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All questions of Electrostatic Potential and Capacitance for Class 12 Exam
Three capacitors, each of capacitance C = 3 mF, are connected as shown in the figure. The equivalent capacitance between points P and S is
- a)9 μF
- b)6 μF
- c)3 μF
- d)1 μF
Correct answer is option 'A'. Can you explain this answer?
Three capacitors, each of capacitance C = 3 mF, are connected as shown in the figure. The equivalent capacitance between points P and S is
a)
9 μF
b)
6 μF
c)
3 μF
d)
1 μF
|
Dr Manju Sen answered |
If P is at positive potential, then Q is at negative potential and R is at positive potential. The system therefore reduces to 3 capacitors in parallel. C= 9μF
The electric field inside a dielectric decreases, when it is placed in an external electric field. This happens due to- a)Electrostatic shielding
- b)Polarization
- c)Electrostatic repulsion between atoms
- d)Electrostatic attraction between atoms
Correct answer is option 'B'. Can you explain this answer?
The electric field inside a dielectric decreases, when it is placed in an external electric field. This happens due to
a)
Electrostatic shielding
b)
Polarization
c)
Electrostatic repulsion between atoms
d)
Electrostatic attraction between atoms
|
Infinity Academy answered |
The external electric field polarizes the dielectric and an electric field is produced.
The net electric field inside the dielectric decreases due to polarization.
The net electric field inside the dielectric decreases due to polarization.
Three different capacitors are connected in series, then:- a)they will have equal charges
- b)they will have same potential
- c)both 1 & 2
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Three different capacitors are connected in series, then:
a)
they will have equal charges
b)
they will have same potential
c)
both 1 & 2
d)
none of these
|
Mira Joshi answered |
C = Q/V
series connection splits the battery potential, hence....
V = V1 + V2
V1 = Q/C1 & V2 = Q/C2
Therefore, both have equal charge
series connection splits the battery potential, hence....
V = V1 + V2
V1 = Q/C1 & V2 = Q/C2
Therefore, both have equal charge
|
Nikita Singh answered |
q=6x10-10 c
Q=-2x10-9c
r1=3x10-2m
r2=4x10-2m
△vQ=w/q
(1/4πεo)-2x10-9/10-2[(1/4)-(1/3)]=W/6x10-3
9x109x2x10-9x6x10-3/12x10-2=W
W=9x10-3x102
W=0.9J
Q=-2x10-9c
r1=3x10-2m
r2=4x10-2m
△vQ=w/q
(1/4πεo)-2x10-9/10-2[(1/4)-(1/3)]=W/6x10-3
9x109x2x10-9x6x10-3/12x10-2=W
W=9x10-3x102
W=0.9J
The amount of work done in moving a charge from one point to another along an equipotential line or surface charge is- a)Zero
- b)Infinity
- c)One
- d)Two
Correct answer is option 'A'. Can you explain this answer?
The amount of work done in moving a charge from one point to another along an equipotential line or surface charge is
a)
Zero
b)
Infinity
c)
One
d)
Two
|
Om Desai answered |
Since Potential difference between two points in equipotential surfaces is zero, the work done between two points in equipotential surface is also zero.
Two equipotential surfaces have a potential of -10V and 90V respectively, what is the difference in potential between these surfaces?- a)90V
- b)80V
- c)0V
- d)100V
Correct answer is 'D'. Can you explain this answer?
Two equipotential surfaces have a potential of -10V and 90V respectively, what is the difference in potential between these surfaces?
a)
90V
b)
80V
c)
0V
d)
100V
|
Krishna Iyer answered |
Potential = high potential - low potential
so our high potential is 90V and low potential is 10V,
potential = 90-(-10)=100V
so our high potential is 90V and low potential is 10V,
potential = 90-(-10)=100V
There are three capacitors with equal capacitance. In series combination, they have a net capacitance of C1 and in parallel combination, a net capacitance of C2.What will be the value of C1 C2?- a)1:9
- b)3:2
- c)2:3
- d)9:1
Correct answer is option 'A'. Can you explain this answer?
There are three capacitors with equal capacitance. In series combination, they have a net capacitance of C1 and in parallel combination, a net capacitance of C2.What will be the value of C1 C2?
a)
1:9
b)
3:2
c)
2:3
d)
9:1
|
Dilip Chaurasiya answered |
In series C= C/3 and in parallel C= 3C divide it we get 1/9
As shown in the figure below, an ellipsoidal cavity is carved within a perfect conductor. A positive charge Q is placed at the centre of the cavity. If points A and b are shown on the cavity surface (see figure), then which among the following choices is correct?
- a)Potential at A = Potential at B
- b)Charge density at A = Charge density at B
- c)Electric field near A = Electric field near B
- d)Total electric flux at the surface of the cavity = zero
Correct answer is option 'A'. Can you explain this answer?
As shown in the figure below, an ellipsoidal cavity is carved within a perfect conductor. A positive charge Q is placed at the centre of the cavity. If points A and b are shown on the cavity surface (see figure), then which among the following choices is correct?
a)
Potential at A = Potential at B
b)
Charge density at A = Charge density at B
c)
Electric field near A = Electric field near B
d)
Total electric flux at the surface of the cavity = zero
|
|
Om Desai answered |
Electric field at A is different from field at B because, E= K/r2
we know that conductor is an equipotential surface, so potential will be the same at A and B.
As charge density, σ∝1/r, then charge density is different at A and B.
we know that conductor is an equipotential surface, so potential will be the same at A and B.
As charge density, σ∝1/r, then charge density is different at A and B.
The electrostatic potential energy between two charges q1 and q2 separated by a distance by r is given by- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The electrostatic potential energy between two charges q1 and q2 separated by a distance by r is given by
a)
b)
c)
d)
|
Khushi Mittal answered |
Energy = force × distance U=Kq1q2/r^2 × r U=Kq1q2/r
A hollow metal sphere of radius 5cm is charged so that the potential on its surface is 10V. The potential at a distance of 2cm from the centre of the sphere is- a)4V
- b)zero
- c)10/3V
- d)10V
Correct answer is option 'D'. Can you explain this answer?
A hollow metal sphere of radius 5cm is charged so that the potential on its surface is 10V. The potential at a distance of 2cm from the centre of the sphere is
a)
4V
b)
zero
c)
10/3V
d)
10V
|
Ayush Joshi answered |
In the case of a hollow metal sphere (spherical shell), the electric field inside the shell is zero. This means that the potential inside the shell is constant. Therefore the potential at the centre of the sphere is the same as that on its surface, i.e. 10 V.
1 microfarad is equal to- a)10-6F
- b)10-12F
- c)10-15F
- d)10-9F
Correct answer is option 'A'. Can you explain this answer?
1 microfarad is equal to
a)
10-6F
b)
10-12F
c)
10-15F
d)
10-9F
|
Lavanya Menon answered |
The microfarad (symbolized µF) is a unit of capacitance, equivalent to 0.000001 (10 to the -6th power) farad.
Work done in carrying 2C charge in a circular path of radius 2m around a charge of 10C is- a)6.67J
- b)60J
- c)Zero
- d)15J
Correct answer is option 'C'. Can you explain this answer?
Work done in carrying 2C charge in a circular path of radius 2m around a charge of 10C is
a)
6.67J
b)
60J
c)
Zero
d)
15J
|
Suresh Iyer answered |
The overall work performed in carrying a 2coulomb charge in a circular orbit of radius 3 m around a charge of 10 coulomb is calculated below.
It is a well-known fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.
The value of dv=0 and hence W=q0=0.
It is a well-known fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.
The value of dv=0 and hence W=q0=0.
Can you explain the answer of this question below:Equal charges are given to two spheres of different radii. The potential will
- A:
be equal on both the spheres
- B:
be more on the smaller sphere
- C:
be more on the bigger sphere
- D:
depend on the material of the sphere
The answer is b.
Equal charges are given to two spheres of different radii. The potential will
be equal on both the spheres
be more on the smaller sphere
be more on the bigger sphere
depend on the material of the sphere
|
Ashish Bhoumick answered |
Because charge distribute surface of the area.here small sphere has high density charge.so the answer 'b' is right...
Electric field intensity at point ‘B’ due to a point charge ‘Q’ kept at a point ‘A’ is 12 NC-1 and the electric potential at a point ‘B’ due to same charge is 6 JC-1. The distance between AB is- a)2 m
- b)1.5 m
- c)1 m
- d)0.5 m
Correct answer is option 'D'. Can you explain this answer?
Electric field intensity at point ‘B’ due to a point charge ‘Q’ kept at a point ‘A’ is 12 NC-1 and the electric potential at a point ‘B’ due to same charge is 6 JC-1. The distance between AB is
a)
2 m
b)
1.5 m
c)
1 m
d)
0.5 m
|
|
Nandini Patel answered |
As we know , E . l = V where,
E = electric field intensity = 12 N/C
V = electric potential = 6 J/C
=> distance between A and B ,
l = ( 6 / 12 ) m or ( 1 / 2 ) m = 0.5 m
Electric potential at a point located far away from the charge is taken to be- a)Unity
- b)May be zero or infinite
- c)Zero
- d)Infinite
Correct answer is option 'C'. Can you explain this answer?
Electric potential at a point located far away from the charge is taken to be
a)
Unity
b)
May be zero or infinite
c)
Zero
d)
Infinite
|
Suman Kumar answered |
Because the distance is much larger so electric potential is 0 in other words there is no effect of other charge on that charge
The polarized dielectric is equivalent to- a)non-zero volume charge density
- b)one charged surface with induced surface charge density
- c)two charged surfaces with induced surface charge densities
- d)zero surface charge density
Correct answer is option 'C'. Can you explain this answer?
The polarized dielectric is equivalent to
a)
non-zero volume charge density
b)
one charged surface with induced surface charge density
c)
two charged surfaces with induced surface charge densities
d)
zero surface charge density
|
Vijay Bansal answered |
The polarization is defined by the dipole moment per unit volume, so the total dipole moment in the volume is: This is equivalent to two charges, each equal to , separated by distance .
If 100 J of work has to be done in moving an electric charge of 4C from a place where potential is -5 V to another place, where potential is V volt. The value of V is- a)15 V
- b)20 V
- c)25 V
- d)10 V
Correct answer is option 'B'. Can you explain this answer?
If 100 J of work has to be done in moving an electric charge of 4C from a place where potential is -5 V to another place, where potential is V volt. The value of V is
a)
15 V
b)
20 V
c)
25 V
d)
10 V
|
|
Suresh Iyer answered |
From the definition, the work done to a test charge ‘q0’ from one place to another place in an electric field is given by the formula
W=q0x[vfinal-vinitial ]
100=4x[v-(-5)]
v+5=25
v=20V
W=q0x[vfinal-vinitial ]
100=4x[v-(-5)]
v+5=25
v=20V
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 volt. The potential at the centre of the sphere is- a)8 volt
- b)zero
- c)800 volt
- d)80 volt
Correct answer is option 'D'. Can you explain this answer?
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 volt. The potential at the centre of the sphere is
a)
8 volt
b)
zero
c)
800 volt
d)
80 volt
|
Sunidhi Pandey answered |
Chage is uniformly distributed in an hollow sphere.
When two capacitors C1 and C2 are connected in series and parallel, their equivalent capacitances comes out to be 3 μƒ and 16 μƒ respectively. Calculate values of C1 and C2.- a)4μƒ and 12μƒ
- b)6μƒ and 10μƒ
- c)4μƒ and 10μƒ
- d)6μƒ and 12μƒ
Correct answer is option 'A'. Can you explain this answer?
When two capacitors C1 and C2 are connected in series and parallel, their equivalent capacitances comes out to be 3 μƒ and 16 μƒ respectively. Calculate values of C1 and C2.
a)
4μƒ and 12μƒ
b)
6μƒ and 10μƒ
c)
4μƒ and 10μƒ
d)
6μƒ and 12μƒ
|
Anonymous answered |
When a positive charge is moved in an electrostatic field from a point at high potential to a low potential, its kinetic energy- a)Remains constant
- b)Decreases
- c)Increases
- d)Either increase or remain constant
Correct answer is option 'C'. Can you explain this answer?
When a positive charge is moved in an electrostatic field from a point at high potential to a low potential, its kinetic energy
a)
Remains constant
b)
Decreases
c)
Increases
d)
Either increase or remain constant
|
Siddhi Bhardwaj answered |
You can generalise it by assuming a positive charge moving away from another positive charge. now both of them are repelling each other with some force. so that positive charge will accelerate which results in the increase in K.E.
Two capacitors of 20 μƒ and 30 μƒ are connected in series to a battery of 40V. Calculate charge on each capacitor.- a)480 C
- b)478 C
- c)450 C
- d)500 C
Correct answer is option 'A'. Can you explain this answer?
Two capacitors of 20 μƒ and 30 μƒ are connected in series to a battery of 40V. Calculate charge on each capacitor.
a)
480 C
b)
478 C
c)
450 C
d)
500 C
|
|
Nikita Singh answered |
C1= 20×10µf
and C2= 30×10µf
in series Ceq = C1C2/(C1+C2)
Ceq = 20×10^(-6)×30×10^(-6)/20×10^(-6)+30^×10(-6)
Ceq= 12×10^(-6)f
As we know that Q = CV
Putting the values of C and V= 40V, we get
Q = (12 * 10^-6) * 40
= 480µC
The shape of equipotential surface for an infinite line charge is:- a)Coaxial cylindrical surfaces
- b)Parallel plane surfaces
- c)Parallel plane surfaces perpendicular to lines of force
- d)None of above
Correct answer is option 'A'. Can you explain this answer?
The shape of equipotential surface for an infinite line charge is:
a)
Coaxial cylindrical surfaces
b)
Parallel plane surfaces
c)
Parallel plane surfaces perpendicular to lines of force
d)
None of above
|
Rajesh Gupta answered |
The shape of equipotential surface for an infinite line charge is coaxial cylindrical because A curved surface on which potential is constant is equipotential curve . If we consider the line charge then the focus of the point should have the same potential hence it is a coaxial cylinder.
Calculate the equivalent capacitance for the following combination between points A and B
- a)12/5 μƒ
- b)5 μƒ
- c)40/9 μƒ
- d)37/5 μƒ
Correct answer is option 'D'. Can you explain this answer?
Calculate the equivalent capacitance for the following combination between points A and B
a)
12/5 μƒ
b)
5 μƒ
c)
40/9 μƒ
d)
37/5 μƒ
|
Divey Sethi answered |
4μf and 6μf are in series (right side of AB) so 6x(4/6)+4=24/10μf
And now 5 μf and the resulting of above two are in parallel (as on different sides of AB) so 5+(24/10) =50+(24/10)=74/10=37/5μf
And now 5 μf and the resulting of above two are in parallel (as on different sides of AB) so 5+(24/10) =50+(24/10)=74/10=37/5μf
Here is a combination of three identical capacitors. If resultant capacitance is 1 μƒ, calculate capacitance of each capacitor.
- a)6 μƒ
- b)3 μƒ
- c)4 μƒ
- d)2 μƒ
Correct answer is option 'B'. Can you explain this answer?
Here is a combination of three identical capacitors. If resultant capacitance is 1 μƒ, calculate capacitance of each capacitor.
a)
6 μƒ
b)
3 μƒ
c)
4 μƒ
d)
2 μƒ
|
|
Lavanya Menon answered |
Combination of Capacitors in Series
C(effective) = C/3 = 1
C(effective) = C/3 = 1
A particle of charge q1 = 3μC is located on x-axis at the point x1 = 6 cm. A second point charge q2 = 2μC is placed on the x-axis at x2 = -4 cm. The absolute electric potential at the origin is- a)9 x 10-6 V
- b)9 x 105 V
- c)9 x 10-4 V
- d)9 x 10-2 V
Correct answer is option 'C'. Can you explain this answer?
A particle of charge q1 = 3μC is located on x-axis at the point x1 = 6 cm. A second point charge q2 = 2μC is placed on the x-axis at x2 = -4 cm. The absolute electric potential at the origin is
a)
9 x 10-6 V
b)
9 x 105 V
c)
9 x 10-4 V
d)
9 x 10-2 V
|
|
Nikita Singh answered |
V=kq/r
Therefore, Vnet=Kx[(2µc/4) + (3µc/6)]
=9c109x10-6[1/2+1/2]x102
=9x105
Therefore, Vnet=Kx[(2µc/4) + (3µc/6)]
=9c109x10-6[1/2+1/2]x102
=9x105
Electric field at the surface of a charged conductor is proportional to- a)Surface charge density
- b)Volume of the sphere
- c)Volume charge density
- d)Area of the sphere
Correct answer is option 'A'. Can you explain this answer?
Electric field at the surface of a charged conductor is proportional to
a)
Surface charge density
b)
Volume of the sphere
c)
Volume charge density
d)
Area of the sphere
|
Ciel Knowledge answered |
Electric field at the surfaces of charged conductors is σ/ε0.n, where n is a unit vector normal to the surface.
We clearly see that the electric field is perpendicular to surface charge density (σ).
We clearly see that the electric field is perpendicular to surface charge density (σ).
Electric potential is- a)scalar and dimensionless
- b)vector and dimensionless
- c)scalar with dimension
- d)vector with dimension
Correct answer is option 'C'. Can you explain this answer?
Electric potential is
a)
scalar and dimensionless
b)
vector and dimensionless
c)
scalar with dimension
d)
vector with dimension
|
Pooja Mehta answered |
He electric potential due to a system of point charges is equal to the sum of the point charges' individual potentials. This fact simplifies calculations significantly, since addition of potential (scalar) fields is much easier than addition of the electric (vector) fields.
Can you explain the answer of this question below:The potential energy of a system containing only one point charge is
- A:
Zero
- B:
Infinity
- C:
Nonzero finite
- D:
None of the above
The answer is a.
The potential energy of a system containing only one point charge is
Zero
Infinity
Nonzero finite
None of the above
|
.mie. answered |
Answer is 0 as there are no other sources of electrostatic potential .... against which an external agent must do work.... in moving the point charge.... from infinity to its final location.... therefore correct opt is A
The dipole moment per unit volume is called- a)Polarization
- b)Surface charge density
- c)Linear charge density
- d)Volume charge density
Correct answer is option 'A'. Can you explain this answer?
The dipole moment per unit volume is called
a)
Polarization
b)
Surface charge density
c)
Linear charge density
d)
Volume charge density
|
Preeti Iyer answered |
The electric dipole moment per unit volume is called polarization or polarization density (vector p).
It is always directed from negative charge to positive charge.
If there are N atoms per unit volume than
vector p = N (vector p)
where p is the electric dipole moment of individual atom.
In bringing an electron towards another electron, the electrostatic potential energy of the system- a)becomes zero
- b)decreases
- c)remains same
- d)increases
Correct answer is option 'D'. Can you explain this answer?
In bringing an electron towards another electron, the electrostatic potential energy of the system
a)
becomes zero
b)
decreases
c)
remains same
d)
increases
|
|
Nandini Patel answered |
The electron has negative charge. When an electron is bringing towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative: Electrostatic potential energy of system of two electrons
U=[1/4πε0][(−e)(−e)/r] = [1/4πε0](e^2/r)
Thus, as r decreases, potential energy U increases.
If a unit charge is taken from one part to another part over an equipotential surface, then what is the change in electrostatic potential energy of the charge?- a)10 J
- b)100 J
- c)1 J
- d)0 J
Correct answer is option 'D'. Can you explain this answer?
If a unit charge is taken from one part to another part over an equipotential surface, then what is the change in electrostatic potential energy of the charge?
a)
10 J
b)
100 J
c)
1 J
d)
0 J
|
Utsav Srivastava answered |
Equipotential surface means the potential on every. point on that surface is constant. it means the change in potential on equipotential surface is zero we know that... ( electrostatic potential energy = change in potential × charge.)... ... according to this electrostatic potential energy is zero
If the potential difference between the plates of a capacitor is increased by 0.1%, the energy stored in the capacitor increases by very nearly- a)0.1%
- b)0.144%
- c)0.11%
- d)0.2%
Correct answer is option 'D'. Can you explain this answer?
If the potential difference between the plates of a capacitor is increased by 0.1%, the energy stored in the capacitor increases by very nearly
a)
0.1%
b)
0.144%
c)
0.11%
d)
0.2%
|
Amar Shah answered |
**Explanation:**
To understand why the correct answer is option 'D', let's consider the equation for the energy stored in a capacitor:
**Energy (E) = 1/2 * C * V^2**
where E is the energy, C is the capacitance, and V is the potential difference (voltage) across the capacitor plates.
According to the question, the potential difference between the plates of the capacitor is increased by 0.1%. Let the initial potential difference be V1 and the increased potential difference be V2.
**Change in Potential Difference = (V2 - V1) = 0.1% of V1**
Using this information, we can write the new potential difference as:
**V2 = V1 + (0.1/100) * V1 = V1 + 0.001 * V1 = (1 + 0.001) * V1 = 1.001 * V1**
Next, we need to find the change in energy stored in the capacitor. Let the initial energy be E1 and the increased energy be E2.
**Change in Energy = (E2 - E1) = ?**
Using the equation for energy, we can write the new energy as:
**E2 = 1/2 * C * V2^2 = 1/2 * C * (1.001 * V1)^2 = 1/2 * C * (1.001^2) * V1^2**
Now, let's calculate the change in energy:
**Change in Energy = (E2 - E1) = 1/2 * C * (1.001^2) * V1^2 - 1/2 * C * V1^2**
Simplifying this expression:
**Change in Energy = 1/2 * C * V1^2 * (1.001^2 - 1) = 1/2 * C * V1^2 * (1.001^2 - 1)**
To find the percentage increase in energy, we need to divide the change in energy by the initial energy and multiply by 100. Let's calculate:
**Percentage Increase in Energy = (Change in Energy / E1) * 100**
Substituting the expression for change in energy:
**Percentage Increase in Energy = [1/2 * C * V1^2 * (1.001^2 - 1)] / [1/2 * C * V1^2] * 100**
Simplifying this expression:
**Percentage Increase in Energy = (1.001^2 - 1) * 100 = (1.001 * 1.001 - 1) * 100 = (1.002001 - 1) * 100 = 0.2001 * 100 = 0.2%**
Therefore, the energy stored in the capacitor increases by approximately 0.2% when the potential difference between the plates is increased by 0.1%. Hence, the correct answer is option 'D' (0.2%).
To understand why the correct answer is option 'D', let's consider the equation for the energy stored in a capacitor:
**Energy (E) = 1/2 * C * V^2**
where E is the energy, C is the capacitance, and V is the potential difference (voltage) across the capacitor plates.
According to the question, the potential difference between the plates of the capacitor is increased by 0.1%. Let the initial potential difference be V1 and the increased potential difference be V2.
**Change in Potential Difference = (V2 - V1) = 0.1% of V1**
Using this information, we can write the new potential difference as:
**V2 = V1 + (0.1/100) * V1 = V1 + 0.001 * V1 = (1 + 0.001) * V1 = 1.001 * V1**
Next, we need to find the change in energy stored in the capacitor. Let the initial energy be E1 and the increased energy be E2.
**Change in Energy = (E2 - E1) = ?**
Using the equation for energy, we can write the new energy as:
**E2 = 1/2 * C * V2^2 = 1/2 * C * (1.001 * V1)^2 = 1/2 * C * (1.001^2) * V1^2**
Now, let's calculate the change in energy:
**Change in Energy = (E2 - E1) = 1/2 * C * (1.001^2) * V1^2 - 1/2 * C * V1^2**
Simplifying this expression:
**Change in Energy = 1/2 * C * V1^2 * (1.001^2 - 1) = 1/2 * C * V1^2 * (1.001^2 - 1)**
To find the percentage increase in energy, we need to divide the change in energy by the initial energy and multiply by 100. Let's calculate:
**Percentage Increase in Energy = (Change in Energy / E1) * 100**
Substituting the expression for change in energy:
**Percentage Increase in Energy = [1/2 * C * V1^2 * (1.001^2 - 1)] / [1/2 * C * V1^2] * 100**
Simplifying this expression:
**Percentage Increase in Energy = (1.001^2 - 1) * 100 = (1.001 * 1.001 - 1) * 100 = (1.002001 - 1) * 100 = 0.2001 * 100 = 0.2%**
Therefore, the energy stored in the capacitor increases by approximately 0.2% when the potential difference between the plates is increased by 0.1%. Hence, the correct answer is option 'D' (0.2%).
Dimensional formula for potential difference is- a)[ML2 T-2 A-1]
- b)[ML2 T-3 A-1]
- c)[ML T-1 A-1]
- d)[M2L T-2 A-1]
Correct answer is option 'B'. Can you explain this answer?
Dimensional formula for potential difference is
a)
[ML2 T-2 A-1]
b)
[ML2 T-3 A-1]
c)
[ML T-1 A-1]
d)
[M2L T-2 A-1]
|
Rohit Shah answered |
Potential Diff = Volts = V = Workdone/Charge(Q)
Workdone = [Mass][Length]^2/[Time]^2
Hence, Potential Diff = [Mass][Length]^2[Time]^−2[Charge(Q)]^−1
Hence, The correct answer is Option A.
Find the equivalent capacitance of the combination.
- a)16/3 μƒ
- b)16 μƒ
- c)19.0 μƒ
- d)8 μƒ
Correct answer is option 'C'. Can you explain this answer?
Find the equivalent capacitance of the combination.
a)
16/3 μƒ
b)
16 μƒ
c)
19.0 μƒ
d)
8 μƒ
|
Kumari Sakshi answered |
none of the optaion match answer should be 4
Equipotential surfaces cannot- a)be parallel
- b)be spherical
- c)Intersect
- d)be irregularly shaped.
Correct answer is option 'C'. Can you explain this answer?
Equipotential surfaces cannot
a)
be parallel
b)
be spherical
c)
Intersect
d)
be irregularly shaped.
|
|
Krishna Iyer answered |
The electric field lines are perpendicular to the equipotential surface. The field lines can not intersect each other because the electric force can not have two directions at a point.
What is the direction of the lines of force at any point on the equipotential surface?- a)Parallel to it
- b)Perpendicular to it.
- c)Inclined at 45 degrees
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
What is the direction of the lines of force at any point on the equipotential surface?
a)
Parallel to it
b)
Perpendicular to it.
c)
Inclined at 45 degrees
d)
None of the above
|
Anjali Iyer answered |
When electric lines of force get perpendicular to equipotential surface then area vector and electric lines of force are parallel to each other.so angle between them is zero. due to that reason they get perpendicular to each other.
Find the band gap energy when a light of wavelength 1240nm is incident on it.- a)1eV
- b)2eV
- c)3eV
- d)4eV
Correct answer is option 'A'. Can you explain this answer?
Find the band gap energy when a light of wavelength 1240nm is incident on it.
a)
1eV
b)
2eV
c)
3eV
d)
4eV
|
|
Khushi Singh answered |
Find the band gap energy when a light of wavelength 1240nm is incident on it. Explanation: The band gap energy in electron volt when wavelength is given is, Eg = 1.24(μm)/λ = 1.24 x 10-6/1240 x 10-9 = 1eV.
The dielectric Constant of a metal is- a)infinite
- b)zero
- c)equal to one
- d)less than one
Correct answer is option 'A'. Can you explain this answer?
The dielectric Constant of a metal is
a)
infinite
b)
zero
c)
equal to one
d)
less than one
|
|
Nabanita Sen answered |
An electric field (E) is generated both inside the metal and outer (plates) which are equal in magnitude but opposite, making the net resultant field (Er) in the metal to be zero. Hence, the derived dielectric constant (K) of the metal is (E/Er = E/0) which is infinity. This is why the dielectric constant is infinity.
When a conductor is placed in an electric field; its free charge carriers adjust itself in order to oppose the electric field. This happen until- a)Both the fields cancel out each other
- b)Induced field become more than external field
- c)External field reach the maximum value
- d)Induced field reach the maximum value
Correct answer is option 'A'. Can you explain this answer?
When a conductor is placed in an electric field; its free charge carriers adjust itself in order to oppose the electric field. This happen until
a)
Both the fields cancel out each other
b)
Induced field become more than external field
c)
External field reach the maximum value
d)
Induced field reach the maximum value
|
|
Lavanya Menon answered |
- When an external electric field is applied to the conductor, the free electrons in the conductor move in an opposite direction to that of the applied electric field.
- This movement of electrons induces another electric field inside the conductor which opposes the original external electric field.
- This continues until the induced electric field cancels out the external field.
when two capacitors are put in series, the equivalent capacitance is - a)the reciprocal of the capacitances
- b)smaller than both capacitances
- c)the sum of the capacitances
- d)the product of the capacitances
Correct answer is option 'B'. Can you explain this answer?
when two capacitors are put in series, the equivalent capacitance is
a)
the reciprocal of the capacitances
b)
smaller than both capacitances
c)
the sum of the capacitances
d)
the product of the capacitances
|
Sreemoyee Sengupta answered |
Explanation:
When two capacitors are put in series, the equivalent capacitance is smaller than both capacitances.
Why is this so? Let's look at the equation for capacitance in series:
1/Ceq = 1/C1 + 1/C2
Where Ceq is the equivalent capacitance and C1 and C2 are the capacitances of the individual capacitors. From this equation, we can see that the sum of the reciprocals of the capacitances is equal to the reciprocal of the equivalent capacitance.
Now, let's consider what happens when we put two capacitors in series. In this case, the capacitors share the same charge, so the voltage across each capacitor is different. This means that the capacitances are not simply added together, but rather combine in a way that reduces the overall capacitance.
To see why this is the case, let's take a simple example. Suppose we have two capacitors, each with a capacitance of 1 microfarad. If we put these capacitors in series, the equivalent capacitance is:
1/Ceq = 1/1 + 1/1 = 1/2
Ceq = 2 microfarads
As we can see, the equivalent capacitance is smaller than the capacitance of either capacitor alone. This is because the capacitors effectively "cancel out" some of each other's capacitance due to the shared charge and differing voltages.
Overall, when two capacitors are put in series, the equivalent capacitance is smaller than both capacitances because the capacitors "cancel out" some of each other's capacitance.
When two capacitors are put in series, the equivalent capacitance is smaller than both capacitances.
Why is this so? Let's look at the equation for capacitance in series:
1/Ceq = 1/C1 + 1/C2
Where Ceq is the equivalent capacitance and C1 and C2 are the capacitances of the individual capacitors. From this equation, we can see that the sum of the reciprocals of the capacitances is equal to the reciprocal of the equivalent capacitance.
Now, let's consider what happens when we put two capacitors in series. In this case, the capacitors share the same charge, so the voltage across each capacitor is different. This means that the capacitances are not simply added together, but rather combine in a way that reduces the overall capacitance.
To see why this is the case, let's take a simple example. Suppose we have two capacitors, each with a capacitance of 1 microfarad. If we put these capacitors in series, the equivalent capacitance is:
1/Ceq = 1/1 + 1/1 = 1/2
Ceq = 2 microfarads
As we can see, the equivalent capacitance is smaller than the capacitance of either capacitor alone. This is because the capacitors effectively "cancel out" some of each other's capacitance due to the shared charge and differing voltages.
Overall, when two capacitors are put in series, the equivalent capacitance is smaller than both capacitances because the capacitors "cancel out" some of each other's capacitance.
Which of the following statements is not true for polar molecules?- a)permanent dipole moment
- b)examples are oxygen and hydrogen molecules
- c)the centres of positive and negative charges are separated
- d)examples are HCl or water molecules
Correct answer is option 'B'. Can you explain this answer?
Which of the following statements is not true for polar molecules?
a)
permanent dipole moment
b)
examples are oxygen and hydrogen molecules
c)
the centres of positive and negative charges are separated
d)
examples are HCl or water molecules
|
Gaurav Kumar answered |
- Polar molecules occur when two atoms do not share electrons equally in a covalent bond.
- The oxygen side of the molecule has a slight negative charge, while the side with the hydrogen atoms has a slight positive charge.
- Ethanol is polar because the oxygen atoms attract electrons because of their higher electronegativity than other atoms in the molecule.
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