505

You visited us 505 times! Enjoying our articles? Unlock Full Access!

Gauss's Law

The electric ...

Question

The electric field in a region is given by $\to E = \frac{E_{o} x}{l} ˆ i$ . Find the charge contained inside a cubical volume bounded by the surfaces $x = 0, x = a, y = 0, y = a, z = o$ and $z = a$ . Take $E_{o} = 5 \times 10^{3} N / C, l = 2 c m$ and $a = 1 c m$ .

Open in App

Solution

For the surfaces $y = 0, y = a, z = 0, z = a$ , the field lines are parallel to the surface.
Therefore total flux through these surface is zero.
Flux through the surface $x = 0 = a^{2} \times \frac{E_{o} 0}{l} = 0$
Flux through the surface x=a = $a^{2} \times \frac{E_{o} a}{l} = 0.25 V m$
By Gauss's LAW,
net flux = $\frac{q_{e n c l o s e d}}{ϵ}$

$\Rightarrow q_{e n c l o s e d} = 0.25 ϵ = 2.2 \times 10^{- 12} C$

Suggest Corrections

Similar questions

\to E = \frac{E_{0} x}{l} \to i

x = 0, x = a, y = 0, y = a, z = 0, z = a

x \times 10^{- 12} C

x

E_{0} = 5 \times 10^{3} N / C, l = 2 c m, a = 1 c m

\vec{E} = \frac{E_{0} x}{l} \vec{i}

x = 0, x = a, y = 0, y = a, z = 0 and z = a

E_{0} = 5 \times 10^{3} N C^{- 1}, l = 2 cm

\to E = E_{0} \frac{x}{L}^i .

x = 0, x = L, y = 0, z = 0

z = L .

E = \frac{5 \times 10^{3} x}{2}^i N C^{- 1} c m^{- 1}

x = 0, x = 1, y = 0, y = 1, z = 0, z = 1

\to E = \frac{E_{0} x}{l}^i

x = 0, x = a, y = 0, y = a,

z = 0

z = a

E_{0} = 5 \times 10^{3} N/C, l = 2 cm,

a = 1 cm, ε_{0} = 8.86 \times 10^{- 12} C^{2} / {Nm}^{2}

Join BYJU'S Learning Program