A point charge -q is placed the centre of a cube of side L- The electric flux emerging from the cube is-frac -q-epsilon-0-Zerofrac -6qL-2-epsilon-0-frac -q-6L-2epsilon-0-

Question

Toppr Admin · Accepted Answer

Applying Gauss-apos-s law- electric flux -x3D5-x222B-x2192-E-nds-Qnet-x3F5-0therefore- -x3D5-q-x3F5-0Ans- -A-

A point charge +q is placed at the centre of a cube of side L. The electric flux emerging from the cube is-
$\frac{q}{ϵ_{0}}$
Zero
$\frac{6 q L^{2}}{ϵ_{0}}$
$\frac{q}{6 L^{2} ϵ_{0}}$

Applying Gauss's law, electric flux $(ϕ) = \int \to E .^n d s = \frac{Q_{n e t}}{ϵ_{0}}$
therefore, $ϕ = \frac{q}{ϵ_{0}}$
Ans: (A)

A point charge +q is placed at the centre of a cube of side L. The electric flux emerging from the cube is-qϵ0Zero6qL2ϵ0q6L2ϵ0

Applying Gauss's law, electric flux (ϕ)=∫→E.^nds=Qnetϵ0therefore, ϕ=qϵ0Ans: (A)

A point charge +q is placed at the centre of a cube of side L. The electric flux emerging from the cube is-
$\frac{q}{ϵ_{0}}$
Zero
$\frac{6 q L^{2}}{ϵ_{0}}$
$\frac{q}{6 L^{2} ϵ_{0}}$

Applying Gauss's law, electric flux $(ϕ) = \int \to E .^n d s = \frac{Q_{n e t}}{ϵ_{0}}$
therefore, $ϕ = \frac{q}{ϵ_{0}}$
Ans: (A)