Question

1. Two -3.00 nC point charges occupy the two lower vertices of a rectangle that is 6.00 cm wide and 8.00 cm tall. Two -1.00 nC point charges occupy the upper two vertices.

1. Determine the magnitude and direction of the electric field at the center of the rectangle (Answer: 11520 N/C downward)
2. What would be the magnitude and direction of the electric force acting on a -3.00 nC point charge placed at the center of the square? (Answer: 3.46 x 10^-5 N upward)

Answer 1

here,

the charges at the lower vertices , qA = qB = - 3 nC = - 3 * 10^-9 C

the chrage at the upper vertices , qC = qD = - 1 nC = - 1 * 10^-9 C

a = 6 cm = 0.06 m

b = 8 cm = 0.08 m

the distance of center from each vertex , r = sqrt(a^2+ b^2)/2

r = sqrt(0.06^2 + 0.08^2)/2 = 0.05 m

the electric feild magnitude due to charges placed diagonally , E = K * qC /r^2 - K * qA * r^2

E = 9 * 10^9 * 10^-9 /0.05^2 * ( 3 - 1)

E = 7200 N/C

due to similarity , other diagonal combination of charge has same electric feild magnitude and direction is 90 degree between them

so, by vector addition

the net electric feild is acted in DOWNward direction

and the magnitude of net electric feild , En = 2 * E * cos(90/2)

En= 2 * 7200 * cos(45) = 11520 N/C

b)

the charge at the center , q = - 3 nC = - 3 * 10^-9 C

as the charge is negative

the direction of force is opposite to the electric feild

so, the direction of force is UPWARDS

the magnitude of force , Fn = |q * En|

Fn = | - 3 * 10^-9 * 11520|

Fn = 3.46 * 10^-5 N