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If an insulated non-conducting sphere of radius R has charge density ρ. The electric field at a distance r from the centre of sphere (r < R) will be
For non-conducting sphere
Infinite charges of magnitude q each are lying at x =1, 2, 4, 8... meter on X-axis. The value of the intensity of the electric field at point x = 0 due to these charges will be
Net field at origin
Using formula for infinite Geometric progression (G.P.) ,
if Sum =
and r<1
then ,sum =
In question, a =1 and r=
Two infinitely long parallel conducting plates having surface charge densities +σ and –σ respectively, are separated by a small distance. The medium between the plates is a vacuum. If ε0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is
In the given scenario of two infinitely long parallel conducting plates with opposite surface charge densities +σ and -σ separated by a small distance in vacuum, the electric field between the plates is given by E = σ/ε0, where ε0 is the permittivity of free space.
An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct ?
In electric dipole, the flux coming out from positive charge is equal to the flux coming in at negative charge i.e. total charge on sphere = 0. From Gauss law, total flux passing through the sphere = 0.
The electric intensity due to an infinite cylinder of radius R and having charge q per unit length at a distance r(r > R) from its axis is
According to Gauss law
(E is constant)
∴
⇒ i.e.
Two equal negative charge – q is fixed at the fixed points (0, a) and (0, –a) on the Y-axis. A positive charge Q is released from rest at the point (2a, 0) on the X-axis. The charge Q will
Two equal negative charges -q are fixed on the Y-axis, and a positive charge Q is released from rest on the X-axis. The net force on Q will be attractive towards the origin, but the force will not be along the line joining Q and the origin, causing oscillatory but not simple harmonic motion.
An electric line of force in the xy plane is given by equation x2 + y2 = 1. A particle with unit positive charge, initially at rest at the point x = 1, y = 0 in the xy plane will -
Charge will move along the circular line of force because x2 + y2 = 1 is the equation of circle in xy-plane.
A positively charged ball hangs from a silk thread. We put a positive test charge q0 at a point and measure F/q0, then it can be predicted that the electric field strength E
Because of the presence of positive test charge q0 in front of positively charged ball, charge on the ball will be redistributed, less charge on the front half surface and more charge on the back half surface. As a result of this net force F between ball and point charge will decrease i.e. actual electric field will be greater than F/q0.
A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shell having charge –Q. The radius of the sphere is a and that of the spherical shell is b (b > a). What is the electric field at a distance R(a < R < b) from the centre
Electric field at a distance R is only due to sphere because electric field due to shell inside it is always zero. Hence electric field =
A point charge q is placed at a distance a/2 directly above the centre of a square of side a. The electric flux through the square is
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.