Question Number 36279 by Cheyboy last updated on 31/May/18
$${Consider}\:{a}\:{particle}\:{in}\:{a}\: \\ $$$${uniformly}\:{charge}\:{electric}\:{field}, \\ $$$${if}\:{the}\:{particle}\:{has}\:{a}\:{charge}\:{of}\:\mathrm{2}{C} \\ $$$${and}\:{is}\:{place}\:\mathrm{3}{m}\:{away}\:{from}\:{the} \\ $$$${charge}\:{plate}. \\ $$$$\boldsymbol{{calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{work}}\:\boldsymbol{{needed}}\:\boldsymbol{{to}} \\ $$$$\boldsymbol{{move}}\:\boldsymbol{{the}}\:\mathrm{2}\boldsymbol{{C}}\:\boldsymbol{{particle}}\:\boldsymbol{{to}}\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{distance}}\:\boldsymbol{{of}}\:\mathrm{1}\boldsymbol{{m}}\:\boldsymbol{{from}}\:\boldsymbol{{the}}\: \\ $$$$\boldsymbol{{plate}}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18
$${W}={qEd} \\ $$$${q}=\mathrm{2}{C} \\ $$$$ \\ $$$$\:\:{E}=\frac{\sigma}{\epsilon_{\mathrm{0}} }\:{for}\:{conducting}\:{pate} \\ $$$${W}={q}×\frac{\sigma}{\epsilon_{\mathrm{0}} }×{d} \\ $$$${W}=\mathrm{2}×\frac{\sigma}{\epsilon_{\mathrm{0}} }×\mathrm{1}=\frac{\mathrm{2}\left(\sigma\right)}{\epsilon_{\mathrm{0}} }\:\:{in}\:{problem}\:{value}\:{of}\:\sigma \\ $$$${nlg}\:{given} \\ $$
Commented by Cheyboy last updated on 31/May/18
$${ok}\:{sir}\:{the}\:{book}\:{made}\:{mistake} \\ $$$$ \\ $$$$ \\ $$