AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

a

$(π + 2) {πa}^{2} E / (2 \sqrt{2})$

b

$π a^{2} E$

c

$π a^{2} E / (2 \sqrt{2})$

d

$π a^{2} E / \sqrt{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Given Data: Electric field is applied to the hemisphere as shown below.

Diagram, engineering drawing

Description automatically generated

Concept Used: Gauss Law

Detailed Solution:

By Gauss law,

$ϕ = \oint E . d S = E \oint d S \cos 45 °$

Here, for the hemisphere, $ϕ_{curved} = ϕ_{circular}$

Thus, $ϕ_{curved} = E π a^{2} \cdot \frac{1}{\sqrt{2}} = \frac{E π a^{2}}{\sqrt{2}}$

Watch 3-min video & get full concept clarity

courses

No courses found

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

gplay

mentor

Download the App

gplay

personalised 1:1 online tutoring