Three charges at the corners of a square of sides s = 3.40 m. Here, q1...

Three charges at the corners of a square of sides s = 3.40 m. Here, q1 = q2 = −q and q3 = −1.5q where q = 9.00 nC. q2 is at the top left of the square, q3 at the bottom left and q1 at the bottom right.

(a) What is the magnitude of the electric field at the center of the square due to these three charges?

(b) You now replace q1 in the square with another point charge q4, while leaving q2 and q3 alone. What value of q4 will produce an electric field at the center directed vertically down?

Solutions

Expert Solution

Part (a)

Part (b)

now for the net electric field to be vertically down ,E₃ and E₄ should be equal in magnitude and both are perpendicular so net electric field will be on the angle bisector i.e 45^o .find the diagram and explanation in image below.q₄=2.5q=2.59nC=22.5nC

genius_generous answered 1 year ago

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