(a) 2 μF
(b) 1 μF
(c) 0.5 μF
(d) 4 μF [2024]
Ans: (a)
Given circuit is balanced Wheatstone bridge
C_{AB} = 1 + 1
= 2 μF
Q3: If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
(a) A, B and E only
(b) A, C and E only
(c) B, D and E only
(d) A, B and C only [2024]
Ans: (b)
Given V′ = V = Constant
Hence, final capacitance greater than initial capacitance,
Hence final energy is greater than initial energy
(iv) Product of charge and voltage
where p is the electric dipole moment, E is the electric field intensity, and θ is the angle between the dipole and the electric field. The electric dipole moment p can be expressed as:
where q is the charge on the dipole, and d is the dipole length.
We are given the following values:
Torque τ = 4 N·m
Electric field intensity E = 2 x 10^{5} NC^{1} Angle G = 30°
Dipole length d = 2 cm = 0.02 m
We need to find the charge q on the dipole. Let's first solve for the electric dipole moment p:
Substituting the given values:
So, the magnitude of the charge on the dipole is 2 mC.
Q2: The equivalent capacitance of the system shown in the following circuit is:
(a) 2 μF
(b) 3 μF
(c) 6 μF
(d) 9 μF
Ans: (a)
For parallel grouping
C_{1} = 3 + 3 = 6 μF
For series grouping
Q3: According to Gauss's law of electrostatics, electric flux through a closed surface depends on:
(a) the area of the surface
(b) the quantity of charges enclosed by the surface
(c) the shape of the surface
(d) the volume enclosed by the surface
Ans: (b)
only depends on the charge enclosed by the surface.
Q4: A charge Q /μC is placed at the center of a cube. The flux coming out from any one of its faces will be (in SI unit):
(a)
(b)
(c)
(d)
Ans: (d)
Total flux from cube =q/ϵ_{0}
So the flux of any one surface of the cube
Q5: If a conducting sphere of radius R is charged. Then the electric field at a distance r(r > R) from the center of the sphere would be, (V = potential on the surface of the sphere)
(a)
(b)
(c)
(d) V/r
Ans: (c)
Q6: if over a surface, then
(a) the magnitude of the electric field on the surface is constant.
(b) all the charges must necessarily be inside the surface.
(c) the electric field inside the surface is necessarily uniform.
(d) the number of flux lines entering the surface must be equal to the number of flux lines leaving it,
Ans: (d)
The number of field lines entering is equal to the number of field lines leaving.
Q7: An electric dipole is placed as shown in the figure.
The electric potential (in 10^{2} V) at point P due to the dipole is (ϵ_{o}  permittivity of free space and
(a) (5/8) qK
(b) (8/5) qK
(c) (8/3) qK
(d) (3/8) qK
Ans: (d)
Q1: Six charges +q, −q, +q, −q, +q, and −q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q_{0} to the centre of the hexagon from infinity is
(a)
(b) Zero
(c)
(d)
Ans: (b)
Q2: Twopoint charges −q and +q is placed at a distance of L, as shown in the figure.
The magnitude of electric field intensity at a distance R(R >> L) varies as:
(a) 1/R^{2}
(b) 1/R^{3}
(c) 1/R^{4}
(d) 1/R^{6}
Ans: (b)
For R » L, arrangement is an electric dipole
Q3: A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in Figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in Figure (b). The electrostatic energy stored by the system (b) is
(a) 3.25 × 10^{–6} J
(b) 2.25 × 10^{–6} J
(c) 1.5 × 10^{–6} J
(d) 4.5 × 10^{–6 }J
Ans: (b)
Q4: Two hollow conducting spheres of radii R1 and R2 (R1 >> R2) have equal charges. The potential would be
(a) More on the smaller sphere
(b) Equal in both the spheres
(c) Dependent on the material property of the sphere
(d) More on the bigger sphere
Ans: (a)
The potential of conducting hollow sphere = KQ/R
Now, Q = same
Q5: The angle between the electric lines of force and the equipotential surface is
(a) 45°
(b) 90°
(c) 180°
(d) 0°
Ans: (b)
Q1: A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A , the energy stored in the capacitor is (ε_{0} = permittivity of free space).
Ans: (c)
Q2: A parallel plate condenser has a uniform electric field E(V/m) in the space between the plates. If the distance between the plates is d(m) and the area of each plate is A(m^{2}), the energy (joule) stored in the condenser is
(a)
(b)
(c)
(d)
Ans: (c)
The capacitance of a parallel plate capacitor is
Potential difference between the plates is V = Ed ........ (ii)
The energy stored in the capacitor is
Q3: The equivalent capacitance of the combination shown in the figure is:
(a) C/2
(b) 3C/2
(c) 3C
(d) 2C
Ans: (d)
The voltage at node A matches the voltage at node B. Likewise, the voltage at node F is identical to the voltage at node E. Consequently, there is no voltage drop across the EF segment, leading to the absence of current (and therefore, no charge movement) within the circuit. Therefore, EF acts as if it were an open circuit. So the capacitor between the arms E and F gets short.
⇒ Ceq = C1 + C2
= C + C
= 2C
Q4: Two charged spherical conductors of radius R_{1} and R_{2} are connected by a wire. Then the ratio of surface charge densities of the spheres (σ_{1}/σ_{2}) is :
(a)
(b)
(c)
(d)
Ans: (d)
When two charged spherical conductors are connected by a wire, charges will flow between them until their potentials are equal. Since they are conductors, the potential at any point on each sphere will be uniform. The potential �V for a charged sphere is given by the formula:
V=kQ/R
When the two spheres are connected by a wire, the charges �1Q_{1} and �2Q_{2} will redistribute between the spheres until �1=�2V_{1}=V_{2}. If we assume �totalQ_{total} is the total charge conserved across both spheres, the equation can be written as:
Surface charge density (�σ) is the charge per unit area. For a sphere, surface area �A is 4��24πR^{2}. Thus, the surface charge density is:
Q5: Twentyseven drops of the same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
(a) 1520 V
(b) 1980 V
(c) 660 V
(d) 1320 V
Ans: (b)
⇒ V_{2} = 220 × 9
= 1980 Volt
Ans: (b)
Since, electric potential remains constant inside the metallic spherical shell and same as the surface of spherical shell.
Outside the spherical shell, V ∝ 1/r
Hence, variation of potential (V ) with distance r is given as
Q2: A short electric dipole has a dipole moment of 16 × 10^{−9} Cm. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is
(a) 200 V
(b) 400 V
(c) zero
(d) 50 V
Ans: (a)
Given, electric dipole moment, p = 16 × 10^{−9} Cm
Distance, r = 0.6 m
Angle, θ = 60° ⇒ cos60° = 1/ 2
Electric potential at a point which is at a distance r at some angle θ from electric dipole is
Hence, correct option is (a).
Q3: In a certain region of space with a volume of 0.2 m^{3}, the electric potential is found to be 5 V throughout. The magnitude of the electric field in this region is:
(a) 1 N/C
(b) 5 N/C
(c) Zero
(d) 0.5 N/C
Ans: (c)
given V = const. (5 volts)
E = 0
Q4: The capacitance of a parallel plate capacitor with air as the medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30 μF. The permittivity of the medium is :
(ε_{0} = 8.85×10^{12} C^{2} N^{}^{1} m^{2})
(a) 0.44×10^{10} C^{2} N^{1} m^{2}
(b) 5.00 C^{2} N^{1} m^{2}
(c) 0.44×10^{13} C^{2} N^{1} m^{2}
(d) 1.77×10^{12} C^{2} N^{1} m^{2}
Ans: (a)
C_{med} = KC_{air}
30μF = K6μF
K = 5
∴ ε = ε_{0}k = 8.85 × 10^{12} × 5
ε = 44.25 × 10^{12}
ε = 0.4425 × 10^{10}
ε = 0.44 × 10^{10} C^{2}N^{1}m^{2}
Q5: A 40 μF capacitor is connected to a 200 V, 50 Hz AC supply. The RMS value of the current in the circuit is, nearly:
(a) 2.5 A
(b) 25.1 A
(c) 1.7 A
(d) 2.05 A
Ans: (a)
Q1: Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?
Ans: (d)
The surface charge density of a closed surface area having charge Q is given by
Thus, the charges on sphere P and Q having same charge density as shown in the figure below is given by
when they are brought in contact with each other, the total charge will be
In connection of two charged conducting bodies, the potential will become same on both, i.e.
So, the charges on the sphere P and Q after separation will be distributed as
After separation, the new surface charge densities on P and Q will be
Q2: Two identical capacitors C_{1} and C_{2} of equal capacitance are connected as shown in the circuit.
Terminals a and b of the key k are connected to charge capacitor C_{1} using battery of emf V volt. Now, disconnecting a and b the terminals b and c are connected.
Due to this, what will be the percentage loss of energy?
(a) 75%
(b) 0%
(c) 50%
(d) 25%
Ans: (c)
When C_{1} is connected to voltage source, it is charged to a potential V and this will be stored as a potential energy in the capacitor given by
When key is disconnected from battery and b and c are connected, the charge will be transformed from the capacitor C_{1} to capacitor C_{2} , then
The loss of energy due to redistribution of charge is given by
Q1: The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is:
(a) independent of the distance between the plates.
(b) linearly proportional to the distance between the plates
(c) proportional to the square root of the distance between the plates.
(d) inversely proportional to the distance between the plates.
Ans: (a)
Solution:
F = QE
∴ The electrostatic force is independent of the distance between plates.
Q1: A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:
(a) Decreases by a factor of 2
(b) Remains the same
(c) Increases by a factor of 2
(d) Increases by a factor of 4
Ans: (a)
Solution:
Charge on capacitor
q = CV
when it is connected to another uncharged capacitor.
Q2: The diagrams below show regions of equipotentials:
A positive charge is moved from A to B in each diagram.
(a) Maximum work is required to move q in Figure (c)
(b) Minimum work is required to move q in Figure (a)
(c) Maximum work is required to move q in Figure (b)
(d) In all four cases the work done is the same
Ans: (d)
Solution:
W = qΔV
as ΔV is the same in all conditions, work will be the same.
Q1: A parallelplate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k_{1}, k_{2}, k_{3} and k_{4} as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by
Ans: (*)
Given capacitor is equivalent to capacitors K_{1} , K_{2} and K_{3} in parallel and part of K_{4} in series with them
No option is matching.
Q2: A capacitor of 2µF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :
(a) 80%
(b) 0%
(c) 20%
(d) 75%
Ans: (a)
Solution:
Consider the figure given above.
When switch S is connected to point 1, the initial energy stored in the capacitor is given as,
When the switch S is connected to point 2, energy dissipated on connection across 8μF will be,
Therefore, the percentage loss of energy
Q1: If potential (in volts) in a region is expressed as V (x, y, z) = 6xy − y + 2yz, the electric field (in N/C) at point (1, 1, 0) is [CBSE AIPMT 2015] (a)
(b)
(c)
(d)
Ans: (b)
Given, potential in a region,
V = 6xy − y + 2yz.
Electric field in a region,
At, (1, 1, 0), electric field can be expressed,
Q2: A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
(a) The charge on the capacitor is not conserved.
(b) The potential difference between the plates decreases K times
(c) The energy stored in the capacitor decreases K times
(d) the change in energy stored is Ans: (a)
Solution:
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it.
The charge on the capacitor is given by
Q = CV
The energy stored in the capacitor is
E = 1/2 CV^{2}
When a dielectric slab of dielectric constant K is inserted in it, the charge Q is conserved.
The capacitance becomes K times the original capacitance. (C' = KC)
The voltage becomes 1/K time the original voltage
V' = V/K
The change in energy stored is
Q1: In a region, the potential is represented by V(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, and z are in meters. The electric force experienced by a charge of 2 coulombs situated at point (1,1,1) is:
(a) 24N
(b)
(c)
(d) 30N
Ans: (b)
Solution:
Q2: A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the center of the sphere respectively are :
(a)
(b) Both are zero
(c)
(d)
Ans: (d)
Solution:
The electric field inside, E_{inside} = 0
Potential, V_{inside} = V_{surface}
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1. What is the formula for calculating the electric potential due to a point charge? 
2. How is the electric potential energy related to the electric potential in a system of charges? 
3. What is the difference between electric potential and electric potential energy? 
4. How does the capacitance of a capacitor affect its ability to store electric charge? 
5. How can the capacitance of a parallel plate capacitor be increased? 

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