A point charge of +12 μ C is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be× 103 Nm 2 / C.

Question

A point charge of +12 μ C is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be× 103 Nm 2 / C. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From symmetry φ =(1/6)(( q /ε0)) =(12 × 10-6/6 × 8.85 × 10-12) =225.98 × 103 ( Nm 2/ s ) ≃ 226 × 103 ( Nm 2/ C)

Q. A point charge of $+ 12 μ C$ is at a distance $6 c m$ vertically above the centre of a square of side $12 c m$ as shown in figure. The magnitude of the electric flux through the square will be_______ $\times 1 0^{3} N m^{2} / C$ .

From symmetry $ϕ = \frac{1}{6} (\frac{q}{ε _{0}})$
$= \frac{12 \times 1 0 ^{- 6}}{6 \times 8.85 \times 1 0 ^{- 12}}$
$= 225.98 \times 1 0^{3} \frac{N m ^{2}}{s}$
$≃ 226 \times 1 0^{3} \frac{N m ^{2}}{C}$

Q. A point charge of +12μC is at a distance 6cm vertically above the centre of a square of side 12cm as shown in figure. The magnitude of the electric flux through the square will be_______×103Nm2/C.

From symmetry ϕ=61​(ε0​q​) =6×8.85×10−1212×10−6​ =225.98×103sNm2​ ≃226×103CNm2​

Q. A point charge of $+ 12 μ C$ is at a distance $6 c m$ vertically above the centre of a square of side $12 c m$ as shown in figure. The magnitude of the electric flux through the square will be_______ $\times 1 0^{3} N m^{2} / C$ .

From symmetry $ϕ = \frac{1}{6} (\frac{q}{ε _{0}})$
$= \frac{12 \times 1 0 ^{- 6}}{6 \times 8.85 \times 1 0 ^{- 12}}$
$= 225.98 \times 1 0^{3} \frac{N m ^{2}}{s}$
$≃ 226 \times 1 0^{3} \frac{N m ^{2}}{C}$