Q. A charge Q is uniformly distributed over a spherical surface of radius R. What is the electric field at a point outside the sphere at distance r from the center?
A.
0
B.
Q/4πε₀r²
C.
Q/4πε₀R²
D.
Q/4πε₀R
Solution
For points outside the sphere, the electric field behaves as if all the charge were concentrated at the center, so E = Q/4πε₀r².
Q. A cylindrical conductor of radius R carries a uniform charge per unit length λ. What is the electric field at a distance r from the axis of the cylinder (r > R)?
A.
0
B.
λ/(2πε₀r)
C.
λ/(2πε₀R)
D.
λ/(4πε₀r²)
Solution
For a point outside the cylinder, the electric field is given by E = λ/(2πε₀r).
Q. A cylindrical Gaussian surface encloses a charge Q. If the height of the cylinder is doubled while keeping the radius constant, what happens to the electric flux through the curved surface?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
Solution
The electric flux through the curved surface is proportional to the charge enclosed, which remains constant, so the flux through the curved surface doubles if the height is doubled.
Q. A cylindrical Gaussian surface encloses a charge Q. If the radius of the cylinder is r and its height is h, what is the electric flux through the curved surface?
A.
Q/ε₀
B.
Q/(2ε₀)
C.
Q/(4ε₀)
D.
0
Solution
The electric flux through the curved surface of a cylinder is given by Φ = Q_enc/ε₀, where Q_enc = Q.
Q. A cylindrical Gaussian surface of length L and radius R encloses a charge Q uniformly distributed along its length. What is the electric field at a distance R from the axis of the cylinder?
A.
Q/(2πε₀R)
B.
Q/(4πε₀R²)
C.
0
D.
Q/(ε₀L)
Solution
Using Gauss's law, the electric field outside the cylinder is E = Q/(2πε₀R).
Q. A cylindrical Gaussian surface of length L and radius R encloses a charge Q. What is the electric field E at a distance R from the axis of the cylinder?
A.
Q/(2πε₀R)
B.
Q/(4πε₀R²)
C.
Q/(ε₀L)
D.
0
Solution
Using Gauss's law, the electric field E at a distance R from the axis of a long charged cylinder is E = Q/(2πε₀L) for points outside the cylinder.
Q. A point charge of +Q is placed at the center of a spherical shell of radius R with surface charge density σ. What is the electric field inside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
σ/ε₀
D.
Q/(4πε₀R)
Solution
According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.
Q. A uniformly charged sphere of radius R has a total charge Q. What is the electric field at a point outside the sphere (r > R)?
A.
0
B.
Q/(4πε₀r²)
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
Solution
For a uniformly charged sphere, the electric field outside the sphere behaves as if all the charge were concentrated at the center, thus E = Q/(4πε₀r²).
Gauss Law is a fundamental principle in electrostatics that plays a crucial role in various exams. Understanding this law not only helps in grasping key concepts but also enhances your ability to tackle objective questions effectively. Practicing MCQs related to Gauss Law can significantly improve your exam preparation and boost your confidence in solving important questions.
What You Will Practise Here
Understanding the statement and mathematical formulation of Gauss Law.
Applications of Gauss Law in calculating electric fields for symmetrical charge distributions.
Deriving Gauss's Law from Coulomb's Law and vice versa.
Key concepts such as electric flux and its significance in Gauss Law.
Solving problems involving spherical, cylindrical, and planar symmetry.
Identifying and correcting common misconceptions related to Gauss Law.
Diagrams illustrating electric field lines and flux through closed surfaces.
Exam Relevance
Gauss Law is frequently featured in CBSE, State Boards, NEET, and JEE examinations. Students can expect questions that require them to apply the law to various charge configurations and calculate electric fields. Common question patterns include direct application of Gauss Law, conceptual questions about electric flux, and problem-solving scenarios that involve symmetry. Mastering this topic is essential for achieving high scores in competitive exams.
Common Mistakes Students Make
Misunderstanding the concept of electric flux and its dependence on the angle of the surface.
Failing to recognize the importance of symmetry in simplifying problems.
Confusing the application of Gauss Law with Coulomb's Law in certain scenarios.
Overlooking the conditions under which Gauss Law is applicable.
FAQs
Question: What is Gauss Law? Answer: Gauss Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space.
Question: How can I apply Gauss Law to find the electric field of a charged sphere? Answer: By using a spherical Gaussian surface, you can apply Gauss Law to derive the electric field outside and inside the charged sphere.
Now is the time to enhance your understanding of Gauss Law! Dive into our practice MCQs and test your knowledge to ensure you are well-prepared for your upcoming exams. Remember, consistent practice is the key to success!
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