Answer:
the total force vector, on test charge is points from origin to point C( 1, 1 )
Explanation:
Given the data in the question, as illustrated in the image below;
from the Image, OA = 1, OB = AC = 1
so using Pythagoras theorem
a² = b² + c²
a = √( b² + c² )
so
OC = √( OB² + AC² )
we substitute
OC = √( OA² + AC² )
OC = √( 1² + 1² )
OC = √( 1 + 1 )
OC = √2
Coordinate of C( 1, 1 )
Hence, the total force vector, on test charge is points from origin to point C( 1, 1 )
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]