Chaukaghat, Varanasi, India - 221002.
Verified 3
Details verified of Abhimanyu Kumar✕
Identity
Education
Know how UrbanPro verifies Tutor details
Identity is verified based on matching the details uploaded by the Tutor with government databases.
Hindi Mother Tongue (Native)
English Proficient
Aktu 2018
Bachelor of Technology (B.Tech.)
Chaukaghat, Varanasi, India - 221002
ID Verified
Education Verified
Phone Verified
Email Verified
Facebook Verified
Report this Profile
Is this listing inaccurate or duplicate? Any other problem?
Please tell us about the problem and we will fix it.
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
State, CBSE, ICSE
CBSE Subjects taught
Mathematics, Science, Computer Practices
ICSE Subjects taught
Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics, Science
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
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Physics, Computer Science, Mathematics
CBSE Subjects taught
Mathematics, Physics, Computer Science
Taught in School or College
No
State Syllabus Subjects taught
Computer Science, Physics, Mathematics
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
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Physics, Mathematics
CBSE Subjects taught
Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Physics, Mathematics
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
Strength of Materials, Heat & Mass Transfer, Internal Combustion Engines and Emissions, Thermodynamics, Material Science and Metallurgy, Mechanics of Machines, Operations Research, Welding Technology, Fluid Mechanics, Manufacturing Technology, Analysis and Design of Machine Components, Kinematics of Machinery, Machine Design, Destructive & Non Destructive Testing/ Fracture Mechanics, Composites: Mechanics & Processing
BTech Branch
BTech Mechanical Engineering
Type of class
Regular Classes, Crash Course
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
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
Thermal Physics, Statistical Physics, Mathematical Physics, Nuclear and Particle Physics, Atomic and Molecular Physics, Solid State Physics, Numerical Analysis, Mechanics, Electricity and Magnetism, Mathematics, Optics, Quantum Mechanics
BSc Computer Science Subjects
Operational Research
Type of class
Regular Classes, Crash Course
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
BSc Branch
BSc Computer Science, BSc Mathematics, BSc Physics
BSc Mathematics Subjects
Numerical Methods and Programming, Calculus, Algebra, Differential Equations and Mathematical Modelling, Physics, Number Theory, Probability and Statistics, Mechanics
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?
State, CBSE and ICSE
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.
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
State, CBSE, ICSE
CBSE Subjects taught
Mathematics, Science, Computer Practices
ICSE Subjects taught
Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics, Science
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
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Physics, Computer Science, Mathematics
CBSE Subjects taught
Mathematics, Physics, Computer Science
Taught in School or College
No
State Syllabus Subjects taught
Computer Science, Physics, Mathematics
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
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Physics, Mathematics
CBSE Subjects taught
Mathematics, Physics
Taught in School or College
No
State Syllabus Subjects taught
Physics, Mathematics
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
Strength of Materials, Heat & Mass Transfer, Internal Combustion Engines and Emissions, Thermodynamics, Material Science and Metallurgy, Mechanics of Machines, Operations Research, Welding Technology, Fluid Mechanics, Manufacturing Technology, Analysis and Design of Machine Components, Kinematics of Machinery, Machine Design, Destructive & Non Destructive Testing/ Fracture Mechanics, Composites: Mechanics & Processing
BTech Branch
BTech Mechanical Engineering
Type of class
Regular Classes, Crash Course
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
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
Thermal Physics, Statistical Physics, Mathematical Physics, Nuclear and Particle Physics, Atomic and Molecular Physics, Solid State Physics, Numerical Analysis, Mechanics, Electricity and Magnetism, Mathematics, Optics, Quantum Mechanics
BSc Computer Science Subjects
Operational Research
Type of class
Regular Classes, Crash Course
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
BSc Branch
BSc Computer Science, BSc Mathematics, BSc Physics
BSc Mathematics Subjects
Numerical Methods and Programming, Calculus, Algebra, Differential Equations and Mathematical Modelling, Physics, Number Theory, Probability and Statistics, Mechanics
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.
Share this Profile
Also have a look at
Reply to 's review
Enter your reply*
Your reply has been successfully submitted.
Certified
The Certified badge indicates that the Tutor has received good amount of positive feedback from Students.