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A spherical piece of radius much less than the radius of a charged spherical shell (change density σ = 1 c/m3) is removed from the shell itself than calculate the electric field intensity at the midpoint of the aperture?
 (in the order of 109)
    Correct answer is '564.9'. Can you explain this answer?
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    A spherical piece of radius much less than the radius of a charged sph...
    The aperfure midpoint will act like the midpoint of a capacitor, thus
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    A spherical piece of radius much less than the radius of a charged sph...
    The aperfure midpoint will act like the midpoint of a capacitor, thus
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    A spherical piece of radius much less than the radius of a charged sph...
    Correct answer is 56.49 ,the given answer is wrong
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    A spherical piece of radius much less than the radius of a charged sphericalshell (change density σ = 1 c/m3) is removed from the shell itself thancalculate the electric field intensity at the midpoint of the aperture?(in the order of 109)Correct answer is '564.9'. Can you explain this answer?
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    A spherical piece of radius much less than the radius of a charged sphericalshell (change density σ = 1 c/m3) is removed from the shell itself thancalculate the electric field intensity at the midpoint of the aperture?(in the order of 109)Correct answer is '564.9'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A spherical piece of radius much less than the radius of a charged sphericalshell (change density σ = 1 c/m3) is removed from the shell itself thancalculate the electric field intensity at the midpoint of the aperture?(in the order of 109)Correct answer is '564.9'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical piece of radius much less than the radius of a charged sphericalshell (change density σ = 1 c/m3) is removed from the shell itself thancalculate the electric field intensity at the midpoint of the aperture?(in the order of 109)Correct answer is '564.9'. Can you explain this answer?.
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