Electrostatics Test Paper

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