Given,
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)

Given,
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Therefore, Flux does not depend on the radius of the sphere but only on the net charge enclosed. Hence, the flux remains the same although radius is doubled.

Given,
q = net charge inside the cube =
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
(Note: Using Gauss's formula, we see that the electric flux through the cube is independent of the position of the charge and dimension of...

Let us assume that the charge is at the centre of cube with edge 10 cm.
Using Gauss's law , we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Therefore, flux through the cube:
Due to symmetry, we can conclude that the flux through each side of the...

Using Gauss's law , we know that
Since flux is zero, q = 0, but this q is the net charge enclosed by the surface.
Hence, we can conclusively say that net charge is zero, but we cannot conclude that there are no charges inside the box.

Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Given,
This is the net charge inside the box.

The net flux of the uniform electric field through a cube oriented so that its faces are parallel to the coordinate planes is zero.
This is because the number of lines entering the cube is the same as the number of lines leaving the cube.
Alternatively,
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e. ...

Now, Since the normal of the square plane makes a angle with the x-axis
Therefore, flux through this surface:

Given,
Area of the square =
Since the square is parallel to the yz plane, Therefore it's normal is in x-direction.(i.e direction )
Therefore, flux through this surface:

Charges 1 and 2 are repelled by the negatively charged plate of the system
Hence 1 and 2 are negatively charged.
Similarly, 3 being repelled by positive plate is positively charged.
(Charge to mass ratio: Charge per unit mass)
Since, 3 is deflected the most, it has the highest charge to mass ratio.

When two spheres of same size are touched, on attaining equipotential state, the total charge of the system is equally distributed on both of them.
Therefore, (i) When uncharged third sphere,C, is touched with A, charge left on A =
and charge attained by C =
(ii) Now, Charge on B + Charge on C = + =
When touched, Charge left...

We know, Force between two charged particles separated by a distance r is:
Now if
New value of force:
Therefore, the force increases 16 times!

Since, the radii of the spheres A and B are negligible compared to the distance of separation, we consider them as point object.
Given,
Charge on each of the spheres =
and Distance between them, r = 50 cm = 0.5 m
We know,
Therefore, the mutual force of electrostatic repulsion(Since they have same sign of charge)

The charge attained by polyethene (and also wool!) is solely due to the transfer of free electrons.
We know, Mass of an electron =
The total mass of electron transferred = Number of electrons transferred x Mass of an electron
=
Yes, there is a transfer of mass but negligible.

Clearly, polyethene being negatively charged implies that it has an excess of electrons(which are negatively charged!). Therefore, electrons were transferred from wool to polyethene.
Given, Charge attained by polyethene = C
We know, Charge on 1 electron =
Therefore, the number of electrons transferred to attain a charge of =
electrons.

Given,
Electric dipole moment, p =
E =
We know , Torque acting on a dipole is given by:
Therefore, magnitude of torque acting on the dipole is

Given,
and
Total charge of the system =
The system is electrically neutral. (All dipole systems have net charge zero!)
Now, distance between the two charges, d = 15 + 15 = 30 cm = 0.3 m
We know, The electric dipole moment of the system, p = x d = x d (i.e, magnitude of charge x distance between the two charges)
The direction of a dipole is towards the positive charge. Hence , in...

Let Q = C
The force experienced by Q when placed at O due to the charges at A and B will be:
where E' is the net Electric field at the point O.
Q being negatively charged will be attracted by positive charge at A and repelled by negative charge at B. Hence direction of force experienced by it will be in the direction of OA.

Given, AB = 20 cm
Since, O is the midpoint of the line AB.
AO = OB = 10 cm = 0.1m
The electric field at a point caused by charge q, is given as,
Where, q is the charge, r is the distance between the charges and the point O
k = 9x109 N m2 C-2
Now,
Due to charge at A, Electric field at O will be and in the direction AO.
Similarly the Electric field at O due to charge at B, also in the...

A tangent drawn at any point on a field line gives the direction of force experienced by a unit positive charge due to the electric field on that point. If two lines intersect at a point, then the tangent drawn there will give two directions of force, which is not possible. Hence two field lines cannot cross each other at any point.

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