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Question : 37
Total: 37
(a) Write two important characteristics of equipotential surfaces.
(b) A thin circular ring of radiusr λ P x ( x > > r )
OR
(a) State Gauss's law on electrostatics and derive an expression for the electric field due to a long, straight and thin uniformly charged wire (linear charge densityλ r
(b) The magnitude of electric field (inNC − 1 r m
E = 1 0 r + 5
By how much, does the electric potential increase in moving from point atr = 1 m r = 10 m
(b) A thin circular ring of radius
OR
(a) State Gauss's law on electrostatics and derive an expression for the electric field due to a long, straight and thin uniformly charged wire (linear charge density
(b) The magnitude of electric field (in
By how much, does the electric potential increase in moving from point at
Solution:
(a) Characteristics of equipotential surface :
(i) Potential remains the same at all the points on equipotential surface.
(ii) No work is required to move a charge within an equipotential surface.
(b) PointP r
The fielddE dl
Linear charge density= λ
So, charge ofdl dq = λ dl
The perpendicular field component for the whole ring will be zero.
So,dE =
cos θ =
×
Or,dE =
Or,E = ∫
= ∫
∴ E =
=
Whenx > > r
E =
=
This reduces to a simple Coulomb field. In this case, the charged ring looks like a point charge.
OR
(a) Gauss's Law: The total of the electric flux out of a closed surface is equal to the chargeenclosed divided by the permittivity.
Φ =
Electric field due to an infinitely long straight uniformly charged wire : charge density= λ
P
r = P
E = P
A cylinder of length 1 , radiusr
d s = a E ( ϕ )
= ∮ Eds cos θ
ϕ = ∮ Eds [ ∵ θ = 0 ∘ ]
ϕ = E ( 2 r π l )
∴ Q = λ l =
∴ By Gauss's law,
ϕ =
Or, E ( 2 r π l ) =
=
∴ E =
(b)E = −
∴ | V | =
E d r
Or, | V | =
E d r
Or, | V | =
( 10 r + 5 ) d r
Or, | V | = [ 5 r 2 + 5 r ] 1 10
Or, | V | = 540 V
Thus, electric potential changes by540 V
(i) Potential remains the same at all the points on equipotential surface.
(ii) No work is required to move a charge within an equipotential surface.
(b) Point
The field
Linear charge density
So, charge of
The perpendicular field component for the whole ring will be zero.
So,
Or,
Or,
When
This reduces to a simple Coulomb field. In this case, the charged ring looks like a point charge.
OR
(a) Gauss's Law: The total of the electric flux out of a closed surface is equal to the chargeenclosed divided by the permittivity.
Electric field due to an infinitely long straight uniformly charged wire : charge density
A cylinder of length 1 , radius
(b)
Thus, electric potential changes by
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