Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

The electric flux for gaussian surface A that enclose the charged particles in free space is (given q1=14nC, q2=78.85nC, q3=56nC ).

seo images

(A) 103(NC1m2)
(B) 103(CN1m2)
(C) 6.32×103(NC1m2)
(D) 6.32×103(CN1m)

Answer
VerifiedVerified
404.7k+ views
1 likes
like imagedislike image
Hint: We know that the electric flux passing through the point P will be dependent on the Gaussian surface and the charges it encloses, consistent with Gauss’s law. Hence, only the charges q1, q2 and q3 will contribute to the electric flux in this region. The formula here for electric flux for gaussian surface used will be: Φelectric=Qε0 where Φelectric is electric flux; Q is the total charges enclosed per permittivity of free space; ε0 is permittivity of free space =8.85×1012NC2m2 .

Complete step by step answer:
As we know, the electric flux due to the charges enclosed in a hypothetical Gaussian surface is always due to the charges enclosed within the surface. There is no contribution to the flux due to charge outside the Gaussian surface.
By calculating the flux by enclosing the charge in a hypothetical surface called Gaussian surface. The Gauss law states that the electric flux is due to some charges enclosed in a Gaussian surface. Thus, the total flux due to charges enclosed per permittivity of free space (ε0) and given by Q=q1+q2+q3=(14+78.8556)nC=8.85nC. Since we know for Charge unit that is nC stands nano Columb and nano is 109. Q=8.85×109C.
Now that we have total charges enclosed per ε0 we have to calculate electric flux for gaussian surface A that enclose the charged particles in free space is given by Φelectric=Qε0
On substitution the values in above equation we get;
 Φelectric=Qε0=8.85×109(C)8.85×1012(NC2m2) .
On further solving and taking denominator to numerator we get;
 Φelectric=109(C)1012(NC2m2)=109(C)×10+12(NC2m+2)
On simplifying we get the value of electric flux:
 Φelectric=10129(NC1m2)=103(NC1m2).
Therefore, correct answer is option A i.e. the electric flux for gaussian surface A that enclose the charged particles in free space is 103(NC1m2) .

Note:
Note that Coulomb's law can be derived from Gauss’s law and vice versa. It is one of Maxwell's four equations of electromagnetic interaction. It can be used to find the electric field due to discrete and continuous charge distributions.
Latest Vedantu courses for you
Grade 11 Science PCM | CBSE | SCHOOL | English
CBSE (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for CBSE students
PhysicsPhysics
ChemistryChemistry
MathsMaths
₹41,848 per year
Select and buy