Ncert Solutions Physics Class 12th

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New answer posted•a year ago
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a year ago

1.34 Velocity of the particle, vx = 2.0 ×106m/s

Separation of the two plates, d = 0.5 cm = 0.005 m

Electric field between two plates, E = 9.1 ×102 N/C

Charge of an electron, q = 1.6 ×10-19 C

Mass of an electron, me = 9.1 ×10-31 kg

Let the electron strike the upper plate at the end of the plate L, when d

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New answer posted•a year ago
S
a year ago

1.32 (a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards t

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

1.30 

Take a long thin wire XY of uniform linear charge density λ. Consider a point A at a perpendicular distance l from the midpoint O of the wire. Let E be the electric field at point A due to the wire XY. Consider a small length element dx on the wire section with OZ = x. Let q be the charge o

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New answer posted•a year ago
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a year ago

1.29 Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E be the electric field outside the conductor, q is the electric charge, σis the charge density and ε0 is the permittivity of free space.

Charge q =σ ×ds 

According to Gauss’s law, fluxφ = E.

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

1.27 Dipole moment of the system, p = q×dl = 10-7 Cm

Rate of increase of electric field per unit length,dEdl =105 NC-1 

Force experienced by the system is given by the relation, F = qE = q×dEdl×dl 

= (q  ×dl)×dEdl =p ×dEdl =-10-7× 105= -10-2 N

The force is  N in the negative z-direction i.e. opposite to the direction of

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New answer posted•a year ago
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a year ago

Ans.1.26

(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

 

(b) The field lines shown in (b) do not represent electrostatic field lines because field lines can not emerge from a negative charge and c

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New answer posted•a year ago
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a year ago

1.25 Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 ×104NC-1

Density of oil, ρ= 1.26 g / cm3 = 1.26×103  g/ m3

Acceleration due to gravity, g = 9.81 m/s2 

Charge of an electron, e = 1.60×10-19C

Let the radius of the oil drop be r

Force (F) due to electric field (E) is equal to the wei

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New answer posted•a year ago
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a year ago

1.24 The given conditions are explained in the adjacent diagram

 

Where A and B represent two large, thin metal plates, parallel and close to each other. The outer surface of A is shown as I, outer surface of B is shown as II and the surface in between A and B is shown as III.

Charge density of pl

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