A point charge + q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is

Question

A point charge + q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is (A)  (q/ε 0)  (B)  (q/6L2ε 0)  (C)  (6qL2/ε 0)  (D) zero

Tardigrade Admin · Accepted Answer

From Gauss theorem, the electric flux (φ ) through any closed surface is equal to (1/ε 0) times the net charge q enclosed by the surface. Thus, φ =∫E.dA=(q/ε 0)

Q. A point charge + q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is

$\frac{q}{ε _{0}}$

$\frac{q}{6 L ^{2} ε _{0}}$

$\frac{6 q L ^{2}}{ε _{0}}$

zero

From Gauss theorem, the electric flux $(ϕ)$ through any closed surface is equal to $\frac{1}{ε _{0}}$ times the net charge q enclosed by the surface. Thus,
$ϕ = \int E . d A = \frac{q}{ε _{0}}$

Q. A point charge + q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is

ε0​q​

6L2ε0​q​

ε0​6qL2​

zero

From Gauss theorem, the electric flux (ϕ) through any closed surface is equal to ε0​1​ times the net charge q enclosed by the surface. Thus, ϕ=∫E.dA=ε0​q​

$\frac{q}{ε _{0}}$

$\frac{q}{6 L ^{2} ε _{0}}$

$\frac{6 q L ^{2}}{ε _{0}}$

From Gauss theorem, the electric flux $(ϕ)$ through any closed surface is equal to $\frac{1}{ε _{0}}$ times the net charge q enclosed by the surface. Thus,
$ϕ = \int E . d A = \frac{q}{ε _{0}}$