SATHEE: Electrostatics Question 748

Electrostatics Question 748

Question: One-fourth of a sphere of radius R is removed as shown in Fig. An electric field E exists parallel to the xy plane. Find the flux through the curved part.

Options:

A) $π R^{2} E$

B) $\sqrt{2} π R^{2} E$

C) $π R^{2} E / \sqrt{2}$

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $ϕ_{p l a i n} + ϕ_{c u r v e} = 0 or ϕ_{p l a i n} = - ϕ_{c u r v e}$

${\vec{A}}_{1} = - \frac{π R^{2}}{2} \hat{i}, {\vec{A}}_{2} = - \frac{π R^{2}}{2} \hat{j}$

$\vec{E} = E \cos 45^{\circ} \hat{i} + E \sin 45^{\circ} \hat{j}$

$= \frac{E}{\sqrt{2}} \hat{i} + \frac{E}{\sqrt{2}} \hat{j} and$

$ϕ = \vec{E} . ({\vec{A}}_{1} + {\vec{A}}_{2})$

$= \frac{- E}{\sqrt{2}} \frac{π r R^{2}}{2} - \frac{E}{\sqrt{2}} \frac{π r R^{2}}{2} = \frac{- π R^{2} E}{\sqrt{2}}$ This is the flux entering. So flux is $\frac{π R^{2} E}{\sqrt{2}}$

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Electrostatics Question 748

Question: One-fourth of a sphere of radius R is removed as shown in Fig. An electric field E exists parallel to the xy plane. Find the flux through the curved part.

Options:

Answer:

Solution:

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