In the figure, the four particles form a square of edge length a = 5.00 cm and have charges q1 = +15.0 nC, q2 = -30.0 nC, q3 = +30.0 nC and q4 = -15.0 nC. What net electric field do the particles produce at the square’s center?
____N/Ci + ___N/Cj
Answer
General guidance
[Section: Concepts and reason]
The question is based upon the electric field due to a dipole at a point on the equatorial line.
Initially, draw the labeled diagram of the electric field due to dipole. Find the expressions for the angle made by the field with the vertical and the distance of the charge from the field point.
Then, find the electric field due to one of the dipole at the center of the square. For this write the expression of the electric field due to one of the charge substitute the values of the distance and the angle in the expression. After that take, the vector sum of the electric field due to charges.
Repeat the same with another dipole and take the vector sum of the field due to two dipoles. After that plug in the values in the derived expression of electric field and solve.
The expression of the electric field is,
Here, E is the electric field, k is the coulomb’s constant, Q is the charge, and R is the distance.
Step-by-step
Step 1 of 2
Consider an electric dipole of charges 
and 
placed at 
distance apart. The electric field at a point on the perpendicular bisector at distance 
from the center of the dipole is shown in the following diagram.
Here, 
is the angle made by the electric field with the vertical, is the length of the rod and 
is the distance between the center of the dipole to the field point.
and 
are the electric field at the field point due to positive charge 
and negative charge 
respectively.
From the figure, the distance of the charge from the field point is,
The angle 
is given below:
Substitute 
for 
as follows:
Let the horizontal direction be taken as 
axis and vertical direction is 
component.
From the given diagram, due to symmetry the horizontal component of electric field cancels out each other and vertical component of electric fields add up. Thus, the net electric field due to charges 
and 
acts in vertically downward direction.
On the same line, the net electric field due to charges 
and 
acts in vertically upward direction.
The 
component of electric field 
due to charge 
is,
Substitute
for
and 
for 
as follows:
Again substitute 
for 
as follows:
So, the net electric field at the center of the square due to charges and is given below:
On the same line, the electric field at the center of the square due to charges 
and 
is given as follows:
Hence, the expression for the net electric field at the center of the square is,
The direction of the electric field due to positive charge is directed radially outward from the positive charge and radially inwards for a point negative charge.
Step 2 of 2
The net electric field at the center of the square is given as follows:
Substitute 
for and ; 
for and ;
for 
and 
for 
as follows:
Hence, the electric field at the center of the square is
.
The electric field at the center of the square is .
The electric field at any point due to the dipole is the vector sum of electric field along 
and 
direction. But here due to symmetry, the horizontal component of the electric field cancels each other.
So, the net electric field at the center of the square is equal to the sum of vertical components of electric field.
Answer
The electric field at the center of the square is .
Answer only
The electric field at the center of the square is .