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A point charge Q is placed at point O as shown in the figure. The potential difference VA – VB is positive. Is the charge Q negative or positive? |
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Answer» We know that, V = \(\cfrac{1}{4\pi\varepsilon_0}\cfrac{Q}r\) ⇒ V \(\propto\cfrac1r\) The potential due to a point charge decreases with increase of distance. VA – VB > 0 ⇒ VA > VB Hence, the charge Q is positive. |
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