Figures \((a)\) and \((b)\) show the field lines of a positive and negative point charge respectively.
Give the signs of the potential difference \(V_{\text P}-V_{\text Q}\) and \(V_{\text B}-V_{\text A} .\)
Give the signs of the potential difference \(V_{\text P}-V_{\text Q}\) and \(V_{\text B}-V_{\text A} .\)
Answer
\(\left. \left({V}_{P}−{V}_{Q}\right)\right.\)- Positive. \(\left(V_B-V_A\right)\)-Negative
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Solution
Solution -
As potential is given by\(V\;=\;\dfrac{1}{4\pi {\varepsilon }_{0}}\dfrac{q}{r}\)
or \(V\;\propto \;\frac{1}{r}\) .
So, \({V}_{P}\;> \;{V}_{Q}\).
Thus, \(\left. \left({V}_{P}−{V}_{Q}\right)\right.\) is positive.
Also \(V_B\) is less negative than \(V_A\)in fig (b)
Thus, \({V}_{B}\;> \;{V}_{A}\)
But in this case the charge is negative
so \(\left(V_B-V_A\right)\) is negative
Hence, \(\left. \left({V}_{P}−{V}_{Q}\right)\right.\) is positive and \(\left(V_B-V_A\right)\) is negative.
As potential is given by\(V\;=\;\dfrac{1}{4\pi {\varepsilon }_{0}}\dfrac{q}{r}\)
or \(V\;\propto \;\frac{1}{r}\) .
So, \({V}_{P}\;> \;{V}_{Q}\).
Thus, \(\left. \left({V}_{P}−{V}_{Q}\right)\right.\) is positive.
Also \(V_B\) is less negative than \(V_A\)in fig (b)
Thus, \({V}_{B}\;> \;{V}_{A}\)
But in this case the charge is negative
so \(\left(V_B-V_A\right)\) is negative
Hence, \(\left. \left({V}_{P}−{V}_{Q}\right)\right.\) is positive and \(\left(V_B-V_A\right)\) is negative.
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