A 4μF conductor is charged to 400 volts andthen its plates are joined through a resistance of1 kΩ. The heat produced in the resistance is [1989]
The energy stored in the capacitor
This energy will be converted into heat in the resistor
A hollow metal sphere of radius 10 cm is chargedsuch that the potential on its surface is 80 V. Thepotential at the centre of the sphere is [1994]
Potential at the centre of the sphere
= potential on the surface = 80 V.
The four capacitors, each of 25μ F are connected as shown in fig. The dc voltmeter reads 200 V. The charge on each plate of capacitor is [1994]
Charge on each plate of each capacitor
Q = CV = 25 x10^{6} x 200
= 5 x10^{3}C
If the potential of a capacitor having capacity 6μF is increased from 10 V to 20 V, then increase in its energy will be [1995]
Capacitance of capacitor (C) = 6μF = 6 ×10^{–6} F; Initial potential (V_{1}) = 10 V and final potential (V_{2}) = 20 V.
The increase in energy (ΔU)
= (3x10^{6} )x300 = 9x10^{4} J .
A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes
When oil is placed between space of plates
When oil is removed ... (2)
On comparing both equations, we get
C ' = C/2
What is the effective capacitance between points X and Y?
Equivalent circuit
Here,
Hence, no charge will flow through 20μF
C_{1} and C_{2} are in series, also C_{3} and C_{4} are in
series.
Hence, C' = 3 μF, C'' = 3 μF
C' and C'' are in parallel.
Hence net capacitance = C' + C'' = 3 + 3
= 6 μF
A capacitor is charged to store an energy U.The charging battery is disconnected. Anidentical capacitor is now connected to the firstcapacitor in parallel. The energy in each of thecapacitor is [2000]
Energy stored in a capacitor is
As the battery is disconnected, total charge Q is shared equally by two capacitors.
So energy of each capacitor
In a parallel plate capacitor, the distance between the plates is d and potential difference across the plates is V. Energy stored per unit volume between the plates of capacitor is
Energy stored per unit volume
(∵ E = V/d)
The capacity of a parallel plate condenser is 10μF when the distance between its plates is 8cm. If the distance between the plates is reducedto 4 cm then the capacity of this parallel platecondenser will be [2001]
C = 10 μF d = 8 cm
C ' = ? d ' = 4 cm
If d is halved then C will be doubled.
Hence, C ' = 2C = 2 × 10μF = 20μF
Energy stored in a capacitor is
Energy stored in capacitor
A capacitor C_{1} is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor C_{2}. The potential difference across the combination is
Charge Q = C_{1}V
Total capacity of combination (parallel)
C = C_{1}+ C_{2}
A solid spherical conductor is given a charge. The electrostatic potential of the conductor is [2002]
Electric potential is constant
within or on the surface of conductor.
Each corner of a cube of side l has a negative charge, –q. The electrostatic potential energy of a charge q at the centre of the cube is [2002]
Length of body diagonal =
∴ Distance of centre of cube from each
corner
P.E at centre
= 8 × Potential Energy due to A
Three capacitors each of capacity 4μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by [2003]
For series,
For parallel, C_{eq} = C '+C_{3} = 2 + 4 = 6μF
An electric dipole has the magnitude of its chargeas q and its dipole moment is p. It is placed inuniform electric field E. If its dipole moment is along the direction of the field, the force on itand its potential energy are respectively [2004]
When the dipole is in the direction of field then net force is qE + (–qE) = 0
and its potential energy is minimum
= – P.E. = –qaE
A bullet of mass 2 g is having a charge of 2μC.Through what potential difference must it beaccelerated, starting from rest, to acquire a speedof 10 m/s? [2004]
∴ V = 50 kV
As per the diagram, a point charge +q is placed at the origin O. Work done in taking another point charge –Q from the point A [coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight path AB is:
We know that potential energy of two charge system is given by
According to question,
ΔU= UB–UA = 0
We know that for conservative force,
W = –ΔU = 0
Two charges q_{1} and q_{2} are placed 30 cm apart, as shown in the figure. A third charge q_{3} is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is where k is
We know that potential energy of discrete system of charges is given by
According to question,
A network of four capacitors of capacity equal to C_{1} = C, C_{2} = 2C, C_{3} = 3C and C_{4} = 4C are conducted to a battery as shown in the figure. The ratio of the charges on C_{2} and C_{4} is: [2005]
Equivalent capacitance for three capacitors (C_{1}, C_{2} & C_{3}) in series is given by
⇒ Charge on capacitors (C_{1}, C_{2} & C_{3}) in series
Charge on capacitor C_{4} = C_{4}V = 4C V
A parallel plate air capacitor is charged to apotential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates [2006]
If we increase the distance between the plates its capacity decreases resulting in higher potential as we know Q = CV. Since Q is constant (battery has been disconnected), on decreasing C, V will increase
Charges +q and –q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is [2007]
Potential at C = V_{C} = 0
Potential at D = V_{D}
Potential difference
Two condensers, one of capacity C and other of capacity C/2 are connected to a Vvolt battery, as shown. [2007]
The work done in charging fully both the condensers is
Work done = Change in energy
The electric potential at a point in free space due to a charge Q coulomb is Q × 10^{11} volts. The electric field at that point is [2008]
Given that, V = Q × 10^{11} volts
Electric potential at point is given by
= 4π∈_{0} Q x 10^{22 }volt m^{1}
The energy required to charge a parallel plate condenser of plate separation d and plate area of crosssection A such that the uniform electric field between the plates is E, is [2008]
The energy required to charge a parallel
plate condenser is given by
As and V = E.d
Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ , – σ and σ respectively. If V_{A}, V_{B} and V_{C} denotes the potentials of the three shells, then for c = a + b, we have [2009]
c = a + b.
V_{A} = V_{C} ≠ V_{B}
Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be [2009]
In series combination of capacitors
Thus, the capacitance and breakdown
voltage of the combination will be C/3 and 3V
The electric potential at a point (x, y, z) is given by V = – x2y – xz3 + 4. The electric field at that point is [2009]
The electric field at a point is equal to
negative of potential gradient at that point.
A series combination of n1 capacitors, each of value C_{1}, is charged by a source of potential difference 4 V. When another parallel combination of n_{2} capacitors, each of value C_{2}, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C_{2} , in terms of C_{1}, is then [2010]
In series,
∴ Energy stored
In parallel, C_{eff} = n_{2} C_{2}
∴ Energy stired ,
A condenser of capacity C is charged to a potential difference of V_{1}. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V_{2} is
q = CV_{1} cos ωt
Also, and V = V_{1} cosωt
At t = t1 , V = V_{2} and i = ωCV_{1} sinωt_{1}
(–ve sign gives direction)
Two parallel metal plates having charges + Qand –Q face each other ata certain distance between them. If the plaves are now dipped inkerosene oil tank, the electric field between the plates will
Electric field
ε of kerosine oil is more than that of air.
As ε increases, E decreases
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 area of each plate is A(m^{2}) the energy (joules) stored in the condenser is [2011]
Four electric charges +q, +q, –q and – q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, midway between the two charges +q and +q, is [2011]
Distance of point A from the two +q
charges = L.
Distance of point A from the two –q
charges =
The potential energy of particle in a force field is , where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is : [2012]
for equilibrium
for stable equilibrium
should be positive for the value of r.
here, is +ve value for r = 2A/B so
Four point charges –Q, –q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is : [2012]
Let the side length of square be 'a' then
potential at centre O is
(given)
= – Q – q + 2q + 2Q = 0 = Q + q = 0
= Q = – q
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 area of each plate is A, the energy stored in the capacitor is : [2012M]
The energy stored by a capacitor
...(1)
V is the p.d. between two plates of the
capacitor.
The capacitance of the parallel plate
capacitor
V = E.d.
Substituting the value of C in equation (i)
A, B and C are three points in a uniform electric field. The electric potential is [NEET 2013]
Potential at B, V_{B} is maximum
V_{B} > V_{C} > V_{A}
As in the direction of electric field
potential decreases.
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 







