Electric Charge, Coulomb's law and Electric Field Test

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Electric Charge, Coulomb's law and Electric Field Test - 11

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Question 1
1 / -0.25

A charged particle of charge q and mass m is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after time 't 'seconds is

A

B

C

D

Solution

Force on particle=F=qE
Hence, acceleration of particle=a=F/m =qE/m
Initial speed=u=0
Hence, final velocity=v=u+at=qEt/m
Kinetic energy=K=(1/² )mv2=(1/2)m(qEt/m )2
⟹K=E² q² t² /2m
Question 2
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Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle with a negative charge starts on the x-axis at a large distance from O, moves along the +x-axis, passes through O and moves far away from O. Its acceleration a is taken as positive along its direction of motion. The particle 's acceleration a is plotted against its x-coordinate. Which of the following best represents the plot ?

A

B

C

D

Solution

Consider the attached free body diagram for the given scenario.
From coulomb 's law,
∣F1 ∣=∣F2 ∣= qQ /4 πε_o r²
In y-direction, forces cancel, F_y =0
In x-direction, forces add up. Also, it is directed in +ve x direction if the position of −q is along −ve x axis and in −ve x direction if the position of −q is along +ve x axis.
Hence,
F_x =2qQcos θi /4 πε_o r²   ∀x <0
F_x =2qQcos θi /4 πε_o r²   ∀x ≥0
But cos θ=x/r
    r= √x² +d²
Substituting in force equation, we get
F_x =2qQx i /4 πε_o ( x² +d² )^3/2 ∀x <0
F_x =2qQx i /4 πε_o ( x² +d² )^3/2 ∀x ≥0

Now, a=F/m by Newton 's second law. It is positive when x is negative and negative when x is positive. This is satisfied only in option (B).
Question 3
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Four equal positive charges are fixed at the vertices of a square of side L. Z-axis is perpendicular to the plane of the square. The point z = 0 is the point where the diagonals of the square intersect each other. The plot of electric field due to the four charges, as one moves on the z-axis.

A

B

C

D
Question 4
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A small circular ring has a uniform charge distribution. On a far-off axial point distance x from the centre of the ring, the electric field is proportional to

A
x^-1

B
x^-3/2

C
x^-2

D
x^5/4

Solution

Electric field due to a charged ring in given by: -
at point p
∣E ∣= KQx /(R² +x² )^3/2 Q =λ(2 πr)
at a large distance, x ≫R, so :- R2+x2 ≃x2
⇒∣E ∣= K &x /(x² )^3/2 =KQ/x² =E αx⁻²
so at a large distance, the ring behaves as a point particle.
Question 5
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A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d <

A
Directed towards the gap, inversely proportional to R³

B
Directed towards the gap, inversely proportional to R²

C
Directed away from the gap, inversely proportional to R³

D
Directed away from the gap, inversely proportional to R² .

Solution

If we fill the gap then there will be no electric field at the center of ring
This signifies that electric field due to rest ring is equal to electric field due to gap and opposite in direction
so electric field due to gap= Kq/R²
q=Qd/2 πR
thus electric field E=KQd / R³
Hence E will be inversely proportional to R³ and direction will be toward the gap.
Question 6
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The electric field intensity at a point situated 4 meters from a point charge is 200 N/C. If the distance is reduced to 2 meters, the field intensity will be

A
400 N/C

B
600 N/C

C
800 N/C

D
1200 N/C
Question 7
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The direction (θ) of E at point P due to uniformly charged finite rod will be -

A
At angle 30 °from x-axis

B
45 °from x-axis

C
60 °from x-axis

D
None of these

Solution

The angle suspended by the finite line charge at P=60 °
So, the resultant electric field due to line charge will be at 60/2 ⇒30 °since we can assume the charge concentrated at the centre of finite line charge.
Question 8
1 / -0.25

When charge of 3 coulomb is placed in a Uniform electric field , it experiences a force of 3000 newton, within this field, potential difference between two points separated by a distance of 1 cm is-

A
10 Volt

B
90 Volt

C
1000 Volt

D
3000 Volt.

Solution

Electric force F=qE
given, F=3000N,q=3C
∴E=F/q =3000/3 =1000 V/m
Potential difference, V=Ed where d=0.01m
∴V=1000 ×0.01=10V
Question 9
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An electric dipole is placed at an angle of 30 °with an electric field intensity 2 ×10⁵ N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is

A
8 mC

B
2 mC

C
5 mC

D
7 μC
Question 10
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Electric potential is a -

A
Vector quantity

B
Scalar quantity

C
Neither vector Nor scalar

D
Fictious quantity

Solution

Electric potential is a pure Scalar quantity, The reason is as follows.
The Electric Potential is defined as the amount of work-done per unit positive charge to bring from infinity to that point under the influence of the primary charge only.
U=W/q
And work done is defined as the dot product of force and displacement which is a scalar quantity.
W=F.S
Thus Electric potential is a scalar quantity.