Download Use Coulomb`s law to determine the magnitude of the electric field at

Use Coulomb's law to determine the magnitude of the electric field at points A and B in the figure due to
the two positive charges (Q = 7 \mu C, a = 5.4 cm) shown.
Part A
Express your answers using two significant figures separated by a comma. E_A,E_B =
Part B
Use Coulomb's law to determine the direction of the electric field at points A and B Express your
answers using two significant figures separated by a comma.
theta_A,theta_B =
SOLUTION: One charge is placed at origin and the other charge is placed at (4a, 0).
Electric field at a distance R is given by E 

Q
4 0 R
2
aR .

Where Q is charge, R is the distance between charge Q and the point where filed is to be calculated aR
is unit vector in the direction of line joining Q and the point. For positive charge, it points away from the
charge.  0 is permittivity of space.  0 
1
Farad / m
36 109
At point B, the field will be vector sum of the field created by charges at origin and at the point (4a,0).
2
 5.4 
3
For charge at origin and point B, R  a  a  2a  2  
  2  2.916  10 .
 100 
2

Unit vector aR 
ai  aj
2a
2

i j
.
2
2
2
2
2
 5.4 
3
  10  2.916 10
 100 
Similarly, for charge at (4a,0) and point B, R 2  9a 2  a 2  10a 2  10  

Unit vector aR 
Hence, EB =
3ai  aj
10a
2

3i  j
.
10
7 106
4 
1
36 109
i j


 2  2.916 103  2  4 
7 106
1
36 109
 3i  j 


10  2.916 10 3  10 
 7.64 106 i  7.64 106 j  2.049 106 i  0.6832 106 j
 5.591106 i  8.3232 106 j
 10.026 106 56.1090
Similarly, at point A,
EA =
7 106
4 
1
36 109
 2i  j 


3 
5
 4 
 5  2.916 10
 1.932 106 (2i  j )  1.932 106 (2i  j )
 3.864 106 j
 3.864 106 900
Hence, part A
EA  3.864 106 N / Coulomb
EB  10.026 106 N / Coulomb
Part B
 A  900
 B  56.1090
7 106
1
36 109
 2i  j 


5 
 5  2.916 103 