Chaukaghat, Varanasi, India - 221002.
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Aktu 2018
Bachelor of Technology (B.Tech.)
Chaukaghat, Varanasi, India - 221002
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3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
ICSE, CBSE, State
CBSE Subjects taught
Science, Mathematics, Computer Practices
ICSE Subjects taught
Physics, Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics, Science
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
1
Board
ISC/ICSE, State, CBSE
ISC/ICSE Subjects taught
Mathematics, Computer Science, Physics
CBSE Subjects taught
Computer Science, Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Computer Science, Physics, Mathematics
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
ISC/ICSE, State, CBSE
ISC/ICSE Subjects taught
Mathematics, Physics
CBSE Subjects taught
Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Physics, Mathematics
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BTech Tuition
1
BTech Mechanical subjects
Composites: Mechanics & Processing, Internal Combustion Engines and Emissions, Material Science and Metallurgy, Analysis and Design of Machine Components, Welding Technology, Heat & Mass Transfer, Strength of Materials, Operations Research, Fluid Mechanics, Mechanics of Machines, Kinematics of Machinery, Destructive & Non Destructive Testing/ Fracture Mechanics, Thermodynamics, Machine Design, Manufacturing Technology
BTech Branch
BTech Mechanical Engineering
Type of class
Crash Course, Regular Classes
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BSc Tuition
2
BSc Physics Subjects
Quantum Mechanics, Nuclear and Particle Physics, Mechanics, Statistical Physics, Thermal Physics, Solid State Physics, Numerical Analysis, Optics, Mathematical Physics, Electricity and Magnetism, Atomic and Molecular Physics, Mathematics
BSc Computer Science Subjects
Operational Research
Type of class
Crash Course, Regular Classes
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
BSc Branch
BSc Physics, BSc Computer Science, BSc Mathematics
BSc Mathematics Subjects
Calculus, Differential Equations and Mathematical Modelling, Probability and Statistics, Mechanics, Number Theory, Physics, Algebra, Numerical Methods and Programming
5 out of 5 1 review
Ayansh Sinha
"Best teacher you will get. His tips and tricks for concepts of physics are awesome. Great Teacher. Thank you very much for your support. Highly recommend. "
1. Which school boards of Class 10 do you teach for?
ICSE, CBSE and State
2. Do you have any prior teaching experience?
No
3. Which classes do you teach?
I teach BSc Tuition, BTech Tuition, Class 10 Tuition, Class 11 Tuition and Class 12 Tuition Classes.
4. Do you provide a demo class?
Yes, I provide a free demo class.
5. How many years of experience do you have?
I have been teaching for 2 years.
Answered on 21/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
We know, capacitance = dielectric constant × C₀
C = KC₀ = 6 × 18pF = 108pF [ see question - 2.8]
(a) if the voltage supply remained connected , then the potential difference across the capacitor will remain the same. e.g., V = 100V and hence, charge on the capacitor becomes Q = CV = 108pF × 100V
Q = 108 × 10⁻¹² × 100 = 1.08 × 10⁻⁸ C
(b) if the voltage supply was disconnected , then charge on the capacitor reamins the same. e.g., Q = 1.8 × 10⁻⁹C [ see question - 2.8]
And hence the potential difference across the capacitor becomes
V = Q/C = 1.8 × 10⁻⁹/1.08 × 10⁻¹⁰ = 16.6V
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
A regular hexagon ABCDEF of side 10cm has a charge 5 μC at each of its vertices as shown in the figure.
Potential due to charge placed at A, at the centre is Kq/r
Potential due to charge placed at B, at the centre is Kq/r
Potential due to Charge placed at F , at centre is Kq/r
so, total potential at centre of regular hexagon , V = V₁ + V₂ + V₃ + V₄ + V₅ + V₆
so, V = kq/r + kq/r + kq/r + kq/r + kq/r + Kq/r
V = 6kq/r
Hence, k = 9 × 10⁹Nm²/C² , q = 5 × 10⁻⁶C and r = 10cm = 0.1 m
Now, V = 6 × 9 × 10⁹ × 5 × 10⁻⁶/0.1 = 2.7 × 10⁶ Volts.
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
initially d,
therefore, C= EA/d=8 -(1)
now d1=d/2, and dielectric is put,
therefore C1=KEA/d1. -(2) here k=6 which is dielectric constant
we know d1=d/2 substituting it in (1)
C1=6EA/(d/2)=12EA/d. -(3)
(3)÷(1)
C1÷8=12EA/d÷EA/d
C1=96 uF
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
a) We know, charge have nature to reside outer surface of the conductor. It means, charge inside the surface equals zero.
According to Gaussian theorem,
Ф = q/ε ₀, here q is charged inclosed the Gaussian surface.
∵ q = 0
so, Ф = 0 and flux , Ф = E.A = 0
so, E = 0
Hence, inside the sphere, the electric field equals zero.
(b) Take a Gaussian surface of radius r > R = 12cm
then, charged inclosed into the Gaussian surface is q = 1.6 × 10⁻⁷ C
So, Ф = q/ε₀
So, EA = q/ε₀
E = q/ε₀A, here A is the surface area of Gaussian spherical surface
e.g., A = 4πr²
So, E = q/4πε₀r² = 9 × 10⁹ × 1.6 × 10⁻⁷/(12 × 10⁻²)²
= 10⁵ N/C
(C) Similarly explanation of (B),
So, E = kq/r²
Here , k = 9 × 10⁹ Nm²/C² , q = 1.6 × 10⁻⁷C and r = 18cm
So, E = 9 × 10⁹ × 1.6 × 10⁻⁷/(18 × 10⁻²)²
= 4.44 × 10⁴ N/C.
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
(a) an equipotential surface is a plane on which potential is zero everywhere.
This plane is normal to the plane in which two point charges are placed.
Let normal plane be placed at C of x distance from +2μC charge.
Then, potential due to 2μC + potential due to -2μC = 0
Kq/x + k(-q)/(6 - x) = 0
x = 3 cm
Hence, the equipotential surface of the system appears at midpoint of the system of two given charges, and it is normal to the plane of charges.
(B) we know, the direction of the electric field from positive to negative charge.
So, the direction of the electric field at every point on the surface is normal to the plane in the direction AB.
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
ICSE, CBSE, State
CBSE Subjects taught
Science, Mathematics, Computer Practices
ICSE Subjects taught
Physics, Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics, Science
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
1
Board
ISC/ICSE, State, CBSE
ISC/ICSE Subjects taught
Mathematics, Computer Science, Physics
CBSE Subjects taught
Computer Science, Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Computer Science, Physics, Mathematics
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
ISC/ICSE, State, CBSE
ISC/ICSE Subjects taught
Mathematics, Physics
CBSE Subjects taught
Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Physics, Mathematics
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BTech Tuition
1
BTech Mechanical subjects
Composites: Mechanics & Processing, Internal Combustion Engines and Emissions, Material Science and Metallurgy, Analysis and Design of Machine Components, Welding Technology, Heat & Mass Transfer, Strength of Materials, Operations Research, Fluid Mechanics, Mechanics of Machines, Kinematics of Machinery, Destructive & Non Destructive Testing/ Fracture Mechanics, Thermodynamics, Machine Design, Manufacturing Technology
BTech Branch
BTech Mechanical Engineering
Type of class
Crash Course, Regular Classes
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
3 Students Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BSc Tuition
2
BSc Physics Subjects
Quantum Mechanics, Nuclear and Particle Physics, Mechanics, Statistical Physics, Thermal Physics, Solid State Physics, Numerical Analysis, Optics, Mathematical Physics, Electricity and Magnetism, Atomic and Molecular Physics, Mathematics
BSc Computer Science Subjects
Operational Research
Type of class
Crash Course, Regular Classes
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
BSc Branch
BSc Physics, BSc Computer Science, BSc Mathematics
BSc Mathematics Subjects
Calculus, Differential Equations and Mathematical Modelling, Probability and Statistics, Mechanics, Number Theory, Physics, Algebra, Numerical Methods and Programming
5 out of 5 1 review
Ayansh Sinha
"Best teacher you will get. His tips and tricks for concepts of physics are awesome. Great Teacher. Thank you very much for your support. Highly recommend. "
Answered on 21/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
We know, capacitance = dielectric constant × C₀
C = KC₀ = 6 × 18pF = 108pF [ see question - 2.8]
(a) if the voltage supply remained connected , then the potential difference across the capacitor will remain the same. e.g., V = 100V and hence, charge on the capacitor becomes Q = CV = 108pF × 100V
Q = 108 × 10⁻¹² × 100 = 1.08 × 10⁻⁸ C
(b) if the voltage supply was disconnected , then charge on the capacitor reamins the same. e.g., Q = 1.8 × 10⁻⁹C [ see question - 2.8]
And hence the potential difference across the capacitor becomes
V = Q/C = 1.8 × 10⁻⁹/1.08 × 10⁻¹⁰ = 16.6V
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
A regular hexagon ABCDEF of side 10cm has a charge 5 μC at each of its vertices as shown in the figure.
Potential due to charge placed at A, at the centre is Kq/r
Potential due to charge placed at B, at the centre is Kq/r
Potential due to Charge placed at F , at centre is Kq/r
so, total potential at centre of regular hexagon , V = V₁ + V₂ + V₃ + V₄ + V₅ + V₆
so, V = kq/r + kq/r + kq/r + kq/r + kq/r + Kq/r
V = 6kq/r
Hence, k = 9 × 10⁹Nm²/C² , q = 5 × 10⁻⁶C and r = 10cm = 0.1 m
Now, V = 6 × 9 × 10⁹ × 5 × 10⁻⁶/0.1 = 2.7 × 10⁶ Volts.
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
initially d,
therefore, C= EA/d=8 -(1)
now d1=d/2, and dielectric is put,
therefore C1=KEA/d1. -(2) here k=6 which is dielectric constant
we know d1=d/2 substituting it in (1)
C1=6EA/(d/2)=12EA/d. -(3)
(3)÷(1)
C1÷8=12EA/d÷EA/d
C1=96 uF
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
a) We know, charge have nature to reside outer surface of the conductor. It means, charge inside the surface equals zero.
According to Gaussian theorem,
Ф = q/ε ₀, here q is charged inclosed the Gaussian surface.
∵ q = 0
so, Ф = 0 and flux , Ф = E.A = 0
so, E = 0
Hence, inside the sphere, the electric field equals zero.
(b) Take a Gaussian surface of radius r > R = 12cm
then, charged inclosed into the Gaussian surface is q = 1.6 × 10⁻⁷ C
So, Ф = q/ε₀
So, EA = q/ε₀
E = q/ε₀A, here A is the surface area of Gaussian spherical surface
e.g., A = 4πr²
So, E = q/4πε₀r² = 9 × 10⁹ × 1.6 × 10⁻⁷/(12 × 10⁻²)²
= 10⁵ N/C
(C) Similarly explanation of (B),
So, E = kq/r²
Here , k = 9 × 10⁹ Nm²/C² , q = 1.6 × 10⁻⁷C and r = 18cm
So, E = 9 × 10⁹ × 1.6 × 10⁻⁷/(18 × 10⁻²)²
= 4.44 × 10⁴ N/C.
Answered on 18/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
(a) an equipotential surface is a plane on which potential is zero everywhere.
This plane is normal to the plane in which two point charges are placed.
Let normal plane be placed at C of x distance from +2μC charge.
Then, potential due to 2μC + potential due to -2μC = 0
Kq/x + k(-q)/(6 - x) = 0
x = 3 cm
Hence, the equipotential surface of the system appears at midpoint of the system of two given charges, and it is normal to the plane of charges.
(B) we know, the direction of the electric field from positive to negative charge.
So, the direction of the electric field at every point on the surface is normal to the plane in the direction AB.
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