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Two point charges 20muC and 80muC are placed 18 cm apart. Find the position of the point where the electric field is zero. |
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Answer» Solution :DATA supplied, `q_(1)=20 xx 10^(-6)C,""q_(2)=80 xx 10^(-6) C, "AP=xcm "=x xx 10^(-2)m` `BP=(18-x) cm=(180x) xx10^(-2)m`  Let .P.be the point where electric field is vero, then `E_(PA)=E_(PB)` `(1)/(4PI epsi_(0)) q_(1)/x^(2)=(1)/(4pi epsi_(0)) q_(2)/((18-x)^(2)) ie, (20 xx 10^(-6))/((x xx 10^(-2))^(2))=(80 xx 10^(-6))/([(18-x) xx 10^(-2)])^(-2)` `1/x^(2)=4/((180 -x)^(2))""(18-x)^(2)=4x^(2)=18 -x=2x therefore x=6` Hence for x=6cm, the field at P is zero. |
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