SATHEE: Electrostatics Question 723

Electrostatics Question 723

Question: The electric field intensity at the center of a uniformly charged hemispherical shell is $E_{0} .$ Now two portions of the remaining portion is shown in Fig. If $α = β = π / 3$ , then the electric field intensity at the center due to the remaining portion is

Options:

A) $E_{0} / 3$

B) $E_{0} / 6$

C) $E_{0} / 2$

D) Information incomplete

Show Answer

Answer:

Correct Answer: C

Solution:

[c] The magnitude of electric field intensity due to each part of the hemispherical surface at the center ?O? is same. Suppose it is E. $E + \frac{E}{2} + \frac{E}{2} = E_{0}$ or $2 E = E_{0} or E = \frac{E_{0}}{2}$

Electrostatics Question 723

Question: The electric field intensity at the center of a uniformly charged hemispherical shell is $E_{0} .$ Now two portions of the remaining portion is shown in Fig. If $α = β = π / 3$ , then the electric field intensity at the center due to the remaining portion is

Options:

Answer:

Solution:

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Electrostatics Question 723

Question: The electric field intensity at the center of a uniformly charged hemispherical shell is E0. Now two portions of the remaining portion is shown in Fig. If α=β=π/3 , then the electric field intensity at the center due to the remaining portion is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The electric field intensity at the center of a uniformly charged hemispherical shell is $E_{0} .$ Now two portions of the remaining portion is shown in Fig. If $α = β = π / 3$ , then the electric field intensity at the center due to the remaining portion is