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All questions of Electrostatic Potential and Capacitance for NEET 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
- a)0.5 J
- b)1.12 J
- c)1.2 J
- d)0.9 J
Correct answer is 'D'. Can you explain this answer?
|
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
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
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.
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.
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
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
|
Learners Habitat answered |
A polarized dielectric is equivalent to two charged surfaces with induced charges (polarization charges).
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
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
|
Husna answered |
V is inversely proportional to radius
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.
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
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
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 |
Can you explain the answer of this question below:It requires 4 J of work to move a charge of 20 C from point A to point B, separated by a distance of 0.2 cm. The potential difference between A and B in volts
- A:
0.2
- B:
16
- C:
5
- D:
80
The answer is a.
It requires 4 J of work to move a charge of 20 C from point A to point B, separated by a distance of 0.2 cm. The potential difference between A and B in volts
0.2
16
5
80
|
Mayank Singh answered |
As we know tha..v=w/q. .....1 so according to question w=4j and q=20c on putting value of w and q in equation 1 we find that v=0.2v
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 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.
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.
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.
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.
Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be - a)1:1
- b)1:2
- c)1:4
- d)1:8
Correct answer is option 'B'. Can you explain this answer?
Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be
a)
1:1
b)
1:2
c)
1:4
d)
1:8
|
|
Sanaya Kumar answered |
By dimensional analysis
but by superposition
Because of the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made
therefore,
Because of the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made
therefore,
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
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.
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.
Two metallic spheres of radii 1 cm and 3 cm are given charges of (-1 x 10-2 C) and 5 x 10-2 C , respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is:
- a)2 x 10-2 C
- b)3 x 10-2 C
- c)4 x 10-2 C
- d)1 x 10-2 C
Correct answer is option 'B'. Can you explain this answer?
Two metallic spheres of radii 1 cm and 3 cm are given charges of (-1 x 10-2 C) and 5 x 10-2 C , respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is:
a)
2 x 10-2 C
b)
3 x 10-2 C
c)
4 x 10-2 C
d)
1 x 10-2 C
|
|
Rajeev Saxena answered |
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.
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.
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
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.
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%).
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.
A hollow metal sphere of radius 20 cm is charged such that the potential on its surface is 120 Volt. The potential at the centre of the sphere is- a)80 V
- b)6 V
- c)120 V
- d)Zero
Correct answer is 'C'. Can you explain this answer?
A hollow metal sphere of radius 20 cm is charged such that the potential on its surface is 120 Volt. The potential at the centre of the sphere is
a)
80 V
b)
6 V
c)
120 V
d)
Zero
|
|
Anjana Sharma answered |
Potential inside the charged sphere is constant and equal to potential on the surface. Hence the potential at the centre of the sphere is 120 V.
The potential energy of a system containing only one point charge is
- a)Zero
- b)Infinity
- c)Non zero finite
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
The potential energy of a system containing only one point charge is
a)
Zero
b)
Infinity
c)
Non zero finite
d)
None of the above
|
|
Sahil Menon answered |
Explanation:
Potential energy is the energy possessed by a system due to the positions of its constituents. In the case of a system containing only one point charge, there is no other charge present to interact with it. Hence, the potential energy of such a system is zero.
To understand this concept, let us consider the formula for the potential energy of a system of two point charges:
U = kq1q2/r
where U is the potential energy, k is the Coulomb constant, q1 and q2 are the charges of the two point charges, and r is the distance between them.
When we have only one point charge in the system, q1 or q2 becomes zero. If we consider q1 to be zero, then the formula becomes:
U = kq1q2/r = 0
Therefore, the potential energy of a system containing only one point charge is zero.
In summary, the potential energy of a system containing only one point charge is zero because there is no other charge present to interact with it.
Potential energy is the energy possessed by a system due to the positions of its constituents. In the case of a system containing only one point charge, there is no other charge present to interact with it. Hence, the potential energy of such a system is zero.
To understand this concept, let us consider the formula for the potential energy of a system of two point charges:
U = kq1q2/r
where U is the potential energy, k is the Coulomb constant, q1 and q2 are the charges of the two point charges, and r is the distance between them.
When we have only one point charge in the system, q1 or q2 becomes zero. If we consider q1 to be zero, then the formula becomes:
U = kq1q2/r = 0
Therefore, the potential energy of a system containing only one point charge is zero.
In summary, the potential energy of a system containing only one point charge is zero because there is no other charge present to interact with it.
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.
Electric potential due to a point charge q at a distance r from the point is _______ (in the air).- a)q/r
- b)q*r
- c)q/r2
- d)-q/r
Correct answer is option 'A'. Can you explain this answer?
Electric potential due to a point charge q at a distance r from the point is _______ (in the air).
a)
q/r
b)
q*r
c)
q/r2
d)
-q/r
|
|
Om Desai answered |
Force on a unit point charge kept at a distance r from the charge = q/r2.
Therefore, work done to bring that point charge through a small distance dr = q/r2 * (-dr). Therefore, the potential of that point is = 
.
.The amount of work done in moving a unit positive charge through distance of 10 cm on an equipotential surface is- a)100 joule
- b)10 cm
- c)1/10 cm
- d)Zero
Correct answer is option 'D'. Can you explain this answer?
The amount of work done in moving a unit positive charge through distance of 10 cm on an equipotential surface is
a)
100 joule
b)
10 cm
c)
1/10 cm
d)
Zero
|
|
Anupam Singh answered |
Zero because at equipotential surface potential difference is zero
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