# Physics Question? Its super hard and I need help pls?

4 positive charges are arranged (diagarm)

Determine the magnitude of electric field in the center of the square.

*Directions matter when adding field strength!

### 1 Answer

- Steve4PhysicsLv 71 year agoFavorite Answer
Square’s diagonal, d = √(0.500² + 0.500²)

The distance (r) from any corner-charge to the centre point is half the diagonal. We will need r².

r² = (diagonal/2)² = (diagonal)²/4 = (0.500² + 0.500²)/4 = 0.125

Field at centre from 2e at top = E1 = kq/r² = 9*10^9*2*1.6*10^-19/0.125

= 2.304x10^-8 N/C downwards (-y direction)

Field at centre from 4e at bottom = E2 = double the magnitude of E1

= 4.608x10^-8 N/C upwards (+y direction)

Field at centre from 3e at right= E3 = kq/r² = 9*10^9*3*1.6*10^-19/0.125

= 3.456x10^-8 N/C to left (-x direction)

Field at centre from 6e on left E4 = double the magnitude of E3

= 6.912x10^-8 N/C to right (+x direction)

Total of x components X = -3.456x10^-8 + 6.912x10^-8 = 3.456x10^-8 N/C

Total of y components Y = -2.304x10^-8 + 4.608x10^-8 = 2.304x10^-8 N/C

(Note we could have simplified the working because the distances are all the same. The fields from charges diagonally opposite the centre partially cancel. The 4e at the bottom and 2e at the top could be replaced by a single charge of 2e at the bottom. And the 6e on the left and 3e on the right could be replaced by a single charge of 3e on the right.)

Magnitude of resultant field is:

|E| = √(X² + Y²) = √(3.456² + 2.304²) x 10^-8 = 4.15x10^-8 N/C

Angle of field anticlockwise from +x axis is:

θ = tan⁻¹(Y/X) = tan⁻¹(2.304/3.456) = 33.7º

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