Vriti Edustore
Home »
Ask & Discuss
»
Physics
« Back to Discussion
Electricity
10 Jun 2012 22:07:54 IST
0 People liked this
1
224
Electric field
Engineering Entrance
,
Medical Entrance
,
AIPMT
,
JEE Main
,
AIIMS
,
JEE Advanced
,
Physics
,
Electricity
the electric field components in the cube are : Ex=a1/2 Ey=0 Ez=0 in which calculate
(a)the flux through the cube ans. in terms of 'a'.......and
(b)the charge within the cube...ans.in terms of numericals asumming that a=0.1 m
Comments (1)
Free Sign Up!
Preparing for JEE?
Kickstart your preparation with new improved study material - Books & Online Test Series for JEE 2014/ 2015
@ INR 5,443/-
For Quick Info
Find Posts by Topics
Physics
Topics
10272
20999
3868
2300
8075
2968
2790
Mathematics
Chemistry
Biology
Institutes
Parents Corner
Board
Fun Zone
|
Vertical Limit 
Top Contributors
All Time This Month Last Week | 1. |
|
Bipin Dubey
|
| 2. |
|
Himanshu
|
| 3. |
|
Hari Shankar
|
| 4. |
|
edison
|
| 5. |
|
Sagar Saxena
|
| 6. |
|
Yagyadutt Mishr..
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
3362 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011
E=a^0.5
Area=a^2 from one side and same from other side
hence the flux from both the surface will cancel out as the electric field is constant hence answer to both question is zero.