· 5/12/2018 · Velumanickam Matriculation Higher Secondary School ± vaani ± Ramanathapuram. 1. ELECTROSTATICS (12 th Standard Physics - Objective Questions and Answers for Centum

Velumanickam Matriculation Higher Secondary School – vaani – Ramanathapuram.

1. ELECTROSTATICS (12

th Standard Physics - Objective Questions and Answers for Centum and NEET Aiming students)

1 | P a g e Prepared By,

G.R.Balaji M.Sc.,M.Phil.,B.Ed.,M.I.S.T.E.

P.G Assistant in The Department Of Physics,

Velumanickam Matric Higher Secondary School, Ramanathapuram, Vaani. [email protected]

1. An electron of mass me , initially at rest , moves through certain distance in a uniform electric

field in time t1. A proton of mass mp, also, initially at rest, takes time t2 to move through an equal

distance in this uniform electric field. Neglecting the effect of gravity, the ration t2/t1 is nearly

equal to

(a)1 (b) (mp/me)1/2

(c) (me/mp)1/2

(d) 1836

2. A charge q is placed at the centre of line joining two equal charges Q. The system of the three

charges will be in equilibrium if q is equal to

(a) – Q/2 (b) -Q/4 (c) +Q/4 (d) +Q/2

Answer : Option (b)

Hint:

The electrostatic force acting on both Particle F = eE

Acceleration of Electron, ae=Fe/me

ae=eE/me -------- (1)

Acceleration of Proton, ap=Fp/mp

ap= eE /mp ---------(2)

According to Equation of Motion, S =ut+1/2 at2

Initially both Particles are at rest, therefore , S = 1/2 at2

For electron, S = 1/2 aet12 -------- (3)

For proton, S = 1/2 apt22 ---------- (4)

Compare (3) and (4)

aet12

= apt22

t22/ t1

2 = ae/ap

t2/ t1 = (ae/ap)1/2

Substitute (1) and (2)

t2/ t1 = (eE/me X mp/ eE)1/2

t2/ t1 = (mp / me)1/2

𝜋𝜀 𝑄𝑄 = 𝜋𝜀 𝑄

Answer : Option (b)

Hint:

Since, q is at the centre of two charges Q and Q, net force on it is zero, whatever the magnitude and

sign of the charge on it. For the equilibrium of Q, q should be negative because other charge Q will repel it, so q

should attract it. Simultaneously these attraction and repulsion should be equal.

By Ele trostati Coulo ’s La ,

= 𝑄 (or) with sign = − 𝑄

Q Q q

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3. Assume that an electric field 𝑬 = 𝟎𝒙 �̂� exists in space. Then, the potential difference is

VA – V0 , Where V0 is the potential at the origin and VA the potential at x = 2 m is

(a) 120 J (b) -120 J (c) -80J (d) 80 J

4. Consider four charges, A, B, C, and D, which exist in a region of space. Charge A attracts B, but B

repels C. Charge C repels D, and D is positively charged. What is the sign of charge A?

(a) Positive (b) Negative (c) electrically neutral (d) both (a) and (b) is correct

5. Two charged spheres of equal size carrying a charge of +6 C and –4 C, respectively. The spheres

are brought in contact with one another for a time sufficient to allow them to reach an equilibrium

charge. They are then separated. What is the final charge on each sphere?

(a) +2 C (b) -24 C (c) -2 C (d) +1 C

𝑉 = − ∫ 𝑥 𝑑𝑥

𝑉 − 𝑉 = − 𝑋 [ − ] 𝑉 − 𝑉 = − 𝐽

Answer : Option (c)

Hint:

As we know that the potential difference VA – V0 is

dv = - E dx

Answer : Option (b)

Hint:

If D is positive and it repels C, C must also be positive. Since C repels B, B must also be positive. A attracts

B, so A must be negatively charged.

-4 +6

Answer : Option (d)

Hint:

When the two spheres come in contact with each other, charge will be transferred, but the total amount

of charge is conserved. The total charge on the two spheres is +6 C + -4 C = +2 C, and this is the magnitude of the

equilibrium charge. When they are separated, they divide the charge evenly, each keeping a charge of +1 C.



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6.

Two point charges q1 = +2 μC and q2 = - 4 μC are separated y a dista e r, as shown above. If the

force between the charges is 2 N, what is the value of r?

(a) 2.5 m (b) 0.89 m (c) 3.6 m (d) 0.19 m

7. The electric field due to a point charge Q at a distance r away from the centre of the charge is

(a) 𝐾𝑄

(b) 𝐾𝑄

(c) 𝐾 𝑄

(d) 𝐾

8. Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both

charges are doubled, and the distance between them is halved, the new force between them is

(a) ¼ F (b) ½ F (c) 4 F (d) 16 F

+ μC -4 μC

r

= ( 𝜋𝜀 𝐹 )

Answer : Option (c)

Hint:

By Coulo ’s Law, 𝐹 = 𝜋𝜀

= √( 𝜋𝜀 𝐹 )

= √ 𝑥 9 𝑥 𝑥 −6 𝑥 𝑥 −6 √

= . 𝑚

Answer: Option (b)

Hint:

By Coulo ’s Law, 𝐹 = 𝐾𝑄 2

2

r

KQ

q

r

KQq

q

FE

Answer: Option (d)

Hint: By Coulo ’s Law, 𝐹 = 𝐾 ; F

r

qqKF 16

)2

1(

)2)(2(

2

21



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9. According to Coulomb’s law, if the electric force between two charges is positive, which of the

following must be true?

(a) One charge is positive and the other charge is negative.

(b) The force between the charges is repulsive.

(c) The force between the charges is attractive

(d) The two charges must be equal in magnitude.

10. A force of 40 N acts on a charge of 0.25 C in a region of space. The electric field at the point of

the charge is

(a) 10 N/C (b) 100 N/C (c) 160 N/C (d) 40 N/C

11. Two charged parallel plates are oriented as shown. The following particles are placed between

the plates, one at a time:

I. electron

II. proton

III. neutron

Which of the particles would move to the right between the plates?

(a) I and II only (b) I and III only (c) II and III only (d) II only

Answer: Option (b)

Hint:

In the equation for electric force, two positive or two negative charges multiplied by each other yields

a positive force, indicating repulsion.

Answer: Option (c)

Hint:

C

N

C

N

q

FE 160

25.0

40

E

Answer: Option (d)

Hint:

Only the positively charged proton would move to the right, toward the negatively charged plate



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12.

An amount of positive charge Q is placed on a conducting sphere. A positive point charge Q is

placed at the exact centre of the sphere and remains there. Which of the following graphs best

represents the of electric field E verses distance r from the centre?

(a) (b)

(c) (d)

R

+

+

+

+

+

+ +

+

+

+

+

+

r

Q

Q

R

E

r

R

E

r

R

E

r

R

E

r

Answer: Option (c)

Hint:

The electric field on the inside is E (inside) = KQ / R2 and on the outside is E (outside) = KQ /r

2. In

both cases, the electric field follows the inverse square law.



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If a +10q C of charge is placed at the centre of the cuboid ,

then the flux linked with the surface of the cuboid is,

13. a /ε0 (b) /ε0 /ε0 (d) /ε0

14. If a +q charge is placed outside of the surface as shown in

the figure, the electric flux linked with the surface is

(a) Zero / ε0 / ε0 d / ε0

15. The given surface consist of a dipole system (-2q , +2q) charge,

the flux linked with the surface is

(a) / ε0 (b) / ε0 (c) / ε0 (d) Zero

Answer: Option (c)

Hint:

By Gauss’s La ɸ = q / ε0

In this question q = +10 C

Therefore , ɸ = 10/ ε0

Answer: Option (a)

Hint:

By Gauss’s La , the total flux of the electric field E over any closed surface is equal to 1/ ε0 times the net

charge enclosed by the surface.

Answer: Option (d)

Hint:

The total charge enclosed by the surface is q = -2 +2

q = 0

By Gauss’s La , ɸ = q / ε0

ɸ = 0 / ε0

ɸ = 0



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16. Suppose that the top plate of a parallel-plate capacitor has an electric potential of 0.0 V and

the bottom plate has a potential of 450 V. The plates are 2.0 cm apart. What is the magnitude

of the electric field between the plates?

(a) 25000 N/C (b) 22500 N/C (c) 76 x 103

N/C (d) 10 x 10-12

N/C

17. Consider a charge configuration and spherical Gaussian

Surface as shown in the figure. When we calculate the flux of

the electric field over the spherical surface, the electric field

will be due to

(a) (b) only the positive charges (c) all the charges (d) +q1 and –q1

18. An electric field line and equipotential surface are

(a) Always parallel al ays at ̊ I li ed at a y a gle d o e of the a o e

Answer: Option (b)

Hint:

E = ΔV / d = V / . = 22500 N/C

Answer: Option (c)

Hint:

At any point over the spherical Gaussian surface, the net electric field is vector sum of electric field due to

+q1,–q1 and q2.

Do not be confused with the net electric flux which is zero passing over the Gaussian the surface as the

net charge enclosing the surface is zero.

Answer: Option (b)

Hint:

From this figure,

It is clear that the direction of electric field E and Equipotntial

surfaces are always perpendicular each other.



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19. A solid insulating sphere, a solid conducting sphere and a hollow conducting sphere are each

given identical charge +Q. The spheres all have the same radius r. At a distance a > r from the

centre of the spheres, the electric field

(a) is greatest for the solid insulating sphere. (b) is greatest for the solid conducting sphere.

(c) is greatest for the hollow conducting sphere. (d) is a non-zero value but the same for all three spheres.

20. An insulating sphere of radius R = 2a / b has a volume charge density ρ r = ar – br2, where a

and b are constants, the total charge contained in the volume of sphere is

(a) - π a5/ 5b

4 (b) π a7

/ b2 (c) - π2

a5/ b

4 (d) π2

a5/ 17b

4

21. Which of the following concepts of capacitor is true?

a) A capacitor consists of two conductors connected in parallel to each other so that it can store

charge in between the plates.

(b) Capacitance is directly proportional to plate area. Hence as plate area increases, the capacitance

also increases

(c) Capacitance is inversely proportional to the distance between the two parallel plates. Hence, as

the distance between the plates decreases, the capacitance increases.

Answer: Option (d)

Hint:

A Gaussian surface of radius a around each of the spheres will enclose the same

amount of charge +Q

Answer: Option (a)

Hint:

2aR

b And 2( )r ar br

2

2 2

0 0

4 53 4

0 0

5 5 5

4 4 4

( ) ( ) 4

( ) ( )4

4 44 5

2 16 128 48

5 5

R R

R R

dQr dQ r dV where dV r dr

dV

Q r dV ar br r dr

aR bRQ ar dr br dr Q

a a a aSince R Q Q

b b b b



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(d) Coulomb is the unit of charge, Volts is the unit of voltage; hence the unit for capacitance is Farad.

(a) Option (a) and (d) is true (b) Option (b) and (c) is true

(c) Option (c) and (d) is true (d) all the options are true

22. What is the combined capacitance of a 10 µF capacitor, a 20 µF capacitor and a 30 µF capacitor

connected (a) in parallel (b) in series?

(a) 600 µF and 0.600 µF (b) 275 µF and 350 µF (c) 60 µF and 5.5 µF (d) 0.4 µF and 350 µF

23. What is the combined capacitance of a 10 µ F capacitor and a 20 µF capacitor connected in

parallel, and then connected in series to a 30 µF capacitor as shown below.

(a) 60 µF (b) 0.45 µF (c) 450 µF (d) 15 µF

Answer: Option (d)

Hint:

Capa ity of the apa ita e C = Aε0 / d and

Capacity of the capacitance C = q/v

Unit: C V-1

(or) Farad

Answer: Option (c)

Hint:

Cp = C1 + C2 + C3

1/Cs = 1/ C1 +1/ C2 +1/ C3

Answer: Option (d)

Hint:

Taking the two in parallel first, we have

Combining this with the other capacitor in

series, we have



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24. The capacitance of a capacitor is not affected by

(a) Distance between the plates (b) area of plates (c) thickness of plates (d) all of them

25. Three capacitors each of capacity C are given. The resultant capacity (2/3) C can obtained by

using them

(a) all in series

(b) all in parallel

(c) two in parallel and third in series with this combination

(d) two in series and third in parallel across this combination

Answer: Option (c)

Hint:

Capa ity of the apa ita e C = Aε / d

Where, ε – Permittivity of dielectric,

A – Area of the plates,

d – Distance between the plates.

So, the distance and area could affect the changes in the capacitance of a capacitor. Hence, thickness of plate

does not affect the capacitance value.

Answer: Option (c)

Hint:

We know that the capacitance value for parallel connected capacitor is Cp = C1 + C2 + C3

And for series connected capacitors is 1/Cs = 1/ C1 + 1/ C2 + 1/ C3

Here, three capacitors have same value. Then, that the capacitance value for two parallel connected

capacitor is ( C1+ C2 ) = (C + C ) = 2C. Now, this parallel connected combination is connected with one

series connected capacitor. Hence, the final value of capacitance is

1/C = 1/ 2C +1/ C

1/C = 3 / 2C

C = 2C / 3



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26. If the dielectric of a capacitor is replaced by conducting material

(a) the capacitance value of the capacitor will shoot upto very high value

(b) the capacitor can store infinite charge.

(c) the plate will get short circuited.

(d) the capacitor will get heated up due to eddy currents

27. If C be the capacitance, V be the potential difference and I be the current, then I / CV will have

the unit of

(a) Frequency (b) power (c) time (d) reactive power

28. for making a capacitor, it is better to select a dielectric having

(a) Low permittivity (b) high permittivity

(c) Permittivity same as that of air (d) permittivity slightly more than of air.

Answer: Option (c)

Hint:

The dielectric material is placed in between the two metal plates. If we replace the dielectric material by a

conducting metal, then it leads to make a direct short circuit between the metal plates.

ICV = 𝑄𝑡𝑄𝑉 𝑥𝑉 = 𝑡 = frequency Hz

Answer: Option (a)

Hint:

We know that one amp is one coulomb per second and one farad is one coulomb per volt. If we apply

these units of electric current and capacitance in the given expression, we will get (1/Second).

Hence, the unit of (I/CV) is frequency.

Answer: Option (b)

Hint:

In order to maximize the charge that a capacitor can hold, the dielectric material needs to have a high

permittivity as possible.



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29. The energy stored by a capacitor is given by

(a) (1/2)CV2

(b) Q2 / 2C (c) (1/2) QV (d) all of them

30. Van de Graaff generator consists of a hollow metallic sphere of diameter 2m. If the potential of

the surface of the sphere is 6 million volt, the charge accumulated over the surface of the sphere is

(a) 0.6 milli coulomb (b) 1 coulomb (c) 8.854 micro coulomb (c) 0.33 micro coulomb

------------------- All the Best -------------------

Answer: Option (d)

Hint:

We know that the basic expression of energy storage by a capacitor is,

If we apply this relationship in the expression of energy, we can get the following expression related with

that expression.

Hence, all of the above expressions are same.

Answer: Option (a)

Hint:

The Electric potential v = πε

V=6 million volt = 6 x 106 v

Diameter d = 2m, radius r = 1 m

πε = x 9 N m2 C

-2

= πε r V

q = x −9x x6x 6

= x − = .66x − C or .66 milli coulomb



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· 5/12/2018 · Velumanickam Matriculation Higher Secondary School ± vaani ± Ramanathapuram. 1. ELECTROSTATICS (12 th Standard Physics - Objective Questions and Answers for Centum