Question 13 Marks
Obtain an expression for equivalent capacity for combination of three capacitors connected in series.
Answer
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Question 23 Marks
Two charges of magnitude \(5 nC\) and \(-2 nC\) are placed at points \((2 cm , 0,0)\) and \((20 cm 0,0)\) in a region of a region of a space where there is a no external field. Find electrostatic potential energy of the system.
Answer
View full question & answer→1. Two charges of magnitude 5nC and - 2nC are placed at 2cm and 20cm from the origin respectively.
We know that,\(U =\frac{1}{4 \pi \epsilon_{ o }} \times \frac{ q _1 q _2}{\Delta r }\)
Substituting the values,
\(\longrightarrow U=9 \times 10^9 \times \frac{5 \times 10^{-9} \times-2 \times 10^{-9}}{20 \times 10^{-2}-2 \times 10^{-2}}\)
\(\longrightarrow U=-\frac{9 \times 10 \times 10^{-9}}{18 \times 10^{-2}}\)
\(\rightarrow U =-0.5 \times 10^{-6} J\)
We know that,\(U =\frac{1}{4 \pi \epsilon_{ o }} \times \frac{ q _1 q _2}{\Delta r }\)
Substituting the values,
\(\longrightarrow U=9 \times 10^9 \times \frac{5 \times 10^{-9} \times-2 \times 10^{-9}}{20 \times 10^{-2}-2 \times 10^{-2}}\)
\(\longrightarrow U=-\frac{9 \times 10 \times 10^{-9}}{18 \times 10^{-2}}\)
\(\rightarrow U =-0.5 \times 10^{-6} J\)
Question 33 Marks
With the help of neat diagrams, explain how the non-polar dielectric material is polarised in external filed of increasing intensity. Define polarisation in dielectrics.
Answer
If a dielectric (non-polar) molecule is placed in an external electric field, a small induced dipole moment is created because the positive charge in each atom is pushed in the direction of the field and negative charge is pushed in the opposite direction as shown in the figure.
Polarization is the amount of induced surface charge per unit area or the surface density of polarization charges appearing at right angles to applied external electric field.
View full question & answer→If a dielectric (non-polar) molecule is placed in an external electric field, a small induced dipole moment is created because the positive charge in each atom is pushed in the direction of the field and negative charge is pushed in the opposite direction as shown in the figure.
Polarization is the amount of induced surface charge per unit area or the surface density of polarization charges appearing at right angles to applied external electric field.
Question 43 Marks
Six capacitors of capacities 5, 5, 5, 5, 10 and \(X \mu F\) are connected as shown in the network given diagram. Find: (a) The value of X if the network is balanced, and (b) The resultant capacitance between A and C.
Answer
View full question & answer→Given balanced network of capacitors with
$C_{AB} = C_{BC} = C_{AD} = C_{BD} = 5 µF, C_{DC} = (10 + X) µF$
The value of X = ?
Resultant capacitance between A and C $(C_{eq}) = ?$
a.Using formula, \(C_s=\frac{C_1 C_2}{C_1+C_2}\) for branch DC of the given network we get
\(C_s=\frac{10 \times X}{10+X}\)...(i)
Now using formula
\(\frac{C_{A B}}{C_{B C}}=\frac{C_{A D}}{C_{D C}}\) (For balance condition)
\(\therefore \frac{5}{5}=\frac{5}{\frac{10 \times X}{10+X}}\)
\(\therefore 1=\frac{10+X}{2 X}\)
\(\therefore 2 X=10+X\)
\(\therefore x=10 \mu F\)
The value of X is 10 µF.
b. As the bridge is balanced, no current flows through branch BD. Hence the network can be reduced as follows:
Here, 5 µF and 5 µF are in series in the branch ABC.
Using formula (i),
\(C_s=\frac{5 \times 5}{5+5}=2.5 \mu F\)
Also, 5 µF and 10 µF are in series in the branch ADC.
\(C_s^{\prime \prime}=\frac{5 \times 10}{10+10}=\frac{50}{20}=2.5 \mu F\)
Now, \(C_s^{\prime}\) and \(C_s^{\prime \prime}\) are in parallel
$\therefore CAC = CP = C_S′ + C_S′′ = 2.5 µF + 2.5 µF$
$= 5 µF$
$\therefore$ The resultant capacitance between A and C is $5 µF.$
$C_{AB} = C_{BC} = C_{AD} = C_{BD} = 5 µF, C_{DC} = (10 + X) µF$
The value of X = ?
Resultant capacitance between A and C $(C_{eq}) = ?$
a.Using formula, \(C_s=\frac{C_1 C_2}{C_1+C_2}\) for branch DC of the given network we get
\(C_s=\frac{10 \times X}{10+X}\)...(i)
Now using formula
\(\frac{C_{A B}}{C_{B C}}=\frac{C_{A D}}{C_{D C}}\) (For balance condition)
\(\therefore \frac{5}{5}=\frac{5}{\frac{10 \times X}{10+X}}\)
\(\therefore 1=\frac{10+X}{2 X}\)
\(\therefore 2 X=10+X\)
\(\therefore x=10 \mu F\)
The value of X is 10 µF.
b. As the bridge is balanced, no current flows through branch BD. Hence the network can be reduced as follows:
Here, 5 µF and 5 µF are in series in the branch ABC.
Using formula (i),
\(C_s=\frac{5 \times 5}{5+5}=2.5 \mu F\)
Also, 5 µF and 10 µF are in series in the branch ADC.
\(C_s^{\prime \prime}=\frac{5 \times 10}{10+10}=\frac{50}{20}=2.5 \mu F\)
Now, \(C_s^{\prime}\) and \(C_s^{\prime \prime}\) are in parallel
$\therefore CAC = CP = C_S′ + C_S′′ = 2.5 µF + 2.5 µF$
$= 5 µF$
$\therefore$ The resultant capacitance between A and C is $5 µF.$
Question 53 Marks
The energy density at a point in a medium of dielectric constant 6 is \(26.55 \times 10^6 J / m ^3\). Calculate electric field intensity at that point. \(\left(\varepsilon_0=8.85 \times 10^{-12}\right.\) SI units) \(\left[\right.\)
Answer
View full question & answer→$k = 6, u = 26.55 \times 10^6 J/m^3$
\(u =\frac{1}{2} \varepsilon_0 kE ^2\)
\(E=\left(\frac{2 u}{\varepsilon_0 k}\right)^{1 / 2}\)
\(=\frac{2 \times 26.55 \times 10^6}{8.85 \times 10^{-12} \times 6}=\left(10^{18}\right)^{1 / 2}\)
\(E =10^9 N / C\)
Electric field intensity at the point in the medium is 109 N/C.
\(u =\frac{1}{2} \varepsilon_0 kE ^2\)
\(E=\left(\frac{2 u}{\varepsilon_0 k}\right)^{1 / 2}\)
\(=\frac{2 \times 26.55 \times 10^6}{8.85 \times 10^{-12} \times 6}=\left(10^{18}\right)^{1 / 2}\)
\(E =10^9 N / C\)
Electric field intensity at the point in the medium is 109 N/C.
Question 63 Marks
A parallel plate capacitor filled with air has an area of \(6 cm ^2\) and plate separation of \(3 mm\). Calculate its capacitance.
Answer
View full question & answer→Given,
$A = 6 cm^2 = 6 \times 10^{-4} m^2$
$d = 3 mm = 3 \times 10^{-3} m$
$k = 1$
$ε_0 = 8.85 \times 10^{-12} c^2/Nm^2$
$C = ?$
Capacitance of parallel plate capacitor
\(C=\frac{A \varepsilon_0 k}{d}\)
\(=\frac{6 \times 10^{-4} \times 8.85 \times 10^{-12} \times 1}{3 \times 10^{-3}}\)
\(=17.7 \times 10^{-13}\)
\(=1.77 \times 10^{-12} F\)
∴ Capacitance of a parallel plate capacitor is \(=1.77 \times 10^{-12} F\)
$A = 6 cm^2 = 6 \times 10^{-4} m^2$
$d = 3 mm = 3 \times 10^{-3} m$
$k = 1$
$ε_0 = 8.85 \times 10^{-12} c^2/Nm^2$
$C = ?$
Capacitance of parallel plate capacitor
\(C=\frac{A \varepsilon_0 k}{d}\)
\(=\frac{6 \times 10^{-4} \times 8.85 \times 10^{-12} \times 1}{3 \times 10^{-3}}\)
\(=17.7 \times 10^{-13}\)
\(=1.77 \times 10^{-12} F\)
∴ Capacitance of a parallel plate capacitor is \(=1.77 \times 10^{-12} F\)
Question 73 Marks
Explain the concept of a parallel plate capacitor. State its any 'two' applications.
Answer
View full question & answer→parallel plate capacitor is a type of capacitor that consists of two parallel conductive plates separated by a small gap or distance. This gap is typically filled with a dielectric material, which helps to increase the capacitor's ability to store electrical charge. The amount of electrical charge that a parallel plate capacitor can store is determined by the surface area of the plates, the distance between the plates, and the type of dielectric material used.
One application of a parallel plate capacitor is in power supplies, where it is used to store electrical energy and then release it as needed to power electrical devices. Another application is in filtering circuits, where the capacitor is used to remove unwanted AC components from an electrical signal.
One application of a parallel plate capacitor is in power supplies, where it is used to store electrical energy and then release it as needed to power electrical devices. Another application is in filtering circuits, where the capacitor is used to remove unwanted AC components from an electrical signal.
Question 83 Marks
A cube of marble having each side $1$ cm is kept in an electric field of intensity $300 V / m$. Determine the energy contained in the cube of dielectric constant $8$ . (Given: $\varepsilon_o=8.85 \times 10^{-12} C ^2 / Nm ^2$ )
Answer
View full question & answer→Given: \(l=1 cm =10^{-2} m , E=300 Vm ^{-11}, k=8\)
To find: Energy contained in the cube (U)
formula: \(u=\frac{U}{V}\)
volume of marble
\(V=\left(i^3\right)=\left(10^{-2}\right) 3=10^{-6} m ^3\)
Energy Density, \(u=\frac{1}{2} \in_0 k E^2\)
\(=\frac{1}{2} \times 8.85 \times 10^{-12} \times 8 \times(300)^2\)
\(=3.185 \times 10^{-6} \frac{ J }{ m ^3}\)
From formula \(U=u \times V=3.185 \times 10^{-6} \times 10^{-6}\)
\(U=3.185 \times 10^{-12} J\)
To find: Energy contained in the cube (U)
formula: \(u=\frac{U}{V}\)
volume of marble
\(V=\left(i^3\right)=\left(10^{-2}\right) 3=10^{-6} m ^3\)
Energy Density, \(u=\frac{1}{2} \in_0 k E^2\)
\(=\frac{1}{2} \times 8.85 \times 10^{-12} \times 8 \times(300)^2\)
\(=3.185 \times 10^{-6} \frac{ J }{ m ^3}\)
From formula \(U=u \times V=3.185 \times 10^{-6} \times 10^{-6}\)
\(U=3.185 \times 10^{-12} J\)
Question 93 Marks
A small particle carrying a negative charge of \(1.6 \times 10^{-19} C\) is suspended in equilibrium between two horizontal metal plates \(10 cm\) apart, having a potential difference of 4000 volts across them. Find the mass of the particle
Answer
View full question & answer→\(6.53 \times 10^{-16} kg\)
Question 103 Marks
Obtain an expression for energy of a charged capacitor and express it in different forms.
Answer
View full question & answer→coming soon…
Question 113 Marks
An electric dipole consists of two opposite charges each of magnitude \(1 \mu C\), separated by \(2 cm\).
The dipole is placed in an external field of \(10^5 N / C\). Calculate:
(a) maximum torque experienced by the dipole and
(b) work done by the external field to turn the dipole through \(180^{\circ}\).
The dipole is placed in an external field of \(10^5 N / C\). Calculate:
(a) maximum torque experienced by the dipole and
(b) work done by the external field to turn the dipole through \(180^{\circ}\).
Answer
View full question & answer→Given:
$q = 1 \mu C = 10^{-6} C, E = 10^5 N/C, 2l = 2 cm = 2 \times 10^{-2}m$
To Find:
$τ_{max} = ?, W = ?$
We know that, $p = q \times 2l = 10^{-6} \times 2 \times 10^{-2} = 2 \times 10^{-8} m$
$τ_{max} = pE = 2 \times 10^{-8} \times 10^5= 2 \times 10^{-3} Nm$
Work done by the external field to turn the dipole through $180^\circ$ is given by,
$W = pE(cos\theta _1 - cos\theta _2)$
$\therefore W = 2 \times 10^{-8} \times 10^5\times (1 + 1)$
$\therefore W = 4 \times 10^{-3} J$
$q = 1 \mu C = 10^{-6} C, E = 10^5 N/C, 2l = 2 cm = 2 \times 10^{-2}m$
To Find:
$τ_{max} = ?, W = ?$
We know that, $p = q \times 2l = 10^{-6} \times 2 \times 10^{-2} = 2 \times 10^{-8} m$
$τ_{max} = pE = 2 \times 10^{-8} \times 10^5= 2 \times 10^{-3} Nm$
Work done by the external field to turn the dipole through $180^\circ$ is given by,
$W = pE(cos\theta _1 - cos\theta _2)$
$\therefore W = 2 \times 10^{-8} \times 10^5\times (1 + 1)$
$\therefore W = 4 \times 10^{-3} J$
Question 123 Marks
A conductor of any shape, having area \(40 cm ^2\) placed in air is uniformly charged with a charge \(0.2 \mu C\). Determine the electric intensity at a point just outside its surface. Also, find the mechanical force per unit area of the charged conductor. $[/ \varepsilon_0=8.85 \times 10^{-12}\ S.I. Units]$
Answer
View full question & answer→Given: $Q=0.2 \mu C=0.2*10^{-6}C$
$A=40cm^2=40*10^{-4}m^2$
$ε_0=8.85*10^{-12}SI units$
The electric field intensity just outside the surface of a charged conductor of any shape is
\(E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{A \varepsilon_0}\)
\(\therefore E=\frac{0.2 \cdot 10^{-6}}{40 \cdot 10^{-4} \cdot 8.85 \cdot 10^{-12}}\)
\(\therefore E=5.65 \cdot 10^6 N / C\)
Now, the mechanical force per unit area of a conductor is
\(f=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \cdot 8.85 \cdot 10^{-12} \cdot\left(5.65 \cdot 10^6\right)^2\)
$\therefore f=141.25N/m^2$
$A=40cm^2=40*10^{-4}m^2$
$ε_0=8.85*10^{-12}SI units$
The electric field intensity just outside the surface of a charged conductor of any shape is
\(E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{A \varepsilon_0}\)
\(\therefore E=\frac{0.2 \cdot 10^{-6}}{40 \cdot 10^{-4} \cdot 8.85 \cdot 10^{-12}}\)
\(\therefore E=5.65 \cdot 10^6 N / C\)
Now, the mechanical force per unit area of a conductor is
\(f=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \cdot 8.85 \cdot 10^{-12} \cdot\left(5.65 \cdot 10^6\right)^2\)
$\therefore f=141.25N/m^2$
Question 133 Marks
A network of four capacitors of \(6 \mu F\) each is connected to a \(240 V\) supply. Determine the charge on each capacitor.
Answer
View full question & answer→$C_1,C_2$ and $C_3$ are in series
\(\therefore \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)
\(=\frac{1}{6 \times 10^{-6}}+\frac{1}{6 \times 10^{-6}}+\frac{1}{6 \times 10^{-6}}\)
\(\frac{1}{C_s}=\frac{3}{6 \times 10^{-6}}\)
$\therefore Cs = 2x10^{-6} F$
$C_s and C_4$ are parallel
$\therefore$ equivalent resistance
$C = C_p = C_s + C_4$
$= 2x10^{-6} + 6x10^{-6}$
$C = 8x10^{-6}$
$\therefore v = v1+ v2+ v3$
$240 = 3v_1 {v_1=v_2=v_3}$
$\therefore v_1 80 volt$
Also charge on $C_1,C_2 ,C_3$ are same.
$Q_1 = Q_2 = Q_3 = Q$
$Q = C_1v_1$
$= 6x10^{-6} x 80$
$Q = 480x10^{-6}C and Q_4 = 6x10^{-6}x 240= 144 \times 10^{-5}$
$Q_4 = 1.44 \times 10^{-3} C$
\(\therefore \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)
\(=\frac{1}{6 \times 10^{-6}}+\frac{1}{6 \times 10^{-6}}+\frac{1}{6 \times 10^{-6}}\)
\(\frac{1}{C_s}=\frac{3}{6 \times 10^{-6}}\)
$\therefore Cs = 2x10^{-6} F$
$C_s and C_4$ are parallel
$\therefore$ equivalent resistance
$C = C_p = C_s + C_4$
$= 2x10^{-6} + 6x10^{-6}$
$C = 8x10^{-6}$
$\therefore v = v1+ v2+ v3$
$240 = 3v_1 {v_1=v_2=v_3}$
$\therefore v_1 80 volt$
Also charge on $C_1,C_2 ,C_3$ are same.
$Q_1 = Q_2 = Q_3 = Q$
$Q = C_1v_1$
$= 6x10^{-6} x 80$
$Q = 480x10^{-6}C and Q_4 = 6x10^{-6}x 240= 144 \times 10^{-5}$
$Q_4 = 1.44 \times 10^{-3} C$