charges of +(10/3)*1/10000000 are placed at each of thefour corners of a square of side 8cm .the potential at the intersection of the diagonals is.PLEASE SOLVE THIS QUESTION.

Dear sunny shah,

V(r) = (1/4 π ε)*   (q1/r1+ q2/r2+ q3/r3+ q4/r4) = (1/4 π ε)*   4*(q/r) = (1/ π ε)*   (q/r)

q1=q2=q3=q4=q 

r=sqrt(4^2+4^2)   because  the total lenght of the side is 8cm . the cahrge is at diagonal of the square from the center

r=sqrt(32)  cm    chane to meter 

r=(1/10)*sqrt(32) m

ε= 8.854187817...×10⁻¹² A·s/(V·m) = 8.854187817...×10⁻¹² F/m

q=+(10/3)*1/10000000 = (10/3)*  10^-6

Π=22/7

now u can use the formula and values given above. do it so u can get practice and solve the otherparts

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