The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field vecE is applied at an angle (π/4) with the vertical direction. The electric flux through the curved surface of the hemisphere is

Question

The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field vecE is applied at an angle (π/4) with the vertical direction. The electric flux through the curved surface of the hemisphere is (A) π a2 E (B) (π a2 E/√2) (C) (π a2 E/2√2) (D) ((π+2 π a2)E/(2√2)2)

Tardigrade Admin · Accepted Answer

We know that, φ= oint E.dS=E oint dS cos 45Â° In case of hemisphere φcurved=φ circuler Therefore, φ curved=Eπ a2. (1/√2)=(E π a2/√2)

Q. The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field $E$ is applied at an angle $\frac{π}{4}$ with the vertical direction. The electric flux through the curved surface of the hemisphere is

$π a^{2} E$

$\frac{π a ^{2} E}{2}$

$\frac{π a ^{2} E}{2 2}$

$\frac{( π + 2 π a ^{2} ) E}{( 2 2 ) ^{2}}$

We know that,
$ϕ = \oint E . d S = E \oint d S cos 4 5^{°}$
In case of hemisphere
$ϕ_{c u r v e d} = ϕ_{c i rc u l er}$
Therefore, $ϕ_{c u r v e d} = E π a^{2} . \frac{1}{2} = \frac{E π a ^{2}}{2}$

Q. The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field E is applied at an angle 4π​ with the vertical direction. The electric flux through the curved surface of the hemisphere is

πa2E

2​πa2E​

22​πa2E​

(22​)2(π+2πa2)E​