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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250×10-7C (charges)
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The electric field at one corner of a square is 1614217 N/C.
Explanation:
The distance between x and y direction diagonals.
As per the given details the distance between diagonals is calculated as
0.5² + 0.5² = c² => c = 0.707 m
Charge to the right: In x direction
In order to find the electric charge towards x direction
we use e = kq/r² formula
As ‘k’ is coulomb’s constant it’s value is 9 x
N m²/C²
e = (9 x
)(250 x 
) / (0.5)²
e = 9 x
N/C
Charge diagonal:
e = kq/r²
e = [(9 x
)(250 x 
) / (0.707)²] cos 45
e = 225000√2 N/C
X direction sum = 1218198 N/C.
Similarly as shown in x direction the charge is same for y direction also
Charge below: For y direction
e = kq/r²
e = (9 x
)(250 x 
) / (0.5)²
e = 9 x
N/C
Charge diagonal:
e = kq/r²
e = [(9 x
)(250 x 
) / (0.5)²] sin 45
e = 159099 N/C
Y direction sum = 1059099 N/C
Resultant electric field strength:
1218198 ² + 1059099² = e²
e = 1614217 N/C [45 degrees below the horizontal]