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.
(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.
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 S1 (ii) surface S2 and (iii) surface S3
Electric field lines are parallel to the direction of normal to the surface S1 i.e. radially outward. For surfaces S2 and S3 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.
The electric field along the x- axis. Hence the electric field passes only right and left of the cube.
(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
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 q1, q2, and q3 have position vectors r1, r2 and r3 with respect to common origin O. the potential energy of the charges q1 and q2,
Similarly, the potential energy between q2 and q3 is given by
And the potential energy between q3 and q1 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
Rate this question :
A. Derive the expression for electric field at a point on the equatorial line of an electric dipole.
B. Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform electric field.
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