Explanation:

The force on  depend on the force acting between  and  and  and  so that the net force acting on  by  and  by  is along the -direction, so the force acting between  and  is attractive force as shown in figure :

The attractive force between these charges states that  is a negative charge (since,  and  are positive).

Then the force acting between  and charge  (positive) is also know as attractive force and then the net force on  by  and  are along the same direction as shown in the figure.

The figure shows that the force on  shall increase along the positive -axis due to the positive charge .

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Explanation:

If a positive point charge is brought near an isolated conducting sphere without touching the sphere, then the free electrons in the sphere are attracted towards the positive Charge and electric field passes through a charged body. This leaves an excess of positive charge on the (right) surface of sphere due to the induction process.

Both type of charges are bound in the (isolated conducting) sphere and cannot escape. They, therefore, reside on the surface. Thus, the left surface of sphere has an excess of negative charge and the right surface of sphere has an excess of positive charge as shown in figure. attracted negative

An electric field lines start from positive charge and ends at negative charge.

Also, electric field line emerges from a positive charge, in case of single charge and ends at infinity shown in figure

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Explanation:

By Gauss's law : The total of the electric flux out of a closed surface is equal to the charge enclosed devided by

the permittivity i.e., .

Thus, electric flux through a surface doesn't depend on the shape, size or area of a surface but it depends on the number of charges enclosed by the surface. So all the given figures have same electric flux as all of them also has same single positive charge

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Explanation:

Gauss's law states that total electric flux of an enclosed surface is given by, , includes the sum of all charges enclosed by the surface.

The charges may be located anywhere inside the surface, and out side the surface. Then, the electric field on the left side of equation is due to all the charges, both inside and outside S.

So, E on LHS of the above equation will have a contribution from all charges while  on the RHS will have a contribution from  and  only.

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Explanation:

The electric field lines, are directed away from positively charged source and directed toward negatively charged source. In electric field force are directly proportional to the electric field strength hence, higher the electric field strength greater the force and vice-versa.

The space between the electric field lines is increasing, from left to right so strength of electric field decreases with the increase in the space between electric field lines. Then the force on charges also decreases from left to right. Thus, the force on charge  is greater than force on charge  in turn dipole will experience a force towards left.

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Explanation:

Consider a point on diameter away from the centre of hemisphere uniformly positively charged, then the electric field is perpendicular to the diameter and the component of electric intensity parallel to the diameter cancel out.

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