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Question Number 39696 by NECx last updated on 09/Jul/18

Commented by NECx last updated on 09/Jul/18

I posted this before but didnt get  any help please help me.I tried  solving it but I got stucked somewhere  maybe as a result of my depth in  the topic.    Thanks in advance.

$${I}\:{posted}\:{this}\:{before}\:{but}\:{didnt}\:{get} \\ $$$${any}\:{help}\:{please}\:{help}\:{me}.{I}\:{tried} \\ $$$${solving}\:{it}\:{but}\:{I}\:{got}\:{stucked}\:{somewhere} \\ $$$${maybe}\:{as}\:{a}\:{result}\:{of}\:{my}\:{depth}\:{in} \\ $$$${the}\:{topic}. \\ $$$$ \\ $$$${Thanks}\:{in}\:{advance}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

wait pls

$${wait}\:{pls} \\ $$

Commented by NECx last updated on 10/Jul/18

ok sir

$${ok}\:{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

electric field at point on x axis due to q_1  is  E_1 =9×10^9 ×((−6×10^(−6) )/((5^ ×10^(−2) )^2 ))    E_2 =9×10^9 ×((−6×10^(−6) )/((5×10^(−2) )^2 ))    both of this electric field has component along  x axis and y axis...yaxis complnents balance  each othe so effective electric field is  =2×9×10^9 ×((−6×10^(−6) )/(25×10^(−4) ))×cosθ  cosθ=(4/5)  9×10^9 ×((−6×10^(−6) )/(25×10^(−4) ))×2×(4/5)=3.456×10^7   direction along −ve x axis  force=qE=3.456×10^7 ×2×10^(−6)   =6.912×10=69.12N  pls check...

$${electric}\:{field}\:{at}\:{point}\:{on}\:{x}\:{axis}\:{due}\:{to}\:{q}_{\mathrm{1}} \:{is} \\ $$$${E}_{\mathrm{1}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{−\mathrm{6}×\mathrm{10}^{−\mathrm{6}} }{\left(\mathrm{5}^{} ×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} }\:\: \\ $$$${E}_{\mathrm{2}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{−\mathrm{6}×\mathrm{10}^{−\mathrm{6}} }{\left(\mathrm{5}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$ \\ $$$${both}\:{of}\:{this}\:{electric}\:{field}\:{has}\:{component}\:{along} \\ $$$${x}\:{axis}\:{and}\:{y}\:{axis}...{yaxis}\:{complnents}\:{balance} \\ $$$${each}\:{othe}\:{so}\:{effective}\:{electric}\:{field}\:{is} \\ $$$$=\mathrm{2}×\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{−\mathrm{6}×\mathrm{10}^{−\mathrm{6}} }{\mathrm{25}×\mathrm{10}^{−\mathrm{4}} }×{cos}\theta \\ $$$${cos}\theta=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{−\mathrm{6}×\mathrm{10}^{−\mathrm{6}} }{\mathrm{25}×\mathrm{10}^{−\mathrm{4}} }×\mathrm{2}×\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{3}.\mathrm{456}×\mathrm{10}^{\mathrm{7}} \\ $$$${direction}\:{along}\:−{ve}\:{x}\:{axis} \\ $$$${force}={qE}=\mathrm{3}.\mathrm{456}×\mathrm{10}^{\mathrm{7}} ×\mathrm{2}×\mathrm{10}^{−\mathrm{6}} \\ $$$$=\mathrm{6}.\mathrm{912}×\mathrm{10}=\mathrm{69}.\mathrm{12}{N} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$

Commented by NECx last updated on 10/Jul/18

yeah...That′s it.Thanks

$${yeah}...{That}'{s}\:{it}.{Thanks} \\ $$

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