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A disc of radius a/4 having a uniformly distributed charge 6C is placed in the xy plane with centre at (-a / 2, 0, 0 ) A rod of length a carrying a uniformly distributed charge 8c is placed on the x-axis from x = a / 4 to x =5a / 42 . Two point Charges -7 c & 3c are placed at (a / 4, -a/ 4 , 0) and ( - 3a / 4, 3a / 4, 0 ) respectively consider a cubical surface formed by 6 surface x=+- a / 2, y=+-a / 2, z =+- a / 2 . The electric flux through the cubicle surface is (a) -2C/E? Please explain in detail

```
6 years ago

Arun Kumar
IIT Delhi
256 Points
```							Hi Ashu,flux=charge enclosed/epsilon$\phi_e=Q/\epsilon$Here if we try to get the charge enclosed then$Q_{1}=a/4*8c/(5a/4-a/4)=2Q$$Q_2=-7C$3C will not be enclosedHalf of the disc will be in the cubeso$Q_3=3C$$Q_T=-2C$Thanks & RegardsArun KumarIIT DelhiAskiitians Faculty
```
6 years ago
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### Course Features

• 731 Video Lectures
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions