33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Electric Charges and Fields
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5
2
∴ n =
q e = 4 πε0Fd
(Using (i))
2
e
F
6. (c) : Fm = 0 K ie ..,decreases K times
7. (a) : The situation is as shown in the figure.
Let two equal charges Q each placed at points A and B at
a distance r apart. C is the centre of AB where charge q
is placed.
For equilibrium, net force on charge Q = 0
1 QQ 1 Qq
∴ +
= 0
4πε
2
0 4 2
r πε0
( r / 2)
2
1 Q 1 4Qq
Q
=− or Q=− 4q or q =−
4πε
2
0 4πε
2
r
0 r
4
8. (c) : Net force on each of the charge due to the other
charges is zero. However, disturbance in any direction
other than along the line on which the charges lie, will
not make the charges return.
9. (a) : Force experienced by a charged particle in an
electric field, F = qE
As F = ma
qE
∴ ma = qE ⇒ a =
....(i)
m
As electron and proton both fall from same height at rest.
Then initial velocity = 0
1 2
From s= ut+ at ( u=
0)
2
1 2 1 qE
∴ h= at ⇒ h=
2
2 m t 2
[Using (i)]
2hm
∴ t = ⇒ t ∝ m as ‘q’ is same for electron and
qE
proton.
As electron has smaller mass so it will take smaller time.
10. (b) :
6
Acceleration, a = − 0 −2
= 6 m s
1
1
Fort = 0tot = 1s s = × = m
2 61 2
, 1 ( ) 3
Fort = 1sto t = 2s s = 6× 1− 1 × = m
2 61 2
, 2
( ) 3
Fort = 2sto t = 3s s = 0 − 1 × =− m
2 61 2
, 3 ( ) 3
Total displacement s = s 1 + s 2 + s 3 = 3 m
3 −
Averagevelocity = = ms
3 1 1
Total distance travelled = 9 m
9 −1
Averagespeed = = 3 ms
3
11. (a) : As v 2 = 0 2 + 2ay = 2 (F/m)y = ⎛ ⎝ ⎜ ⎞
2 qE ⎠
⎟
m y
K.E. = 1 2
mv
2
1 ⎡ ( ) ⎤
∴ K.E. = m⎢2
qE ⎥ ⇒ K.E =
2 ⎣ m y ⎦
. qEy
12. (c) : Electric lines of force start from the positive
charge and end at the negative charge. Since the electric
lines for both the charges are ending, therefore both q 1
and q 2 are negative charges.
13. (d)
14. (d) : Electric flux, φ E = ∫ E
⋅dS
∫
∫
= EdScosθ = EdS cos 90°=
0.
The lines are parallel to the surface.
15. (a) : This consists of two dipoles, –q and +q with
dipole moment along with the +y-direction and –q and
+q along the x-direction.
O
2 2 2 2
\ The resultant moment = qa + qa = 2 qa.
Along the direction 45° that is along OP, where P is
(+a, +a, 0).
16. (d) : The electric field at a point on equatorial line
(perpendicular bisector) of dipole at a distance r is given
by
p 1
E = ⋅
4πε 2 2 32
0 ( r + a ) /
where 2a = length of dipole
For r > > a
p 1
∴ E = ⋅ ie E∝ p E∝ r
− 3
.., and
4πε
3
0 r
17. (b) : Here, q = 30°, E = 2 × 10 5 N C –1 , t = 4 N m,
l = 2 cm = 0.02 m, q = ?
t = pE sinq = (ql)E sinq
τ
4
∴ q = =
El sinθ 5
2× 10 × 0.
02 ×
1
4
2
−3 = = 2× 10 C = 2 mC
2 ×
3
10
P