1) Four point charges placed at the corners of a square were given. The total electric potential at the center of the square was calculated to be 4.5 x 10^4 V. 2) The electric field and potential due to a point charge were given. Using these, the distance of the point from the charge and the magnitude of the charge were calculated. 3) An oil drop carrying a charge between the plates of a capacitor was given. The voltage required to balance the drop, given the mass and distance between plates, was calculated to be 9.19 V.

1. Physics Helpline
L K Satapathy
Electrostatics-4
Electric Field / Potential

A
B
C
E
-q
+q
O
F1
F2

2. Physics Helpline
L K Satapathy
Cells in Parallel
Electrostatics-4
Q1. Four point charges 4C, -2C, -3C and 6C are placed at the four corners of
a square ABCD of side 2 m . Determine the electric potential at the point of
intersection of the diagonals of the square.
Answer :
V = 4.5 x 10 4 V [ Ans ]
9
9 10 [ 1 ]
4 o
q
V q r m
r
    
9 6 3
1 9 10 4 10 36 10V V
      
9 6 3
3 9 10 ( 3) 10 27 10V V
       
9 6 3
4 9 10 6 10 54 10V V
     
3 3
(36 18 27 54) 10 45 10V V V       
A
D
B
C
O
2
1
(-2C)
(-3C)
(4C )
(6C)
AB=BC=CD=DA=2m
AO=BO=CO=DO=1m
9 6 3
2 9 10 ( 2) 10 18 10V V
       

3. Physics Helpline
L K Satapathy
Cells in Parallel
Electrostatics-4
Q2. The electric field at a point due to a point charge is 20N/C and the electric
potential at that point is 10 J/C. Determine the distance of the point from the charge
and the magnitude of the charge.
Answer :
2
. . . (1) . . . (2)
44 oo
q q
E V
rr 
 
10
(1) & (2) 0.5 [ ]
20
V
r m Ans
E
   
9 9
(2) 10 9 10 18 10
4 0.5o
q q
V q
r
        
9 10 1010 50
10 10 5.56 10 [ ]
18 9
q C Ans  
      
Electric field and potential due to a charge (q) at distance (r) are given by
[ E = 20N/C & V = 10 J/C ]

4. Physics Helpline
L K Satapathy
Cells in Parallel
Electrostatics-4
Q3. An oil drop carrying 10 electrons is located between the plates of a capacitor
which are 5mm apart. Given that mass of the drop is 3x10 –16 kg. Determine the
voltage applied across the capacitor to balance the drop.
Answer :
16
18
3
3 10
10 1.6 10
5 5 10
m kg
q e C
d mm m



 
  
  
V qV
E F qE
d d
qV
W mg F mg
d
   
   
16 3
18
(3 10 )(9.8)(5 10 )
1.6 10
3 4.9
9.1875
1.6
mgd
V
q
 

 
  


 
9.19 [ ]V V Ans 

5. Physics Helpline
L K Satapathy
Cells in Parallel
Electrostatics-4
Q4. Two particles A and B of masses m and 2m having equal charge q on them,
fall through a potential difference of V volts. If the speed of A is v , then find the
speed of B .
Answer :
21
. . . (1)
2
mv qV
2 2 1 1
2mv qV const v v
m m
      
1
2 2
v m m
v m m

   

[ ]
2
v
v Ans 
When a charge q , falls through a potential difference of V volts ,
the kinetic energy acquired by it = qV

6. Physics Helpline
L K Satapathy
Cells in Parallel
Electrostatics-4
Q5. An electric dipole of dipole moment 4 x10 – 9 Cm is aligned at 30 with the
direction of a uniform electric field E of strength 5 x104 N/C. Determine the
magnitude of torque acting on the dipole.
Concepts :
sinpE  
Dipole moment : p = 2aq (along BA)
Dipole : (+q) at A and (–q) at B , AB = 2a

A
B
C
E
-q
+q
O
F1
F2
F1 = F2 = qE (equal & opposite forces)
Angle between ( p ) and ( E ) =   BC = 2a sin [ dist. between F1 & F2 ]
Magnitude of torque on the dipole :  = qE  2a sin
[ p = 2aq ]

7. Physics Helpline
L K Satapathy
Cells in Parallel
Electrostatics-4
Answer : Given : Dipole moment
9
4 10 .p C m
 
Angle between p and E : 30 sin 0.5o
   
Electric field : 4
5 10E N C 
 Torque sinpE 
4
10 . [ ]N m Ans 
 
5
(4 5 0.5) 10
   
9 4
(4 10 )(5 10 )(0.5)
  

8. Physics Helpline
L K Satapathy
For More details:
www.physics-helpline.com
Subscribe our channel:
youtube.com/physics-helpline
Follow us on Facebook and Twitter:
facebook.com/physics-helpline
twitter.com/physics-helpline