7 V
See lessIn a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻3 m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Area of each plate of the parallel plate capacitor, A = 6 x 10-3 m2 Distance between the plates, d = 3 mm = 3 x 10-3 m Supply voltage, V = 100 V Capacitance C of a parallel plate capacitor is given by, C = ε0 A/d Where, ε0 = Permittivity of free space = 8.854 x 10-12 N-1m-2 C-2 Therefore C = (8.85Read more
Area of each plate of the parallel plate capacitor, A = 6 x 10-3 m2
Distance between the plates, d = 3 mm = 3 x 10–3 m
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by, C = ε0 A/d
Where,
ε0 = Permittivity of free space = 8.854 x 10-12 N-1m-2 C-2
Therefore C = (8.854 x 10-12 x 6 x 10–3)/3 x 10–3
= 17.71 x 10-12 F = 17.71 pF
So, charge on each plate of the capacitor
q = VC = 100 x 17.71 x 10-12 C = 1.771 x 10–9 C
See lessTherefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 x 10–9 C.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Ans (a). Capacitances of the given capacitors: C₁ = 2 pF, C2 = 3 pF and C3 = 4 pF For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum, Therefore Ceq = C₁+C2 + C3 = 2 + 3+ 4 = 9pF Therefore, total capacitance of the combination is 9 pF. Ans (b). SuppRead more
Ans (a).
Capacitances of the given capacitors: C₁ = 2 pF, C2 = 3 pF and C3 = 4 pF
For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,
Therefore Ceq = C₁+C2 + C3 = 2 + 3+ 4 = 9pF
Therefore, total capacitance of the combination is 9 pF.
Ans (b).
Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation,
q=VC… (i)
For C = 2 pF, charge = VC = 100 x 2 = 200 pC = 2 x lO⁻10 C
For C = 3 pF, charge = VC = 100 x 3 = 300 pC = 3 x lO⁻10 C
For C = 4 pF, charge = VC = 100 x 4 = 400 pC = 4 x 10⁻10 C
See lessThree capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
(a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (Ceq) of the combination of the capacitors is given by the relation, 1/Ceq =1/C + 1/C + 1/C = 3/C = 3/9 = 1/3 => 1/Ceq = 1/3 => Ceq = 3 pF Therefore, total capacitance of the combination is 3 pF. (b) Supply voltaRead more
(a) Capacitance of each of the three capacitors, C = 9 pF
Equivalent capacitance (Ceq) of the combination of the capacitors is given by the relation,
1/Ceq =1/C + 1/C + 1/C = 3/C = 3/9 = 1/3
=> 1/Ceq = 1/3 => Ceq = 3 pF
Therefore, total capacitance of the combination is 3 pF.
(b) Supply voltage, V = 100 V
Potential difference (V₁) across each capacitor is equal to one-third of the supply voltage.
Therefore V1=V/3=120/3=40V
Therefore, the potential difference across each capacitor is 40 V.
See lessA parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10⁻12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Capacitance between the parallel plates of the capacitor, C = 8 pF Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1 Capacitance, C, is given by the formula, C = (kε0 A)/d = ε0 A/d -----------------------Eq-1 Where, A = Area of each pRead more
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air.
Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
C = (kε0 A)/d = ε0 A/d ———————–Eq-1
Where,
A = Area of each plate
ε0 = Permittivity of free space
If distance between the plates is reduced to half, then new distance, di = d/2 Dielectric constant of the substance filled in between the plates, k₁= 6 Hence, capacitance of the capacitor becomes
C₁ = k₁ ε0 A/d₁ = (6 ε0 A)/(d/2) = (12 ε0 A)/d………………Eq-2
Taking ratios of equations (1) and (2), we obtain C₁ = 2 x 6 C = 12 C = 12 x 8 pF = 96 pF Therefore, the capacitance between the plates is 96 pF.
See less