CLASS - XII PHYSICS (Unit – Electrostatics)

[ANSWERS]

FV

force between the charges in vaccuum

=

FM force between two charges in medium

FV

⇒

Fm =

k

⇒ if k increases, Fm decreases.

Ans 01.

Since K=

Ans 02.

Suppose rod P be negatively charged since it attracts rod R ⇒ R is positively charged since it repels rod Q ⇒ Q is negatively charged. So force between Q and R is attractive in nature.

Ans 03.

Since F = q E and a = F/m

Since charge on proton and electron are same but mass for electron is smaller, hence force and acceleration experienced by an electron is greater.

Ans 04.

Two electric lines of force never intersect

each other because if they intersect then

at the point of intersection there will be

two tangents which is not possible as the

two tangents represents two directions

for electric field lines.

Ans 05.

Since F = qE

F qE

∴ a= =

---------1

mm

Using third equation of motion

v2 – u2 = 2as

Initially charged particle is at rest ∴ u = o

⇒ v2 = 2as

1

1

KE = mv 2 = m (2 as) = mas ----- 2

2

2

Substituting 1 in eq. 2

qE

KE = m x

×S

m

KE = qES

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E1

E2

Ans 06.

Let P be the pt where test

charge (+qo) is present

then electric field at pt. P

will be zero if Field at pt. P

due to +q = field at p+. P

due to + 9q------------1

EB

+q

A

x

P

+qv

EA

10a-x

+9q

B

10a

E

⇒ EA =

K (+ q)

x

2

EB =

K (+9q )

(10a − x) 2

Substituting in eq. 1

K (+ q )

x2

=

K ( +9 q )

(10a − x) 2

(10a − x) 2 = 9x 2

⇒ 10a − x = 3x

10a = 4x

x=

10

9

4

x = 2.5 a from change (+q)

⇒

Or

10 a - x = 10a - 2.5a = 7.5a from change (+9q)

Ans 07.

Electric dipole moment is defined as the product of the magnitude of either charge and the length of dipole.

P = q (2l )

Its S.I. unit is coulomb meter (cm)

Consider a dipole placed in uniform electric field and makes an angle ( θ ) with the electric field E Since two forces acts on the charges constituting an electric

()

dipole which are equal and opposite in direction, thus a torque acts on the dipole which makes the dipole rotate.

And Torque τ = Ethier force X ⊥ distance

Here force (F) = qE

And

BN

AB

= sinθ

⇒ BN = AB sinθ = 2ℓ sin θ

( τ ) = qE x 2 ℓ sin θ

( τ ) = PE Sin θ (∵ P = q(2l ))

In vector form

τ = P×E

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Ans 08.

(1) The Significance of writing

lim

means the test charge

q→o

should be vanishingly small so

that it should not disturb the

presence of source charge.

Z

(0,0,10)cm

-7

(-2x10

P

2l=10+10=20cm

(2) (i) Total charge of the

system

X

-7

(2x10 )

(0,0,-10)cm

= 2 x 10 -7 + (-2 x 10 -7)

= zero. P

Y

Z

(ii) P = q × 2 l

P= 2 x 10-7 x 20 x 10-2

P = 4 x 10-8 cm

Direction of Dipole moment – Along negative x-axis.

Ans 09. (a)

(b)

E+q

Kq

E+q = 2

(r + a 2 )

Kq

E−q = 2

(r + a 2 )

θ

2

P (+qo) E= ?

E net

Since E + q = E − q

r +a

2

r 2 + a2

E -q

2

r

(-q)

∴ Enet can be calculated by

parallelogram law of vector

A

θ

θ

a

o

a

using a

addition.

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B(+q)

Enet = E + q 2 + E− q 2 + 2 E+ q E -q cos2θ

Enet = 2 E + q 2 +2 E+ q 2 cos 2θ

Enet = 2 E + q 2 (1 + cos 2θ )

Enet = 2 E + q 2 2 cos 2 θ = 4 E + q 2 cos 2 θ

Enet = 2 E + cos θ

Enet = 2 E +

cos θ =

a

r + a2

2

a

r 2 + a2

kq

a

Enet = 2 2

2

r + a r 2 + a2

Enet =

k 2aq

KP

=2

2 3/ 2

(r + a )

(r + a 2 )3/2

2

For r >>> a (a can be neglected)

Enet =

KP

r3

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