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Ask QuestionPosted by Abhishek Yadav 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
Considera dipole with charges +q and -q placed in a uniform eletric field as shown in figure.
In unifor electric field, dipole experiences a torque τ, which is given by
τ = p×E
where p is electric dipole moment. Torque τ will tend to rotate the dipole if p is not parallel to E.
Suppose an external torque τext is applied to neutalize this torque and rotates the dipole from an angle θo to θ1
at an infinitesimal angular speed and without angular acceleration. The amount of workdon by the external torque will be given by
W =
This work is stored as the potential energy of the system.
We can then assosciate potential energy U(θ) with an inclination θ of the dipole. Inorder to choose the
reference angular position where potential Energy is considered as zero, we select θo as π/2 as the reference angular position.
We can then write U(θ) = pE( cos(π/2) - cosθ) = -pE cosθ = -p·E
when θ = 0 , i.e., p is aligned in the direction of E, potential energy is minimum.
when θ = π , i.e., p is aligned in the oposite direction of E, potential energy is maximum
Posted by Harsh Yadav 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
Given:-
Magnetic filed of the earth
Equatorial magnetic field of the earth is given by,
Where,
Therefore,
Earth's magnetic dipole moment is given by,
_____________
Posted by Thakur Ajab Singh 5 years, 2 months ago
- 2 answers
Yogita Ingle 5 years, 2 months ago
Kirchhoff’s First law: Junction law
- Junction Law is also known as Kirchhoff’s First Law.
- It states that at the junction,sum of current entering the junction is equal to the sum of current leaving the junction.
- Junction is any point in the circuit.
- Consider a case where I1 and I2are the currententering the junction and,currentI4 and I5 are exiting out of the junction.
- According to Kirchhoff’s law; I1+ I2 =I3+ I4+ I5
Ayush Vishwakarma?? 5 years, 2 months ago
Posted by Abhishek Kumar Yadav 5 years, 2 months ago
- 2 answers
Vishal Rajput 5 years, 2 months ago
Gaurav Seth 5 years, 2 months ago
On July 7, HRD Minister Ramesh Pokhriyal announced a major CBSE syllabus reduction with 30% of the syllabus slashed for the year 2020-21 for classes 9 to 12 because of the reduction in classroom teaching time due to the Covid-19 pandemic and lockdown.
Deleted syllabus of CBSE Class 12 Physics
Posted by Chetna Bisht 5 years, 2 months ago
- 2 answers
Gaurav Seth 5 years, 2 months ago
The electric flux through a closed Gaussian surface depends upon Net charge enclosed and permittivity of the medium
The electric flux through a closed Gaussian surface is given by:
∮E.−→ds→=q∈∮E.→ds→=q∈
Where, q is the net charge enclosed by the Gaussian and∈∈is the permittivity of the medium.
Posted by Abhishek Kumar Yadav 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
CBSE stated in its notification that though 30% of the syllabus has been cut from the new academic year, teachers should ensure that students understand these topics so that they can use the information in connection with other topics.
Deleted syllabus of CBSE Class 12 Physics
Posted by Bhavika Singh 5 years, 2 months ago
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Posted by Rahul Yadav 5 years, 2 months ago
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Posted by Bharat Kumar 5 years, 2 months ago
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Ayush Vishwakarma?? 5 years, 2 months ago
Shivam Verma 5 years, 2 months ago
Posted by Devil ? 5 years, 2 months ago
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Devil ? 5 years, 2 months ago
Gaurav Seth 5 years, 2 months ago
On July 7, HRD Minister Ramesh Pokhriyal announced a major CBSE syllabus reduction with 30% of the syllabus slashed for the year 2020-21 for classes 9 to 12 because of the reduction in classroom teaching time due to the Covid-19 pandemic and lockdown.
Deleted topics
Posted by Supriya Tiwari 5 years, 2 months ago
- 3 answers
Rohtash Dagar 5 years, 2 months ago
Posted by Supriya Tiwari 5 years, 2 months ago
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Aditya Mishra 5 years, 2 months ago
Posted by Nisha Yadav 5 years, 2 months ago
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Nargis Ahmed 5 years, 2 months ago
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Nargis Ahmed 5 years, 2 months ago
Posted by Anchal Chaudhary 5 years, 2 months ago
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Ayush Vishwakarma?? 5 years, 2 months ago
Posted by Pavi Nisha Nisha 5 years, 2 months ago
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Posted by Jatin Agarwal 5 years, 2 months ago
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Meghna Thapar 5 years, 2 months ago
Parallel Plate Capacitors are formed by an arrangement of electrodes and insulating material or dielectric. ... When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor. A parallel plate air core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduce between the two plates, then. A . some charge from the capacitor will flow back into the source.
Posted by Vîđhî Hööđã 5 years, 2 months ago
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Posted by Bhavika Singh 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
Factor on which the potential gradient of the potentiometer wire depend
-current passing through the potentio meter wire.
-specific resistance of the material of the potentiometer wire.
-area of cross section of the wire
Posted by Nargis Ahmed 5 years, 2 months ago
- 3 answers
Posted by Sudhanshu Yadav 5 years, 2 months ago
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Meghna Thapar 5 years, 2 months ago
In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of electric field. Relation between Electric Intensity and Potential Gradient. The change of electric potential with respect to distance is called potential gradient. It is denoted by dv/dx. hence, the negative of potential gradient is equal with electric field intensity.
Posted by Ayush Vishwakarma?? 5 years, 2 months ago
- 1 answers
Mr. Badmash 5 years, 2 months ago
Posted by Ayush Vishwakarma?? 5 years, 2 months ago
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Mr. Badmash 5 years, 2 months ago
Posted by Mr. Badmash 5 years, 2 months ago
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Ayush Vishwakarma?? 5 years, 2 months ago
Posted by Chimik Danme Sangma 5 years, 2 months ago
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Posted by Mr. Badmash 5 years, 2 months ago
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Ayush Vishwakarma?? 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
According to Gauss’ Theorem, the total electric flux through all the six faces of the cube is,
φ = Charge enclosed / ε 0
φ = q /ε 0
Each face of the cube comprises 1/6 of the total surface of the cube. Therefore, the flux passing through each face is,
φ = 1/6 ? q/ε 0 = q /6 ε 0
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