The electric field in a region is given by E- - E0 xL- - Find the charge contained inside a cubical volume bounded by the surface x - 0 - x - L - y - 0 - z - 0 and z - L-

According to Gauss theorem, it can be written as, #10; #10; ∮E⃗· ds⃗ #160; = q/ε _0 #10; #10; ∮E_0( x/L) · #10;dx⃗ #160; = q/ε _0 #10; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of Field Lines

The electric ...

Question

The electric field in a region is given by $\to E = E_{0} \frac{x}{L}^i .$ Find the charge contained inside a cubical volume bounded by the surface $x = 0, x = L, y = 0, z = 0$ and $z = L .$

Open in App

Solution

Suggest Corrections

Similar questions

\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}

\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

\to E = \frac{E_{o} x}{l} ˆ i

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

z = a

E_{o} = 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

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

Join BYJU'S Learning Program