The four cases below all show a ball with a charge of 3 q at the top right corne

Question

The four cases below all show a ball with a charge of 3q at the top right corner of a square. Then there is either one, two, or three additional charged balls, each ball located at a different corner of the square.

(a) Rank the cases based on the magnitude of the net electric field at the center of the square, from largest to smallest. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)

(b) In Case D, calculate the magnitude of the net electric field at the center of the square.
_____ N/C

(c) In Case A, calculate the magnitude of the net electric field at the bottom left corner of the square.
_____ N/C

Explanation / Answer

a)
IN A field is due to q and 3q at an angle of 90
in B field is due to q
in C field is due to 3q
in D field is due to q

C > A > B = D

b)
In D, field due to both +q cancels out and so that field due to -2q
Whole system can be seen as -q charge placed in the spot of -3q
E = K*q/r^2
r = L/sqrt(2) = 0.3 m /sqrt (2) = 0.212 m

E = K*q/r^2
= (8.99*10^9)*(3*10^-6) / (0.212)^2
= 2*10^5 N/C
Answer: 2*10^5 N/C

c)
IN A field is due to q and 3q at an angle of 90
E(q) = K*q/r^2
= (8.99*10^9)*(3*10^-6) / (0.212)^2
= 2*10^5 N/C
E(3q) = K*3q/r^2
= (8.99*10^9)*(3*3*10^-6) / (0.212)^2
= 6*10^5 N/C

Now use sum of 2 vectors formula
If a and b are 2 vectors then
sum = sqrt (a^2 + b^2 - 2*a*b*cos thetha)
= sqrt ((2*10^5)^2 + (6*10^5)^2 -2*(2*10^5)*(6*10^5)*cos 90)
=sqrt (4*10^10 + 36*10^10 - 0)
= 6.32*10^5 N/C
Answer: 6.32*10^5 N/C

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