2019-03-01_Physics_Times

[2015]

(a) (50 10 N / C) j^3 ˆ (b) (50 10 N / C)i^3 ˆ

(c) (25 10 N / C) j^3 ˆ (d) (25 10 N / C)i^3 ˆ

An electric field E (25i 30 j)NC ˆ ˆ ^1



exists in a

region of space. If the potential at the origin is

taken to be zero then the potential at

x 2 m, y 2m  is:

(a) -110V (b) -140 V (c) -120 V (d) -130 V

b c d d d

a

Energy of a sphere

2

(^80)
Q
R


12
0
16 10
4.5
8 R
 


12
0
16 10 1
9 4
R
 
 
 = 16 mm
The potential difference across the sphere of
diameter 2R is 0.8 V. The potential at a point P whose
radius vector makes an angle 600 is
The potential difference between A and P is
A P
V R
V V R
2R 2
  
    
  
3 3
V 0.8V
4 4
   
VAVP0.6V VP589.2V
The net charge enclosed inside the gaussian
surfaces is same in all cases.
The differential form of Gauss law is
0
dE
dz



For z 1m V 30 5z ^2
z
dV
E 10z
dz

  
z
0
dE
10
dz

 
    ^100
For z 1m V 35 10 z 
For z > 1 m V = 35 - 10z
z
dV
E 10
dz

 
z
0
dE
0 0
dz

    

For z < -1 mV = 35 -10 z
z
dV
E 10
dz

 
z
0
dE
0 0
dz

    

Length of wire L = 20 cm
charge Q 10^3  0
We know, electric field at the centre of the
semicircular arc
2K
E
r

 and quarter circular arc
is
E
2
^
2Q
2K
r 2Q
E As
r r
 

 
 
    
 
2
3
2 2 2
4KQ 4KQ 4 KQ ˆ
25 10 N / Ci
r L L
 
 
    
The potential difference between two points is
1.Sol:
2.Sol:
3.Sol:
4.Sol:
5.Sol:

6. Sol:
   2
   2 1
   1

VV E d r.

2 2

0 0

V 0 E dxx yE dy

 

    

 

 

   

2 2

V 25 x 0 30 y 0

   

 =-110V