Question-60135

Question Number 60135 by Tawa1 last updated on 18/May/19

Answered by tanmay last updated on 18/May/19

q_1 =8.5×10^(−6) C at x_1 =3.0×10^(−2) meter q_2 =−21×10^(−6) C at x_2 =9.0×10^(−2) meter q_0 (test charge=1C at x_o =6.0×10^(−2) meter foce on q_o by q_1 is repulsive directed towards +ve x axis F_1 =9×10^9 ×((8.5×10^(−6) )/((6×10^(−2) −3×10^(−2) )^2 )) F_1 =9×10^9 ×((8.5×10^(−6) )/(9×10^(−4) ))=8.5×10^7 N force on q_o by q_2 attractiveF directed +ve x axis F_2 =9×10^9 ×((21×10^(−6) )/((9.0×10^(−2) −6.0×10^(−2) )^2 )) F_2 =9×10^9 ×((21×10^(−6) )/(9×10^(−4) ))=21×10^7 N so net electric field=8.5×10^7 +21×10^7 =29.5×10^7 and direction +ve x[axis pls check

$${q}_{\mathrm{1}} =\mathrm{8}.\mathrm{5}×\mathrm{10}^{−\mathrm{6}} {C}\:{at}\:{x}_{\mathrm{1}} =\mathrm{3}.\mathrm{0}×\mathrm{10}^{−\mathrm{2}} {meter} \\ $$$${q}_{\mathrm{2}} =−\mathrm{21}×\mathrm{10}^{−\mathrm{6}} {C}\:{at}\:{x}_{\mathrm{2}} =\mathrm{9}.\mathrm{0}×\mathrm{10}^{−\mathrm{2}} {meter} \\ $$$${q}_{\mathrm{0}} \left({test}\:{charge}=\mathrm{1}{C}\:{at}\:{x}_{{o}} =\mathrm{6}.\mathrm{0}×\mathrm{10}^{−\mathrm{2}} {meter}\right. \\ $$$${foce}\:{on}\:{q}_{{o}} \:{by}\:{q}_{\mathrm{1}} \:{is}\:{repulsive}\:{directed}\:{towards}\:+{ve}\:{x}\:{axis} \\ $$$${F}_{\mathrm{1}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{8}.\mathrm{5}×\mathrm{10}^{−\mathrm{6}} }{\left(\mathrm{6}×\mathrm{10}^{−\mathrm{2}} −\mathrm{3}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}_{\mathrm{1}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{8}.\mathrm{5}×\mathrm{10}^{−\mathrm{6}} }{\mathrm{9}×\mathrm{10}^{−\mathrm{4}} }=\mathrm{8}.\mathrm{5}×\mathrm{10}^{\mathrm{7}} {N} \\ $$$${force}\:{on}\:{q}_{{o}} \:{by}\:{q}_{\mathrm{2}} \:\:{attractiveF}\:{directed}\:+{ve}\:{x}\:{axis} \\ $$$${F}_{\mathrm{2}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{21}×\mathrm{10}^{−\mathrm{6}} }{\left(\mathrm{9}.\mathrm{0}×\mathrm{10}^{−\mathrm{2}} −\mathrm{6}.\mathrm{0}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}_{\mathrm{2}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{21}×\mathrm{10}^{−\mathrm{6}} }{\mathrm{9}×\mathrm{10}^{−\mathrm{4}} }=\mathrm{21}×\mathrm{10}^{\mathrm{7}} {N} \\ $$$${so}\:{net}\:{electric}\:{field}=\mathrm{8}.\mathrm{5}×\mathrm{10}^{\mathrm{7}} +\mathrm{21}×\mathrm{10}^{\mathrm{7}} \\ $$$$=\mathrm{29}.\mathrm{5}×\mathrm{10}^{\mathrm{7}} \:\:{and}\:{direction}\:+{ve}\:\:{x}\left[{axis}\right. \\ $$$${pls}\:{check} \\ $$

Commented by Tawa1 last updated on 18/May/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by peter frank last updated on 19/May/19

$${thanks} \\ $$

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