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1. Electric Charges and Fields
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Two free positive charges $4q$ and $q$ are a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed form charge $q$ ?
A
$Q = \frac{4}{9} q\,(negative)\, at \frac{l}{3}$ distance
B
$Q = \frac{4}{9} q\,(positive)\, at \frac{l}{3}$ distance
C
$Q = q\,(positive)\, at \frac{l}{3}$ distance
D
$Q = q\,(negative)\, at \frac{l}{3}$ distance
Solution
For charge $q:-\frac{K(4 q)(q)}{\ell^{2}}-\frac{K Q q}{x^{2}} \Rightarrow x=\frac{\ell}{3}$
For charge $(4 q):-\frac{\mathrm{k}(4 q)(q)}{\ell^{2}}=\frac{\mathrm{k}(4 \mathrm{q}) \mathrm{Q}}{(\ell-\mathrm{x})^{2}}$
$\Rightarrow \mathrm{Q}=\frac{4}{9} \mathrm{q}$ (negative)
Standard 12
Physics
