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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.
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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^{-19}C

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^{-19}coulomb and g = 9.8 m/s^{2}. 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^{-31}x9.8x10^{19}/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=E_{1}−E_{2} rightward

⟹E=(kq/x^{2})−(kq/(l−x)^{2}) where k=1/4πϵo

⟹E=kq (l(l−2x)/x^{2}(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 = E_{0} sinwt. It will undergo simple harmonic motion of amplitude

Solution:

Due to verifying electric field, it experiences an verifying force :-

F=QE=QE_{0}sinωt

at maximum amplitude A, it experience a maximum force of:-

F_{max}=QE_{0}

also, Restoring force in SHM is given by: - F=mω^{2}x

for amplitude, x=A OR,

mω^{2}A=QE_{0}

⇒A= QE_{0}/mω^{2}

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=E^{2}q^{2}t^{2}/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 /(R^{2}+x^{2})^{3/2} Q =λ(2πr)

at a large distance, x≫R, so :- R2+x2≃x2

⇒∣E∣= K&x/(x^{2})^{3/2}=KQ/x^{2}=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

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 × 10^{5} 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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