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3 years ago

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A given 60uc piont charge located at origion

2 answers:

goblinko [34]3 years ago

7 0

For my first part it like that as we all know that flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc M I RIGHT PLZ tell me

Morgarella [4.7K]3 years ago

6 0

Answer:

flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc

Explanation:

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Answer:

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Explanation:

Given:

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densk [106]

Answer:

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Explanation:

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$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

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$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

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