Three particles with charges of -2, 4 and -3 μC are positioned at (0, 1.2), (0, 0), and (1.2)?
Three particles with charges of -2, 4 and -3 μC are positioned in the xy-plane at (0, 1.2), (0, 0), and (1.2, 0),
where the coordinates are all in meters, and there are no other charges around. What is the potential at (1.2,1.2) in the SI units, assuming that the potential at infinity is zero?
A) -7500 B) 7500 C) 15400 D) -15400 E) -16200
Please help! For some reason, I can't seem to solve this problem. It was on my midterm earlier and i had gotten the question right. But, now I can't remember how i did it.
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Charge,q1=-2x10^-6 C
Distance,r=√(1.2^2 +0)=1.2 m
Electric potential,V1=k q1/r
=9x10^9x(-2x10^-6)/1.2
=-15000 V
Charge,q2=4x10^-6 C
Distance,r=√(1.2^2 +1.2^2)=1.69 m
Electric potential,V2=k q2/r
=9x10^9x4x10^-6/1.69
=2.13x10^4 V
Charge,q3=-3x10^-6 C
Distance,r=√(0+1.2^2)=1.2 m
Electric potential,V3=k q3/r
=9x10^9x(-3x10^-6)/1.2
=-22500 V
Total electric potential,V=V1+V2+V3
=(-15000+2.13x10^4-22500) V
=-16200 V
The correct answer is E)-16200
in this series the extra moderen numbers are the previous style circumstances the sum of the the two numbers in basic terms before it: 4=2(a million+a million), 12=4(a million+2), so 12(4+2)=seventy two, that could be the subsequent style interior the series.