H C VERMA Solutions for Class 12-science Physics Chapter 8 - Gauss' Law
Chapter 8 - Gauss' Law Exercise 141
The electric field in a region is given by Find the flux of this field through a rectangular surface of area 0.2m2 parallel to the Y - Z plane.
(As given plane is parallel to YZ plane)
And flux is given as :
A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.
Linear charge density,
And charge enclose inside is given as:
Show that there can be no net charge in a region in which the electric field is uniform at all points.
Electric field is uniform and given plane is perpendicular to it. Thus it is an equipotential surface with no net current on that surface. So, net charge is :
Q = 0
The electric field in a region is given by Find the charge contained inside a cubical volume bounded by the surfaces x=0, x=a, y=a, z=0 and z=a. Take E0=5×103N/c, l=2cm and a=1cm
A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.
By Gauss Law:
At centre of cube:
(For six surfaces)
A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge 𝛼 as shown in figure. Find the flux of the electric field through the square surface.
Let us assume a cubical surface of side 'a'. So charge will be at centre of the cube and flux through one surface of cube is given as:
Find the flux of the electric field through a spherical surface of radius R due to a charge of 10-7C at the centre and another equal charge at a point 2R away from the centre.
By gauss's law sphere experiences flux only by charge present inside sphere and not by charge present outside sphere.
A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and the Gauss's law, find the flux of the electric field due to this charge through the surface of the hemisphere.
In case of sphere flux is given as:
In case of hemisphere,
A spherical volume contains a uniformly distributed charge of density.2.0×10-4C/m3 Find the electric field at a point inside the volume at a distance 4.0cm from the centre.
Volume charge density =
The radius of a gold nucleus (Z = 79) is about Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?
(a) We know,
A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r1and r2 Find the electric field at a point P a distance away from the centre for r1 < x < r2 Draw a rough graph showing the electric field as a function of for 0 < x < 2r2.
Charge enclosed by sphere of radius is :
By Gauss law,
Chapter 8 - Gauss' Law Exercise 142
A charge Q is placed at the centre of an uncharged, hollow metallic sphere of radius α (a) Find the surface charge density on the inner surface and on the outer surface. (b) If a charge q is put on the sphere, what would be the surface charge densities on the inner and the outer surfaces? (c) Find the electric field inside the sphere at a distance from the centre in the situations (a) and (b).
By Property of induction charge induced at inner surface and outer surface .→ +Q
(a) Surface charge density at inner surface
Surface charge density at outer surface
(b) When q is added then there is no effect on inner surface charge density
But when q is added outer surface charge density becomes,
(c) Let us assume an area inside sphere which is at distance from centre of sphere.
Consider the following very rough model of a beryllium atom. The nucleus has four protons and four neutrons confined to a small volume of radius The two 1s electrons make a spherical charge cloud at an average distance of from the nucleus, whereas the two 2s electrons make another spherical cloud at an average distance of from the nucleus. Find the electric field at (a) a point just inside the 1s cloud and (b) a point just inside the 2s cloud.
(a) Electric field ,
(b) Total Charge Q =
Electric field ,
Find the magnitude of the electric field at a point 4cm away from a line charge of density
A long cylindrical wire carries a positive charge of linear density An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius.
A long cylindrical volume contains a uniformly distributed charge of density p. Find the electric field at a point P inside the cylindrical volume at a distance from its axis.
(Total charge As nQ = where, is volume charge density)
A nonconducting sheet of large surface area and thickness d contains uniform charge distribution of density p. Find the electric field at a point P inside the plate, at a distance from the central plane. Draw a qualitative graph of E against x for 0 < x < d.
A charged particle having a charge of is placed close to a nonconducting plate having a surface charge density Find the force of attraction between the particle and the plate.
We know in case of charge conducting sheet:
And Force is :
One end of a 10cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate and the other end is fastened to a small ball having a mass of 10g and a charge of . In equilibrium, the thread makes an angle of 60° with the vertical. Find the surface charge density on the plate.
According to given figure:
From (1) and (2)
Consider the situation of the previous problem. (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations.
(a) In equilibrium,
(b) For this, acceleration is included. Thus,
Two large conducting plates are placed parallel to each other with a separation of 2.00cm between them. An electron starting from rest near one of the plates reaches the other plate in 2.00 microseconds. Find the surface charge density on the inner surfaces.
Now, electric field is given as:
Two large conducting plates are placed parallel to each other and they carry equal and opposite charges with surface density as shown in figure. Find the electric field (a) at the left of the plates, (b) in between the plates and (c) at the right of the plates.
(a) For any point on left plate, electric flux is not present as it is outside the system. Hence, electric field,
E = 0
(b) For charge present in between plates, it experiences force towards negative plate, which is given as:
(c) For any point of right plate no electric flux is present as it is outside the system. Hence, electric field is given as:
E = 0
Two conducting Plates X and Y, each having large surface area A (on one side), are placed parallel to each other shown in figure. The plate X is given a charge Q whereas the other is neutral. Find (a) the surface charge density at the inner surface of the plate X, (b) the electric field at a point to the left of the plates, (c) the electric field at a point in between the plates and (d) the electric field at a point to the right of the plates.
(a) Electric field at left end
And Electric field at right end
As the electric field is balanced
(b) From above part,
(for left part)
As is charged plate electric field is directed towards left.
(c) Here, is charged plate so it acts as only source of electric field. Thus, electric field is directed from left to right and is given as:
(d) Similar to above part, Y acts as positive plate thus it repels, so electric field is directed towards right and is given as:
Three identical metal plates with large surface areas are kept parallel to each other as shown in figure. The leftmost plate is given a charge Q, the rightmost a charge -2Q and the middle one remains neutral. Find the charge appearing on the outer surface of the Rightmost plate.
Net electric field at point P is:
Solving we get,
And, Net charge on right of right plate is:
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