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2. Electric Potential and Capacitance
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Two metallic charged spheres whose radii are $20\,cm$ and $10\,cm$ respectively, have each $150\,micro - coulomb$ positive charge. The common potential after they are connected by a conducting wire is
A
$9 \times {10^6}\;volts$
B
$4.5 \times {10^6}\;volts$
C
$1.8 \times {10^7}\;volts$
D
$13.5 \times {10^6}\;volts$
Solution
(a) Common potential $V = \frac{{{\rm{Total \,charge }}}}{{{\rm{Total \,capacitance}}}}$
$V = \frac{{150 \times {{10}^{ – 6}} \times 2}}{{4\pi {\varepsilon _0}(10 \times {{10}^{ – 2}} + 20 \times {{10}^{ – 2}})}} = 9 \times {10^6}\,V$
Standard 12
Physics
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