Question

In the figure, four charges, given in multiples of 2.00×10^-6 C form the corners of a square and four more charges lie at the midpoints of the sides of the square. The distance between adjacent charges on the perimeter of the square is d = 4.70×10^-2 m. What are the magnitude and direction of the electric field at the center of the square? The magnitude of E? E_x? E_y? (Answers are E= 6.05e7 N/C, Ex= 5.32e7 N/C, Ey= 2.88e7 N/C) How do I work this problem to get those answers?

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Answer 1

From the figure above you can see electric field due to charges q3 and q7 and q2 and q6 is equal in magnitude but opposite in the direction hence cancel. So the net electric field is just sum of fields due to charges q1,q5,q4 and q8.

In the diagram you can see E4 and E8 are directed along X axis and E5 and E1 in the XY plane.

Let us first find magnitudes of E4,E8, E1 and E5.

Use equation E= kq1/r^2

E4= kq4/d^2 = kq/d^2= (9*10^9*2*10^-6)/(4.7*10^-2)^2 = 8.15*10^6 N/C

E8= kq8/d^2 = k(2q)/d^2= (9*10^9*(2*-2*10^-6)/(4.7*10^-2)^2 = 1.63*10^7 N/C

E1= kq1/d^2 = k(5q)/(sqrt2d)^2= (9*10^9*(5*2*10^-6)/(sqrt2*4.7*10^-2)^2 = 2.04*10^7 N/C

E5= kq5/d^2 = k(5q)/(sqrt2d)^2= (9*10^9*(5*2*10^-6)/(sqrt2*4.7*10^-2)^2 = 2.04*10^7 N/C

Note that I have just taken the magnitude of charges and electric fields so negative or positive sign ignored for simplicity.

Thus

Fnet = E4+E8+E1+E5

Fnetx = E4x+E8x+E1x+E5x

Fnetx = E4cos0+E8cos0+E1cos45+E5cos45     ………..(angle between E4 and X axis is 45 deg)

Enetx=(8.15*10^6)cos0+(1.63*10^7)cos0+(2.04*10^7)cos45+(2.04*10^7)cos45

Enetx= 5.32*10^7 N/C

Fnety = E4y+E8y+E1y+E5y

Fnety = E4sin0+E8 sin0+E1 sin45+E5 sin45     ………..(angle between E4 and X axis is 45 deg)

Enety=(8.15*10^6)sin0+(1.63*10^7)sin0+(2.04*10^7)sin45+(2.04*10^7)sin45

Enety= 2.88*10^7 N/C

Enet= sqrt(Enetx^2+Enety^2) = sqrt((5.33*10^7)^2+(2.9*10^7)^2)

Enet=6.1*10^7 N/C