Home
Class 12
PHYSICS
Answer the following: State Gauss' law...

Answer the following:
State Gauss' law in electrostatics. Derive an expression for the electric field due to an infinitely long straight uniformly charged wire.

Text Solution

AI Generated Solution

### Step-by-Step Solution **Step 1: State Gauss's Law in Electrostatics** Gauss's Law states that the total electric flux (Φ_E) through a closed surface is equal to the charge (Q) enclosed by that surface divided by the permittivity of free space (ε₀). Mathematically, it can be expressed as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0} ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise PHYSICS (Theory) DELHI BOARD -2016 [SET -I][COMPTT.]|11 Videos
  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise C.B.S.E.CLASS-XII PHYSICS (THEORY) [SET-I] (SECTION-A)|5 Videos
  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise SECTION-B|52 Videos
  • SAMPLE PAPER 2019

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise SECTION D|6 Videos

Similar Questions

Explore conceptually related problems

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

State Gauss's law in electrostatics. Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density lamda cm^(-1)

Answer the following: (a) State Gauss' law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density lamda . (b) A wire AB of length L has linear charge density lamda = kx , where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for the electric flux through this surface.

State Gauss's law in electrostatics. Using this law derive an expression for the electric field due to a uniformly charged infinite plane sheet.

Derive an expression for electric field intensity at a point due to point charge.

State Gauss's theorem in electrostatics. Apply this theorem to derive an expression for electric field intensity at a point outside a uniformly charged thin spherical shell.

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.

Answer the following: (i) Use Gauss' law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities? (ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C_(1) and C_(2) with their capacitance in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.

(a) Using Gauss's law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius R and charge density sigma C//m^(2) . Draw the field lines when the charge density of the sphere is (i) positive, (ii) negative. (b) A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of 100 mu C//m^(2) . Calculate the (i) charge on the sphere (ii) total electric flux passing through the sphere.

(a) Use Gauss's law to derive the expression for the electric filed (vecE) due to straight uniformaly charges infinite line of charges density lambda C//m . (b) Draw a graph to show the variation of E with perpendicular from the line of charge. (c) Find the work done in brining a charge q from prependicular distance r_(1) to r_(2) (r_(2) gt r_(1)) .