Question Number 25254 by NECx last updated on 07/Dec/17
$${A}\:{charged}\:{sphere}\:{of}\:{mass}\:\mathrm{3}×\mathrm{10}^{−\mathrm{4}} {kg}\:{is} \\ $$$${suspended}\:{from}\:{a}\:{string}.{An} \\ $$$${electrical}\:{force}\:{acting}\:{horizontally} \\ $$$${on}\:{the}\:{sphere}\:{so}\:{that}\:{the}\:{string} \\ $$$${makes}\:{an}\:{angle}\:\mathrm{37}°\:{with}\:{the}\:{vertical} \\ $$$${when}\:{at}\:{rest}.\:{Find} \\ $$$$\left.{a}\right){magnitude}\:{of}\:{the}\:{electric}\:{force} \\ $$$$\left.{b}\right){the}\:{tension}\:{in}\:{the}\:{string}. \\ $$
Answered by mrW1 last updated on 07/Dec/17
$$\left.{a}\right) \\ $$$${F}_{{e}} ={mg}\mathrm{sin}\:\theta=\mathrm{3}×\mathrm{10}^{−\mathrm{4}} ×\mathrm{10}×\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{2}.\mathrm{25}×\mathrm{10}^{−\mathrm{3}} \:{N} \\ $$$$\left.{b}\right) \\ $$$${T}=\frac{{mg}}{\mathrm{cos}\:\theta}=\frac{\mathrm{3}×\mathrm{10}^{−\mathrm{4}} ×\mathrm{10}}{\frac{\mathrm{4}}{\mathrm{5}}}=\mathrm{3}.\mathrm{75}×\mathrm{10}^{−\mathrm{3}} \:{N} \\ $$