Assuming that \(q_{1}\) and \(q_{2}\) remain constant, what happens to the magnitude of \(\bar{V}\) if the separation, \(d\), is increased?

Short Answer

Expert verified
If the separation (\(d\)) is increased, the magnitude of the electric potential (\(\bar{V}\)) decreases.

Step by step solution

01

Understanding the Basic Formula

The formula to find the electric potential due to a point charge is presented by \(V=k*\frac{q}{r}\). This formula indicates that the electric potential \(V\) is proportional to the charge \(q\) and inversely proportional to the distance \(r\) from the charge.
02

Analyzing the Effect of Increasing Separation

Given that \(q_{1}\) and \(q_{2}\) (which might be considered as \(q\) in this scenario) remain constant, we are only changing the value of the separation \(d\), denoted by \(r\) in our formula. Increasing the separation or distance \(d\) would mean increasing the denominator in the formula. As the denominator of a fraction increases, while the numerator remains constant, the overall value of the fraction (and subsequently \(V\)) decreases.
03

Formulating the Final Implication

With the understanding of the behaviour of the electric potential formula, an increase in the separation \(d\) would lead to a decrease in the magnitude of the electric potential \(\bar{V}\). This is due to the inverse relationship between the electric potential and the distance from the charge.

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