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2. Electric Potential and Capacitance
easy
The potential at a point, due to a positive charge of $100\,\mu C$ at a distance of $9\,m$, is
A
${10^4}\,V$
B
${10^5}\,V$
C
${10^6}\,V$
D
${10^7}\,V$
Solution
(b) By using $V = 9 \times {10^9} \times \frac{Q}{r}$$ = 9 \times {10^9} \times \frac{{100 \times {{10}^{ – 6}}}}{9} = {10^5}\,V$
Standard 12
Physics
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