Q. 354.6( 34 Votes )

# (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 Ex = αx, Ey = 0, Ez= 0; in which α = 400 N/C m. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a = 0.1m.

**OR**

(a) Define electrostatic potential at a point. Write its SI unit.

Three charges q1, q2 and q3 are kept respectively at points A, B and C as shown in figures. Write the expression for electrostatic potential energy of the system.

(b) Depict the equipotential surfaces due to

(i) an electric dipole

(ii) two identical negative charges separated by a small distance.

Answer :

According to Gauss’s law, the net electric flux through the closed surface is equal to 1/ times of the charge enclosed in the surface i.e.,

Let us consider the infinitely long wire carrying uniform charge. Now, we consider a hypothetical gaussian surface which is cylindrical in shape. The surface has radius r and length l and it placed coaxial with the current carrying wire as shown in the figure

We have three parts in the gaussian surface, (i) surface S_{1} (ii) surface S_{2} and (iii) surface S_{3}

Electric field lines are parallel to the direction of normal to the surface S_{1} i.e. radially outward. For surfaces S_{2} and S_{3} the electric field makes right angle with the normal of the surface.

Now, let us consider small elements from each surface and then formulate it for whole gaussian surface.

By using to Gauss’s law

First solving L. H. S.

This implies

The electric field along the x- axis. Hence the electric field passes only right and left of the cube.

ϕ=E.S

(i) Electric flux is the total number of electric field lines passing through the surface which is perpendicular to the electric field. According to gauss’s theorem, flux is

It is given to us that Ex = αx, Ey = 0, Ez= 0 and α = 400 N/C m

The electric field is present in only x- direction, hence flux is

for the faces of cube facing Y- axis and Z- axis

Now, the electric field from left face of cube facing negative X- direction is E= αa where α is 400N/Cm and a is the side of the cube. Flux through this side,

Electric field through the right side of the cube, E= α (2a)

Flux through the right side of the cube,

Net flux through the cube =

(ii) the charge within the cube

**OR**

Electrostatic potential at any point, is defined as the amount of work done to bring the single test charge from infinity to that point.

It can also be defined as the difference in potential energy per unit charge between the two points in the presence of electric field. The S.I. unit of electric potential is volt (V) i.e. Joule per coulomb J/C

To determine the electrostatic potential energy of the system, let three charges q_{1}, q_{2}, and q_{3} have position vectors r_{1}, r_{2} and r_{3} with respect to common origin O. the potential energy of the charges q_{1} and q_{2},

Similarly, the potential energy between q_{2} and q_{3} is given by

And the potential energy between q_{3} and q_{1} is given by

Hence, net electric potential energy is given by

(b) An equipotential surface is a surface where the value of potential is having constant value at all the points on the surface. Therefore, no work is done when we move the charge from one point to another on equipotential surface.

(i) due to an electric dipole

(ii) due to two identical negative charges separated by a small distance

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A sample of HCl gas is placed in an electric field of 2.5 × 10^{4} N C^{–1}. The dipole moment of each HCl molecule is 3.4 × 10^{–30} Cm. Find the maximum torque that can act on a molecule.

Three charges are arranged on the vertices of n equilateral triangle as shown in figure. Find the dipole moment of the combination.

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HC Verma - Concepts of Physics Part 2