When the electric field pierces outward through the section.
Answer:
Explanation:
We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .
∫ E ds = q / ε
Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so
∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .
This also represents total flux coming out of the charge q on all sides .
This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .
If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .
It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors
1 ) charge q and
2 ) the permittivity of medium ε around .
Answer:
1.5 * 10^-2 Tm^2
Explanation:
Electric Flux = B.A cos(theta)
B = 0.055 T
A = 0.32 m^2
theta = 30
Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2
