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All questions of Electrostatics Potential and Capacitance for NEET Exam
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]
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
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]
a)
b)
c)
d)
|
Gaurav Kumar answered |
Potential at C = VC = 0
Potential at D = VD
Potential at D = VD
Potential difference
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]- a)5 μF
- b)10 μF
- c)20 μF
- d)40 μF
Correct answer is option 'C'. Can you explain this answer?
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]
a)
5 μF
b)
10 μF
c)
20 μF
d)
40 μF
|
Krish Khanna answered |
C = 10 μF d = 8 cm
C ' = ? d ' = 4 cm
C ' = ? d ' = 4 cm
If d is halved then C will be doubled.
Hence, C ' = 2C = 2 × 10μF = 20μF
A solid spherical conductor is given a charge.The electrostatic potential of the conductor is [2002]- a)constant throughout the conductor
- b)largest at the centre
- c)largest on the surface
- d)largest somewhere between the centre andthe surface
Correct answer is option 'A'. Can you explain this answer?
A solid spherical conductor is given a charge.The electrostatic potential of the conductor is [2002]
a)
constant throughout the conductor
b)
largest at the centre
c)
largest on the surface
d)
largest somewhere between the centre andthe surface
|
Ishani Nambiar answered |
Electric potential is constant
within or on the surface of conductor.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]- a)zero
- b)80 V
- c)800 V
- d)8 V
Correct answer is option 'B'. Can you explain this answer?
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]
a)
zero
b)
80 V
c)
800 V
d)
8 V
|
Vaibhav Basu answered |
Potential at the centre of the sphere
= potential on the surface = 80 V.
= potential on the surface = 80 V.
A capacitor C1 is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor C2. The potential difference across the combination is
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A capacitor C1 is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor C2. The potential difference across the combination is
a)
b)
c)
d)
|
Nilanjan Chakraborty answered |
Charge Q = C1V
Total capacity of combination (parallel)
Total capacity of combination (parallel)
C = C1+ C2
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]
- a)U / 4
- b)3U / 2
- c)U
- d)U / 2
Correct answer is option 'A'. Can you explain this answer?
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]
a)
U / 4
b)
3U / 2
c)
U
d)
U / 2
|
Ayush Choudhury answered |
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
As the battery is disconnected, total charge Q is shared equally by two capacitors.
So energy of each capacitor
Three capacitors each of capacity 4μF are to beconnected in such a way that the effectivecapacitance is 6 μF. This can be done by [2003]- a)connecting two in parallel and one in series
- b)connecting all of them in series
- c)connecting them in parallel
- d)connecting two in series and one in parallel
Correct answer is option 'D'. Can you explain this answer?
Three capacitors each of capacity 4μF are to beconnected in such a way that the effectivecapacitance is 6 μF. This can be done by [2003]
a)
connecting two in parallel and one in series
b)
connecting all of them in series
c)
connecting them in parallel
d)
connecting two in series and one in parallel
|
Abhiram Nair answered |
For series, 
For parallel, Ceq = C '+C3 = 2 + 4 = 6μF
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]
- a)
2x10-3C - b)5x10-3C
- c)2x10-2C
- d)5x10-2C
Correct answer is option 'B'. Can you explain this answer?
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]
a)
2x10-3Cb)
5x10-3Cc)
2x10-2Cd)
5x10-2C
|
Rajat Roy answered |
Charge on each plate of each capacitor
Q = 
CV = 25 x10-6 x 200
= 5 x10-3C
CV = 25 x10-6 x 200=
5 x10-3CA 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]- a)50 V
- b)5 kV
- c)50 KV
- d)5 V
Correct answer is option 'C'. Can you explain this answer?
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]
a)
50 V
b)
5 kV
c)
50 KV
d)
5 V
|
Anu Sharma answered |
×10^-6 C. It is fired horizontally with a velocity of 400 m/s. If the bullet enters a uniform magnetic field of 0.5 T, find the radius of the circular path traced by the bullet.
Solution:
Given:
Mass of the bullet, m = 2 g = 2×10^-3 kg
Charge on the bullet, q = 2×10^-6 C
Velocity of the bullet, v = 400 m/s
Magnetic field, B = 0.5 T
Formula:
The equation of the motion of a charged particle in a magnetic field is given by:
F = qvBsinθ, where F is the magnetic force acting on the particle, v is the velocity of the particle, B is the magnetic field, q is the charge on the particle, and θ is the angle between v and B.
The magnetic force acts perpendicular to the velocity of the charged particle, causing it to move in a circular path with a radius given by:
r = mv/|q|B
Substituting the given values, we get:
r = (2×10^-3×400)/|2×10^-6|×0.5
r = 4 m
Therefore, the radius of the circular path traced by the bullet is 4 meters.
Solution:
Given:
Mass of the bullet, m = 2 g = 2×10^-3 kg
Charge on the bullet, q = 2×10^-6 C
Velocity of the bullet, v = 400 m/s
Magnetic field, B = 0.5 T
Formula:
The equation of the motion of a charged particle in a magnetic field is given by:
F = qvBsinθ, where F is the magnetic force acting on the particle, v is the velocity of the particle, B is the magnetic field, q is the charge on the particle, and θ is the angle between v and B.
The magnetic force acts perpendicular to the velocity of the charged particle, causing it to move in a circular path with a radius given by:
r = mv/|q|B
Substituting the given values, we get:
r = (2×10^-3×400)/|2×10^-6|×0.5
r = 4 m
Therefore, the radius of the circular path traced by the bullet is 4 meters.
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- a)√2 C
- b)2C
- c)c/√2
- d)C/2
Correct answer is option 'D'. Can you explain this answer?
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
a)
√2 C
b)
2C
c)
c/√2
d)
C/2
|
Moumita Khanna answered |
When oil is placed between space of plates
When oil is removed 
... (2)
... (2)On comparing both equations, we get
C ' = C/2
C ' = C/2
A parallel plate air capacitor is charged to apotential difference of V volts. Afterdisconnecting the charging battery the distancebetween the plates of the capacitor is increasedusing an insulating handle. As a result thepotential difference between the plates [2006]- a)does not change
- b)becomes zero
- c)increases
- d)decreases
Correct answer is option 'C'. Can you explain this answer?
A parallel plate air capacitor is charged to apotential difference of V volts. Afterdisconnecting the charging battery the distancebetween the plates of the capacitor is increasedusing an insulating handle. As a result thepotential difference between the plates [2006]
a)
does not change
b)
becomes zero
c)
increases
d)
decreases
|
Krish Patel answered |
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
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- a)0.16 J
- b)1.28 J [1989]
- c)0.64 J
- d)0.32 J
Correct answer is option 'D'. Can you explain this answer?
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
a)
0.16 J
b)
1.28 J [1989]
c)
0.64 J
d)
0.32 J
|
|
Abhiram Nair answered |
The energy stored in the capacitor = 
This energy will be converted into heat in the resistor
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]- a)Q = – q
- b)Q = – 1/q
- c)Q = q
- d)Q = 1/q
Correct answer is option 'A'. Can you explain this answer?
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]
a)
Q = – q
b)
Q = – 1/q
c)
Q = q
d)
Q = 1/q
|
|
Ayush Choudhury answered |
Let the side length of square be 'a' then
potential at centre O is
potential at centre O is
(given)
= – Q – q + 2q + 2Q = 0 = Q + q = 0
= Q = – q
= Q = – q
The electric potential at a point in free space dueto a charge Q coulomb is Q × 1011 volts.The electric field at that point is [2008]- a)4 πε0 Q×1022 volt/m
- b)12 πε0 Q×1020 volt/m
- c)4 πε0 Q×1020 volt/m
- d)12 πε0 Q×1022 volt/m
Correct answer is option 'A'. Can you explain this answer?
The electric potential at a point in free space dueto a charge Q coulomb is Q × 1011 volts.The electric field at that point is [2008]
a)
4 πε0 Q×1022 volt/m
b)
12 πε0 Q×1020 volt/m
c)
4 πε0 Q×1020 volt/m
d)
12 πε0 Q×1022 volt/m
|
Deepak Joshi answered |
Given that, V = Q × 1011 volts
Electric potential at point is given by
Electric potential at point is given by
= 4π∈0 Q x 1022 volt m-1
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field betwen the plates is E, is [2008]- a)1/2 ∈0 E2 /Ad
- b)∈0 E2 /Ad
- c)∈0 E2 Ad
- d)
Correct answer is option 'D'. Can you explain this answer?
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field betwen the plates is E, is [2008]
a)
1/2 ∈0 E2 /Ad
b)
∈0 E2 /Ad
c)
∈0 E2 Ad
d)
|
Anirudh Datta answered |
The energy required to charge a parallel
plate condenser is given by
plate condenser is given by
As 
and V = E.d
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 VA, VB and VC denotes the potentials of the three shells, then for c = a + b, we have [2009]- a)VC = VB ≠ VA
- b)VC ≠ VB ≠ VA
- c)VC = VB = VA
- d)VC = VA ≠ VB
Correct answer is option 'D'. Can you explain this answer?
Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ , – σ and σ respectively. If VA, VB and VC denotes the potentials of the three shells, then for c = a + b, we have [2009]
a)
VC = VB ≠ VA
b)
VC ≠ VB ≠ VA
c)
VC = VB = VA
d)
VC = VA ≠ VB
|
|
Abhiram Nair answered |
c = a + b.
VA = VC ≠ VB
A, B and C are three points in a uniform electric field. The electric potential is [NEET 2013]
- a)maximum at B
- b)maximum at C
- c)same at all the three points A, B and C
- d)maximum at A
Correct answer is option 'A'. Can you explain this answer?
A, B and C are three points in a uniform electric field. The electric potential is [NEET 2013]
a)
maximum at B
b)
maximum at C
c)
same at all the three points A, B and C
d)
maximum at A
|
Yash Saha answered |
Potential at B, VB is maximum
VB > VC > VA
As in the direction of electric field
potential decreases.
VB > VC > VA
As in the direction of electric field
potential decreases.
A network of four capacitors of capacity equal to C1 = C, C2 = 2C, C3 = 3C and C4 = 4C are
conducted to a battery as shown in the figure. The ratio of the charges on C2 and C4 is: [2005]
- a)4/7
- b)3/22
- c)7/4
- d)22/3
Correct answer is option 'B'. Can you explain this answer?
A network of four capacitors of capacity equal to C1 = C, C2 = 2C, C3 = 3C and C4 = 4C are
conducted to a battery as shown in the figure. The ratio of the charges on C2 and C4 is: [2005]
conducted to a battery as shown in the figure. The ratio of the charges on C2 and C4 is: [2005]
a)
4/7
b)
3/22
c)
7/4
d)
22/3
|
Subham Chavan answered |
Equivalent capacitance for three capacitors (C1, C2 & C3) in series is given by
⇒ Charge on capacitors (C1, C2 & C3) in series
Charge on capacitor C4 = C4V = 4C V
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]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
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]
a)
b)
c)
d)
|
|
Yash Saha answered |
Length of body diagonal = 
∴ Distance of centre of cube from each
corner
corner
P.E at centre
= 8 × Potential Energy due to A
= 8 × Potential Energy due to A
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]- a)B / 2A
- b)2A / B
- c)A / B
- d)B / A
Correct answer is option 'B'. Can you explain this answer?
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]
, 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]a)
B / 2A
b)
2A / B
c)
A / B
d)
B / A
|
Prasenjit Pillai answered |
for equilibrium
for stable equilibrium
should be positive for the value of r.here, 
is +ve value for r = 2A/B so
is +ve value for r = 2A/B soIf the potential of a capacitor having capacity 6μF is increased from 10 V to 20 V, then increasein its energy will be [1995]- a)4x10-4 J
- b)4x10-6 J
- c)9x10-4 J
- d)12x10-6 J
Correct answer is option 'C'. Can you explain this answer?
If the potential of a capacitor having capacity 6μF is increased from 10 V to 20 V, then increasein its energy will be [1995]
a)
4x10-4 J
b)
4x10-6 J
c)
9x10-4 J
d)
12x10-6 J
|
Dipanjan Chawla answered |
Capacitance of capacitor (C) = 6μF = 6 ×10–6 F; Initial potential (V1) = 10 V and final potential (V2) = 20 V.
The increase in energy (ΔU)
The increase in energy (ΔU)
= (3x10-6 )x300 = 9x10-4 J .
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(m2) the energy (joules) stored in the condenser is [2011]- a)E2 AD/∈0
- b)
- c)∈0E AD
- d)
Correct answer is option 'D'. Can you explain this answer?
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(m2) the energy (joules) stored in the condenser is [2011]
a)
E2 AD/∈0
b)
c)
∈0E AD
d)
|
Pooja Saha answered |
A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4 V. When another parallel combination of n2 capacitors, each of value C2, 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 C2 , in terms of C1, is then [2010]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4 V. When another parallel combination of n2 capacitors, each of value C2, 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 C2 , in terms of C1, is then [2010]
it has the same (total) energy stored in it, as the first combination has. The value of C2 , in terms of C1, is then [2010]
a)
b)
c)
d)
|
|
Subham Chavan answered |
In series, 
∴ Energy stored
In parallel, Ceff = n2 C2
∴ Energy stired , 
A condenser of capacity C is charged to a potential difference of V1. 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 V2 is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A condenser of capacity C is charged to a potential difference of V1. 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 V2 is
a)
b)
c)
d)
|
|
Dipanjan Chawla answered |
q = CV1 cos ωt
Also, 
and V = V1 cosωt
and V = V1 cosωtAt t = t1 , V = V2 and i = -ωCV1 sinωt1
(–ve sign gives direction)
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 isalong the direction of the field, the force on itand its potential energy are respectively [2004]- a)q.E and max.
- b)2 q.E and min.
- c)q.E and p.E
- d)zero and min
Correct answer is option 'D'. Can you explain this answer?
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 isalong the direction of the field, the force on itand its potential energy are respectively [2004]
a)
q.E and max.
b)
2 q.E and min.
c)
q.E and p.E
d)
zero and min
|
|
Anirudh Datta answered |
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
= – P.E. = –qaE
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:
- a)zero
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
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:
a)
zero
b)
c)
d)
|
Kunal Rane answered |
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
We know that for conservative force,
W = –ΔU = 0
Two charges q1 and q2 are placed 30 cm apart, as shown in the figure. A third charge q3 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
- a)8q1
- b)6q1
- c)8q2
- d)6q2
Correct answer is option 'C'. Can you explain this answer?
Two charges q1 and q2 are placed 30 cm apart, as shown in the figure. A third charge q3 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
where k isa)
8q1
b)
6q1
c)
8q2
d)
6q2
|
|
Pooja Saha answered |
We know that potential energy of discrete system of charges is given by
According to question,
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]- a)
- b)
- c)3C, 3V
- d)
Correct answer is option 'B'. Can you explain this answer?
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]
a)
b)
c)
3C, 3V
d)
|
Lekshmi Banerjee answered |
In series combination of capacitors
Thus, the capacitance and breakdown
voltage of the combination will be C/3 and 3V
voltage of the combination will be C/3 and 3V
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]- a)
- b)E2Ad/∈0
- c)
- d)∈0EAd
Correct answer is option 'C'. Can you explain this answer?
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]
a)
b)
E2Ad/∈0
c)
d)
∈0EAd
|
|
Nilanjan Chakraborty answered |
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.
capacitor.
The capacitance of the parallel plate
capacitor
V = E.d.
Substituting the value of C in equation (i)
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]
- a)
- b)
- c)
- d)zero
Correct answer is option 'C'. Can you explain this answer?
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]
a)
b)
c)
d)
zero
|
Maya Sengupta answered |
Distance of point A from the two +q
charges = L.
Distance of point A from the two –q
charges =
charges = L.
Distance of point A from the two –q
charges =
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- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
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
a)
b)
c)
d)
|
|
Maya Sengupta answered |
Energy stored per unit volume
(∵ E = V/d)
The electric potential at a point (x, y, z) is given by V = – x2y – xz3 + 4. The electric field 
at that
point is [2009]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The electric potential at a point (x, y, z) is given by V = – x2y – xz3 + 4. The electric field at that
point is [2009]
at thatpoint is [2009]
a)
b)
c)
d)
|
|
Vaibhav Basu answered |
The electric field at a point is equal to
negative of potential gradient at that point.
negative of potential gradient at that point.
Two parallel metal plates having charges + Qand –Q face each other ata certain distancebetween them. If the plaves are now dipped inkerosene oil tank, the electric field between theplates will- a)remain same
- b)become zero
- c)increases
- d)decrease
Correct answer is option 'D'. Can you explain this answer?
Two parallel metal plates having charges + Qand –Q face each other ata certain distancebetween them. If the plaves are now dipped inkerosene oil tank, the electric field between theplates will
a)
remain same
b)
become zero
c)
increases
d)
decrease
|
Harshitha Dey answered |
Electric field
ε of kerosine oil is more than that of air.
As ε increases, E decreases
As ε increases, E decreases
What is the effective capacitance betweenpoints X and Y?
- a)24 μF
- b)18 μF
- c)12 μF
- d)6 μF
Correct answer is option 'D'. Can you explain this answer?
What is the effective capacitance betweenpoints X and Y?
a)
24 μF
b)
18 μF
c)
12 μF
d)
6 μF
|
Anand Jain answered |
Equivalent circuit
Here, 
Hence, no charge will flow through 20μF
C1 and C2 are in series, also C3 and C4 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
series.
Hence, C' = 3 μF, C'' = 3 μF
C' and C'' are in parallel.
Hence net capacitance = C' + C'' = 3 + 3
= 6 μF
Energy stored in a capacitor is- a)
- b)QV
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Energy stored in a capacitor is
a)
b)
QV
c)
d)
|
|
Krish Patel answered |
Energy stored in capacitor
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