A capacitor of is connected as shown in the circuit. The internal resistance of the battery is . The amount of charge on the capacitor plates will be
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As capacitor offers infinite resistance in dc-circuit. So, current flows through resistance from left to right, given by
So, the potential difference across resistance
Since, capacitor is in parallel with resistance, so it also has potential difference across it.
As current does not flow through capacitor branch so no potential drop will be accross resistance. The charge on capacitor
A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge
remains a constant because the electric field is uniform
increases because the charge moves along the electric field
decreases because the charge moves along the electric field
decreases because the charge moves opposite to the electric field
The direction of electric field is always perpendicular to the direction of electric field and equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.
The positively charged particle experiences the electrostatic force in the direction of electric field i.e., from high electrostatic potential to low electrostatic potential. Thus, the work done by the electric field on the positive charge, so electrostatic potential energy of the positive charge decreases because speed of charged particle moves in the direction of field due to force .
Figure shows some equipotential lines distributed in space. A charged object is moved from point to point .
The work done in Fig. (i) is the greatest
The work done in Fig. (ii) is least
The work done is the same in Fig. (i), Fig.(ii) and Fig. (iii)
The work done in Fig. (iii) is greater than Fig. (ii) but equal to that in
The work done (in displacing a charge particle) by a electric force is given by . Here initial and final potentials are same in all three cases are equal (20V) and same charge is moving from A to B, so work done is same in all three cases.
The electrostatic potential on the surface of a charged conducting sphere is . Two statements are made in this regard at any point inside the sphere, electric intensity is zero. at any point inside the sphere, the electrostatic potential is . Which of the following is a correct statement?
is true but is false
Both and are false
is true, is also true and is the cause of
is true, is also true but the statements are independant
As we know that the relation between electric field intensity and electric potential is
Electric field intensity then
This imply that constant
Thus, inside the charged conducting sphere then the constant electrostatic potential at every where inside the sphere and it verifies the shielding effect also.
Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately
spheres
planes
paraboloids
ellipsoids
Here we have to findout the shape of equipotential surface, these surface are perpendicular to the field lines; so there must be electric field which can not be without charge.
So, the collection of charges, whose total sum is not zero, with regard to great distance can be considered as a point charge. The equipotentials due to point charge are spherical in shape as electric potential due to point charge is given by
This suggest that electric potentials due to point charge is same for all equidistant points. The locus of these equidistant points which are at same potential, form spherical surface.
The lines of field from point charge are radial. So the equipotential surface perpendicular to field lines from a sphere.
A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness and dielectric constant and the other has thickness and dielectric constant as shown in figure. This arrangement can be thought as a dielectric slab of thickness and effective dielectric constant . The is
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