39 14. (a) As shown in figure, direction of electric field at P will be along + ve y-axis.
Clearly resultant field is along angle bisector of field towards 9 and 10.
y
6E1
6E1
–Q
6E1
E
E1
E2
E 6E1
x
P
6E1 6E1
Q
Hence time shown by clock in the direction of electric field is 9 : 30. F - eE 16. (a) a = = m m - 1.6 ´ 10-19 ´ 1 ´ 103 = 9.1 ´ 10-31
(b) Positive x-axis. y E2
+Q
P
E
x
E1
+Q
(c) Positive y-axis. E E1
E2
–Q
+Q
1 q × 4pe0 R2 11 E11
u 12 E12
10 E10
1 E1 E2
2
E3 3 E4
E8 7
E7 E 6 6
E
q
9 E9 8
(b) v = u + at 5 ´ 106 t= = 2.8 ´ 10-8 = 28 ns. 1.76 ´ 1014 (c) Dk = work done by electric field. = F × x = - eEx = - 1.6 ´ 10-19 ´ 1 ´ 103 ´ 8 ´ 10-3 = - 1.28 ´ 10-18 J Loss of KE = 1.28 ´ 10-18 J 25 ms -1 17. Here, ux = u cos 45° = 2
P
15. Let E1 =
= - 1.76 ´ 1014 ms -2 u = 5.00 ´ 108 cm/s = 5 ´ 106 ms -1 v=0 v2 - u 2 = 2as (5 ´ 106 )2 s= = 1.4 ´ 10-2 = 1.4 cm 2 ´ 1.7 ´ 1014
E5 5
4
Resultant fields of two opposite charges can be shown as given in figure.
u y = u sin 45° =
25 ms -1 2
a x = qE = 2 ´ 10-6 ´ 2 ´ 107 = 40 ms -1 a y = - 10 ms -1 1 2 y = u yt + t ay y=
25 t - 5t 2 2