The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4
The bigger the object the greater the gravitational pull, so the farther away the big object is its gravitational force begins to decrease. Refer to the picture for more explanation.
Well i think the best answer would be A
In series connection in orde to find the total resistance we add all the resistance capacity but in parallel, we take the 1/R and resistance = V hence the series connection takes up more power,
Like charges repel, while opposite charges attract.
