Home
Class 12
PHYSICS
(a) Using Gauss's law, derive expression...

(a) Using Gauss's law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire.
(b) The electric field components in the following figure are `E_(x)=alphax, E_(y)=0, E_(z)=0`, in which `alpha=400N//Cm`. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that `a=0.1 m`.

Text Solution

Verified by Experts

(a) Consider an infinitely long straight charged wire of linear charge density `lambda`. To find electric field at a point P situated at a distance r from the wire by using Gauss. law consider a cylinder of length l and radius r as the Gaussian surface.

From symmetry consideration electric field at each point of its curved surface is `vecE` and is pointed outwards normally. Therefore, electric flux over the curved surface
`=intvecE.hatn ds=E 2pirl`
On the side faces 1 and 2 of the cylinder normal drawn on the surface is perpendicular to electric field E and hence these surfaces do not contribute towards the total electric flux.
`therefore `Net electric flux over the entire Gaussian surface `phi_(E)=E.2pirl" ....(i)"`
By Gauss law electric flux `phi_(E)=(1)/(in_(0))" (charged enclosed)"=(lambdal)/(in_(0))" ...(ii)"`
Comparing (i) and (ii), we have
`E.2pirl =(lambdal)/(in_(0))`
`rArr" "E=E=(lambda)/(2pi in_(0)r) and vecE=(lambda)/(2pi in_(0)r)hatr`
(b) (i) Here, `E_(x)=alpha. x=400x, E_(y)=E_(z)=0 and a=0.1m`. So the electric flux is linked with only two fa c es of the cube lying in Y -Z plane (i.e., perpendicular to `E_(x)`).
As position of left face of cube is at `x=a=0.1m`, hence `(E_(x))_(1)=400xx0.1=40NC^(-1)` (inward into the Cube)
Now position of right face of cube is `x=a+a=2a=0.2m`,
hence `(E_(x))_(2)=400xx0.2=80NC^(-1)`.
`therefore" flux "phi_(2)=(E_(x))_(2).A^(2)=80xx0.01=0.8Nm^(2)C^(-1)" (outward)"`
`therefore` Net flux `phi_(2)=(E_(x))_(2.)A^(2)=80xx0.01=0.8Nm^(2)C^(-1)" (outward)"`
`therefore` Net flux through the cube `phi_(E)=phi_(2)-phi_(1)=0.8-0.4=0.4Nm^(2)C^(-1)` outward
(ii) Charge within the cube `q=in_(0).phi_(E)=(8.854xx10^(-12))xx0.4=3.54xx10^(-12)C`
Promotional Banner

Topper's Solved these Questions

  • CBSE SAMPLE QUESTION PAPER 2019-20 (SOLVED)

    U-LIKE SERIES|Exercise SECTION C|8 Videos

  • CBSE EXAMINATION PAPER 2020 (SOLVED)

    U-LIKE SERIES|Exercise SECTION - D|6 Videos

  • CURRENT ELECTRICITY

    U-LIKE SERIES|Exercise SELF ASSESSMENT TEST (SECTION C )|3 Videos

Similar Questions

Explore conceptually related problems

Using Gauss’s law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire. The electric field components in the following figure are E_(x) = alphax, E_(y) = 0, E_(z) = 0, " in which " alpha = 400 N//C m. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a = 0.1m.

Using Gauss’ law, derive an expression for the electric field at a point near an infinitely long straight uniformly charged wire.

Using Gauss's law, derive an expression for the electric field intensity at any point near a uniformly charged thin wire of charg e//l eng th = lambda C//m .

The electric components in the figure are E_(x)=alphax^(1//2) ,E_(y)=0,E_(z)=0 where alpha=800N//m^(2) if a=0.1m is the side of cube then the charge with in the cube is

Define electric flux and write its SI unit . The electric field components in the figure shown are : E_(x)=alphax,E_(y)=0,E_(z)=0 where alpha=(100N)/(Cm) . Calculate the charge within the cube , assuming a = 0.1 m .

(a) Define electric flux. Write its SI units. (b) The electric field components due to a charge inside the cube of side 0.1 m are as shown : E_(x)=alpha x, where alpha=500N//C-m E_(y)=0, E_(z)=0 . Calculate (i) the flux through the cube, and (ii) the charge inside the cube.

The electric field componets due to a charge inside the cube of side 0.1m are E_(x) = alpha x, where alpha = 500 (N//C) m^(-1) , E_(y) = 0, E_(z) = 0 . Calculate the flux through the cube and the charge inside the cube.