Physics, asked by yashraj7653, 1 year ago

Two charges +q and –3q are separated by a distance of 1 m. At what point in between the charges on its axis is the potential zero?

Answers

Answered by Astrobolt
3
Hey there!
Let's assume that the point in between the point charges is at a distance X metres from the +q charge and (1 - X) metres from the -3q charge.

Potential at a distance from a point charge is directly proportional to the magnitude of the charge, with its sign and it is inversely proportional to the distance of the point from the charge.

Thus,

V (Potential)
Q (Charge)
R (Distance)

Are related as follows:

V = (kQ)/R

Where k is a constant whose value is approximately 9 × 10^9. However this value is not required in this question.

V1 is the potential due to the +q charge and V2 is the potential due to the -3q charge.

V1 + V2 = 0 (Given in the question)

 \frac{k(q)}{x}  +  \frac{k( - 3q)}{1 - x}  = 0 \\  \frac{kq}{x}  =  \frac{3kq}{1 - x}  \\  \frac{1}{x}  =  \frac{3}{1 - x}  \\ 1 - x = 3x \\ 4x = 1 \\ x =  \frac{1}{4}
Thus X is 1/4 m or 25 cm

Hence the answer is 25 cm from the +q charge and 75 cm from the -3q charge on the line connecting the two charges. At this point the potential due to the two charges is zero.
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