Identify the wrong statement in the following. Coulomb’slaw correctly describes the electric force that

binds the electrons of an atom to its nucleus
binds the protons and neutrons in the nucleus of an atom
binds atoms together to form molecules
binds atoms and molecules together to form solids

Solution

Nuclear force binds the protons and neutrons in the nucleus of an atom.

The charge given to any conductor resides on its outer surface, because

the free charge tends to be in its minimum potential energy state
the free charge tends to be in its minimum kinetic energy state
the free charge tends to be in its maximum potential energy state.
the free charge tends to be in its maximum kinetic energy state

If a charge is moved against the coulomb force of an electric field, then

work is done by the electric field
energy is used from some outside source
the strength of the field is decreased
the energy of the system is decreased

Solution

To overcome electrostatic repulsion forces, work will have to be done by external agency.

At the centre of a cubical box + Q charge is placed. The value of total flux that is coming out a wall is

Q/ε_o
Q/3ε_o
Q/4ε_o
Q/6ε_o

Solution

An electric dipole when placed in a uniform electric field E will have minimum potential energy if the dipole moment makes the following angle with E.

p
p/2
zero
3 p/2

Solution

U_p = –p•E = –p E cos θ

=(U_p)_minimum= –pE, when θ= 0°

A cylinder of radius R and length l is placed in a uniform electric field E parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is

zero
πR²E
2πR²E
E/πR²

Solution

For the curved surface, θ = 90º

∴ φ = E ds cos 90º = 0

In a region of space having a uniform electric field E, a hemispherical bowl of radius r is placed. The electric flux f through the bowl is

2πRE
4πR²E
2πR²E
πR²

Solution

φ = E(ds) cos θ = E(2 π r²)cos 0° = 2 π r² E.

What is the angle between the electric dipole moment and the electric field due to it on the axial line?

0º
90º
180º
None of these

Solution

Electric field \(\underset{E}{\rightarrow}\) is along the dipole axis.

∴ θ = 0º

The surface density on the copper sphere is s. The electric field strength on the surface of the sphere is

σ
σ/2
Q/2ε₀
Q/ε₀

Solution

According to Gauss's theorem,

The electric field strength at a distancer from a charge q is E. What will be electric field strength if the distance of the observation point is increased to 2 r?

E/2
E/4
E/6
None of these

Solution

As new distance = 2 r and electric field due to single

charge, E α ¹⁄_r²

therefore, new intensity = E/4.

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