Question

A hollow cylindrical box of length 1m and area of cross-section 25cm²  is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by vector E =50x cap i, where E is in NC^–1 and x is in metres. Find 

(i) Net flux through the cylinder. 

(ii) Charge enclosed by the cylinder. 

Answer 1

(i) Electric flux through a surface \(\phi = \vec E.\vec S\)

Flux through the left surface, \(\phi _L =-|E||S|\)

\(= -50 x, |S|\)

Since x = 1m

\(\phi _L \) = -50 × 1 × 25 × 10^-4

= -1250 × 10^-4 

= -0.125 Nm²C^-1

Flux through the right surface,

\(\phi _R =|E||S|\)

Since x = 2m,

\(\phi _R \) = 50 × 2 × |s|

= 50 × 2 × 25 × 10^-4

= 2500 × 10^-4

= 0.250 Nm²C^-1

Now, flux through the cylinder

\(\phi_{Ace} = \phi_R + \phi _L\)

= 0.250 - 0.125

= 0.125 Nm²C^-1

(ii) Using Gauss Theorem, we can calculate the charge inside the cylinder 

\(\phi_{net} = \frac q{\in_0}\)

⇒ \(q = \in_0\phi _{net}\)

\(= 8.854 \times 10^{-12} \times 0.125\)

\(= 8.854 \times 10^{-12} \times \frac 18\)

\(= 8.854 \times 10^{-12} \times \frac 18\)

Answer 2

(i) Electric flux through a surface

Net flux through the cylinder