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2. Electric Potential and Capacitance
hard
Point charge ${q_1} = 2\,\mu C$ and ${q_2} = - 1\,\mu C$ are kept at points $x = 0$ and $x = 6$ respectively. Electrical potential will be zero at points
A
$x = 2$ and $x = 9$
B
$x = 1$ and $x = 5$
C
$x = 4$ and $x = 12$
D
$x = - 2$ and $x = 2$
Solution
(c) Potential will be zero at two points
At internal point $(M)$ :$\frac{1}{{4\pi {\varepsilon _0}}} \times \left[ {\frac{{2 \times {{10}^{ – 6}}}}{{(6 – l)}} + \frac{{( – 1 \times {{10}^{ – 6}})}}{l}} \right] = 0$
$==>$ $l = 2$
So distance of $M$ from origin; $x = 6 -2 = 4$
At exterior point $(N)$ :$\frac{1}{{4\pi {\varepsilon _0}}}$ $×$ $\left[ {\frac{{2 \times {{10}^{ – 6}}}}{{6 – l'}} + \frac{{ – 1 \times {{10}^{ – 6}}}}{{l'}}} \right] =0 $
$l'=6$
So distance of $N$ from origin, $x = 6 + 6 = 12$
Standard 12
Physics
