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1. Electric Charges and Fields
medium
A conducting sphere of radius $R$, and carrying a charge $q$ is joined to a conducting sphere of radius $2R$, and carrying a charge $-2q$. The charge flowing between them will be
A
$\frac{q}{3}$
B
$\frac{{2q}}{3}$
C
$q$
D
$\frac{{4q}}{3}$
Solution
(d) Initial charge on sphere of radius $R = q$
Charge on this sphere after joining $q' = \frac{{(q + ( – 2q) \times R}}{{R + 2R}}$ $ = \frac{{ – q \times R}}{{3R}} = – \frac{q}{3}$
Now charge flowing between them $ = q – \left( { – \frac{q}{3}} \right) = \frac{{4q}}{3}$
Standard 12
Physics