Electrostatic Test and Answer Paper

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CBSE TEST PAPER-01
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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