Test Level 2: Electrostatics


30 Questions MCQ Test Physics Class 12 | Test Level 2: Electrostatics


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This mock test of Test Level 2: Electrostatics for Class 12 helps you for every Class 12 entrance exam. This contains 30 Multiple Choice Questions for Class 12 Test Level 2: Electrostatics (mcq) to study with solutions a complete question bank. The solved questions answers in this Test Level 2: Electrostatics quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this Test Level 2: Electrostatics exercise for a better result in the exam. You can find other Test Level 2: Electrostatics extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.
QUESTION: 1

One quantum of charge should be at least be equal to the charge in coulomb:

Solution:

Quantum of charge is represented by q=ne... where e is the charge of an electron. for 1 quantum of charge the value is 1.6 × 10-19C

QUESTION: 2

The force between two charges is 120 N. If the distance between the charges is doubled, the force will be

Solution:

QUESTION: 3

Which one of the following statement regarding electrostatics is wrong ?

Solution:

A stationary charge doesn't produce a magnetic field , it only produces an electric field.

QUESTION: 4

Two similar charge of +Q , as shown in figure are placed at A and B. _q charge is placed at point C midway between A and B. _q charge will oscillate if

Solution:

When -q is displaced left or right of the force is increased so -q move towards that force hence will not perform SHM but when its displaced to AB it will perform SHM because it will return to its mean position.

QUESTION: 5

When the distance between two charged particle is halved, the force between them becomes

Solution:
QUESTION: 6

 Two point charges in air at a distance of 20 cm. from each other interact with a certain force. At what distance from each other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction

Solution:

QUESTION: 7

A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then

Solution:

The force between q & Q-q is given by :-
F = kq(Q-q)/r²
As it is given that force between these two is Maximum....so it will be max. when dF/dq is zero
i.e, dF/dq = {k(Q-2q)}/r² = 0
so, Q = 2q
Q/q = 2/1

QUESTION: 8

Two small balls having equal positive charge Q (Coulomb) on each are suspended by two insulating strings of equal length 'L' metre, from a hook fixed to a stand. The whole set up is taken in a satellite in to space where there is no gravity (state of weight lessness). Then the angle (q) between the two strings is

Solution:

There is a condition of weightlessness in a satellite.
Therefore, mg=0
Due to electrostatic force of repulsion between the balls, the string would become horizontal.
Therefore, angle between the string =180
 

QUESTION: 9

Two point charges -1 × 10-6 C and 4 × 10-6 C separated by 3 m distance have zero field location of

Solution:

 

*Multiple options can be correct
QUESTION: 10

Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero

Solution:

Where E=0 this point is called neutral point.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.

 

QUESTION: 11

If Q =2 coloumb and force on it is F=100 newtons , Then the value of field intensity will be

Solution:
QUESTION: 12

Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square due to any one charge is E, then the resultant electric field intensity at centre of square will be

Solution:
QUESTION: 13

If mass of the electron = 9.1 × 10-31 Kg. Charge on the electron = 1.6 × 10-19coulomb and g = 9.8 m/s2. Then the intensity of the electric field required to balance the weight of an electron is

Solution:

We know that, qE=mg
So,E=mg/q
E=9.1x10-31x9.8x1019/1.6
E=9.1x9.8x10-12/1.6
E=5.6x10-11

QUESTION: 14

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

Solution:

Field intensity is proportional to the inverse of the square of the distance separating the point charges:

F = k*q1*q2/d^2

Since k, q1, and q2 are assumed to be constant, their product can be combined to form a single constant:

K = k*q1*q2, so

F = K/d^2 (and by algebraic manipulation we have K=F*d^2).

Therefore we know that

F(at d=4) = K/4^2, and

F(at d=2) = K/2^2.

Since K is a constant, we can equate K=F*d^2 for each case:

K=F(at d=4)*4^2 = F(at d=2)*2^2

so

F(at d=2) = (F(at d=4)*4^2)/2^2

F(at d=2) = 200N/C * (4^2)/(2^2) = 200N/C *16/4 = 800N/C.

QUESTION: 15

At position of (4i + 3j) by a point charge 5 × 10-6 C placed at origin will produce electric field of

Solution:

QUESTION: 16

Two identical point charges are placed at a separation of l.P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than l. Which of the following best represents the resulting curve ?

Solution:


At point P, 
E=E1​−E2​ rightward
⟹E=(kq/x2​)−(kq/(l−x)2​)   where k=1/4πϵo​ 
⟹E=kq (l(l−2x)​/x2(l−x)2
From graph we can see that it can't be straight line and E=0 at x=l/2​
At x=0 ,,E→∞  and at x=l, E→−∞
Hence, from the shown graphs the correct answer is (D).
 

QUESTION: 17

A particle of mass m and charge Q is placed in an electric field E which varies with time t ass E = E0 sinwt. It will undergo simple harmonic motion of amplitude

Solution:

Due to verifying electric field, it experiences an verifying force :-
F=QE=QE0​sinωt
at maximum amplitude A, it experience a maximum force of:-
Fmax​=QE0​
also, Restoring force in SHM is given by: - F=mω2x
for amplitude, x=A OR,
2A=QE0​
⇒A= QE0/mω2

QUESTION: 18

Four charges are arranged at the corners of a square ABCD, as shown. The force on +ve charge kept at the centre of the square is

Solution:
QUESTION: 19

Two free positive charges 4q and q are a distance l apart. What charge Q is needed to achieve equilibrium for the entire system and where should it be placed form charge q ?

Solution:

Let"r" be the distance from the charge q where Q is in equilibrium .Total Force acting on Charge q and 4q:
F=kqQ/r² + k4qQ/(l-r)²
For Q to be in equilibrium , F should be equated to zero.
kqQ/r² + k4qQ/(l-r)²=0
(l-r)²=4r²
⇒l-r=2r
⇒l=3r
⇒r=l/3
Taking the third charge to be -Q (say) and then on applying the condition of equilibrium on + q charge
kQ/(L/3)² =k(4q)/L²
9kQ/L²=4kq/L²
9Q=4q
Q=4q/9
Therefore a point charge -4q/9 should be placed at a distance of L/3 rightwards from the point charge +q on the line joining the 2 charges.

QUESTION: 20

Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be

Solution:

The charges will be balanced by their counterparts on the opposite side. So, eventually the charges remaining will be 2q and b and 3q on D.
Since the charge distribution is asymmetrical, the net force on charge would be skewed towards D.
 

QUESTION: 21

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

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=E2q2t2​/2m

QUESTION: 22

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 ?

Solution:
QUESTION: 23

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.

Solution:
QUESTION: 24

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

Solution:

Electric field due to a charged ring in given by: -
at point p
∣​E∣​= KQx​ /(R2+x2)3/2        Q =λ(2πr)
at a large distance, x≫R, so :- R2+x2≃x2
⇒∣​E∣​= K&x​/(x2)3/2=KQ/x2​=Eαx−2
so at a large distance, the ring behaves as a point particle.

QUESTION: 25

 A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d<<R). The electric field at the centre of the ring will now be

Solution:
QUESTION: 26

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

Solution:
QUESTION: 27

The direction (θ) of at point P due to uniformly charged finite rod will be -

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: 28

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-

Solution:
QUESTION: 29

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

Solution:
QUESTION: 30

Electric potential is a -

Solution:

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