Test Paper 1 Electrostatics
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ELECTROSTATICS-1 Single Correct Choice Type(+3, 1) 1. A uniform el electric field E a ˆi b ˆj , intersects a surface of area A. What is the flux through this area if the surface lies in the yz plane? (A) aA 2.
(B) 0
(D) A a2
(C) bA
b2
y
The The fie field ld line line to the the rig right ht is a fie field ld line line of of the the ele elect ctri ric c fie field ld,, the then n its representation can be: (A) (A ) E ( x, y) ˆi sin ( x)ˆj
(B) (B ) E ( x, y) ˆi cos ( x)ˆj
0,0
(C) E( x, y) ˆi sin( x)ˆj
x
(D) E( x, y) ˆi cos( x)ˆj
3.
The net electric field E due to the uniformly charged rod at P makes angles 1 and respectively. Then
2 with AP and BP
1 / 2 : E P
1 2
A
B + + + + + + + + + + + + + + + + + +
( A) = 1
(B) > 1
(C) < 1
(D)
AP PB
4.
A particle of charge q and mass m moves in a circle around a long straight wire of linear charge density +. If r = radius of the circular path and T = time period of the motion in circular path. Then 1/2 2 2 3 (A) T = 2r(m/2Kq) (B) T = 4 mr /2qK 1/2 1/2 (C) (C) T = 1/2 1/2r(2Kq/m) (D) (D) T = 1/2 1/2r(m/Kq) , where K = 1/4 0
5.
A ring ring cha charge rge Q dist distri rib buted uted unif unifor orml mly y alo along ng its its len lengt gth. h. A sma small ll portion of the ring is cut so that a little charge Q is removed. The electric field at the centre is: KQ ˆ K (Q Q ) ˆ i i (A) (A ) (B) 2 R R2 K Q (C) 2 ˆi (D) none of these these R
Q
y O
x R
6.
Elec Electri tric c char charge ges s +q, +q, +q and and –2q –2q are are hel held d fixe fixed d at at (0, (0, 0), 0), (d, (d, 0) and and (0, (0, d) d) respe respecti ctive vely ly of a (x, y) coordinate system. The electric dipole moment of the system is (A) qd ( ˆi 2ˆj) (B) qd ( ˆi 2ˆj) (C) qd ( ˆi 2ˆj) (D) qd ( ˆi 2ˆj)
7.
The The potential V is va varyi rying wi with x and y as V (A) 2ˆi ˆj V / m
8.
(B)
2ˆi ˆjV / m
1 2 ( y 4 x ) volt. The field at x = 1m, y = 1m, is: 2 (C) 2ˆi ˆj V / m (D) 2ˆi 2ˆj V / m
A neg negat ativ ive e cha charg rge e – q move moves s slo slow wly in a ci circul rcula ar pa path from from position 1 to position 2. The work done by the external agent in the electric field of + Q fixed at the origin is (take 1 K 40 (A) zero
(B)
4KQq 3a
(C)
2KQq 3a
2 –a
Q +
1 –q a
(D)
2a
3a
2KQq 3a
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ELECTROSTATICS-1 9.
T he E-x patt er n for th e give n V -x patt er n is
v 0 V 10 V
20 V
30 V
60°
x
O
E
E
60°
30°
x
(A) E
E
x
(C) 10.
x
(B)
150° x
(D)
Between two infinitely long wires having linear charge densities and – there are two points A and B as shown in the figure. The amount of work done by the electric field in moving a point charge q0 from A to B is equal to
q0 In 2 (A) 20 2q0 In 2 (C) 0
(B) (D)
2q0
0
A a
r
B a
a
In 2
q0 In 2 0
Paragraph for Questions 11 to 13 The potential at any point can be given as V
–
E dr. This helps us to find V if E is given. If V is given at
any point, E in any direction can be found by taking the derivative of V in that direction which is given as V Ex . Using the above formulae, answer the following questi ons. x 2
11.
E varies along x as E = 3x . If the potential at x = –1 is +3 volt, the potential at x = + 2 is (A) –10 volt (B) + 4 volt (C) – 6 volt (D) – 12 volt
12.
In the previous question, the volume charge density at x = 1 is 40 (A) 6 0 (B) (C) 3 0 (D)
13.
none of these
If the potential at any region varies in x-y plane as V a (x 2 y 2 ), which of the following field patterns is correct? y y (A) (B)
x
(C)
x
y
(D)
x
y
x
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ELECTROSTATICS-1 Multiple Correct Choice Type(+4, 1) 14.
A point charge q is placed at origin. Let E A .EB and EC be the electric field at three points A(1, 2, 3), B(1, 1, –1) and C (2, 2, 2) due to charge q. Then
(A) E A
EB
(D) | EB | 8 | EC |
A particle of mass m and charge + q has y been projected from ground as shown in mg the figure, such that tan . Mark out qE the correct statement (s). (A) The path of motion of the particle is parabolic (B) The path of motion of the particle is a straight line. 2u sin (C) Time of flight of the particle is g (D) Range of motion of the particle is equal to
16.
(C) | EB | 4 | EC | 15.
(B) E A || EB
u
E
(A uniform electric field)
x
u2 sin2 . g
Electrostatic lines of forces: (A) cannot be closed (B) can be circular (C) never intersect (D) terminate on –ve charges and originate from +ve charges
Integer Type(+5, 1) 17.
Two identical helium filled balloons A and B fastened to a weight of 15 gram by threads float in equilibrium as shown in the figure. –6 If the charge on each balloon is Q×10 Coulomb, then find the value of Q. Assuming that they carry equal charges (take g = 10 2 m/s ).
1.2m A
B +
+
1m
18.
1m
A ring of radius R = 3m carries a charge q 10 6 C uniformly distributed over it. A long, thin wire carrying charge 10 C / m per unit length of it is held along its axis with one end coinciding with –3 the centre of the ring. If the interaction force between the ring and the thread is M × 10 Newton, then find the value of M. 6
19.
The bob of a pendulum has mass m = 1 kg and charge q = 40 C. Length of pendulum is = 0.9 m. The point of suspension also has the same charge 40 C. What the minimum speed u(in m/s) should be imparted to the bob so that it can complete vertical circle?
q
q m
u
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ELECTROSTATICS-1 Answers 1. A
2.
S Aiˆ S = E S Aa A Graph of field line represents the graph of ( cos x), so y = cos x
3.
4.
dy coefficient of E along y axis sinx E ˆi sin xjˆ dx coefficient of E along x axis A As we know that direction of field at any point due to uniformly charged rod is along the angle bisector of the angle formed by the ends of rod at the point, So 1 = 2 A + 2kq0 mv 2 mv 2 FC = + r r r +
1 v
T= 5.
+ + + r + + +
m 2kq0
2r v
2r
m 2kq0
C Net electrostatic field at the centre of uniformly charged ring is zero so,
Q
6.
O
k Q ˆ (i ) R2
x R
C
ˆ P (q)(0iˆ 0 ˆj) (q)(diˆ 0 ˆj) ( 2q)(0iˆ dj)
7.
Q
y
EC E C( Q) E C(Q) 0
EC(Q) E c( Q)
q0
ˆ qd(iˆ 2j)
C
V ˆ V ˆ 1 1 i j = ( 4)iˆ (2y)jˆ 2iˆ yjˆ E x 1m 2iˆ ˆj 2 y 2 x y 2m
E 8.
D
8.
Ui UB
KQq 3a
, Uf U A
KQq a
Work done by external filed U Uf Ui 9.
10. 10.
KQq a
2KQq KQ q = 3a 3a
–a A
+3a B
D As we know that field lines are perpendicular to the equi-potential line or surface, and its direction is in the direction of decreasing potential. D 2K ˆ 2K ˆ 1 ˆ + – 1 E i i 2K (i) x 3a x x 3a x + –
dU qE dxiˆ U 2K q UB UA U
2a
a
1 1 x 3a x dx
q q 2a ln(2) ln | x | ln | 3a x | a 20 0
q ln(2) Work done by electrostatic force U 0
+
–
+
3a – x
+ +
x
P
– – –
+
–
+
–
+
–
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ELECTROSTATICS-1 11.
C
11.
dV
Edx 3x 2 dx 2
2
13x2dx V(2) – V(–1) = – x3 1 V(2) – 3 = – [8 – (–1)] = –9 V(2) = –9 + 3 = –6 Volt dV
12.
A
1 3bx 2 2 3b(x dx)2 d = 1 2 3b[(dx)2 2xdx] bdx 6bxdx = 6 0x 0 Second Method dE 6x dx 0
13.
ˆ E 2a[xiˆ yj]
13. 14.
B A, C
E AO 15.
0 dy y dx x
E dE
E
O
= 6 0x
b
x 1 60
dy dx y x
Kq (14)
ˆ E , E (iˆ 2jˆ 3k) A BO 3/ 2
x+dx
x
lny = – lnx + lnc
Kq
y
ˆ E , and E (iˆ ˆj k) B CO 3/ 2 (3)
c x
xy = C
Kq 8(3)
ˆ E (2iˆ 2ˆj 2k) 3/ 2
B, C
16.
17.
qE ˆi gjˆ F mgjˆ qEiˆ a m Hence direction of velocity and acceleration will be along sam e line but opposite, so particle will m ove along straight line. A, C, D Basic definition of field lines
3 2T cos T sin
= mg
=
tan =
...(i)
2
Kq r 2
...(ii)
2kq2 r 2mg
0.6 2 9 10 9 Q 2 10 12 Q=3 0.8 1.2 1.2 15 10 3 10 18.
3
F 19.
k Q R
3
6
9 10 9 1600 10 12 160 N 2 0.9 0.9 9 mg = 1 10 = 10 N kq2
kQ2
2
mg , so string behave as a rod
u 4g
4 10 0.9 6 m/s
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