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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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## Answers ( )

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 directionIn 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 directione = 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]