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Two charges 3 xx 10^(-8) C and - 2 xx 10^(-8) C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero. |
Answer» Solution :Suppose charge of `3 xx 10^(-8)` C be TAKEN on the origin O and charge of `- 2 xx 10^(-8)` C be taken on point A along positive x-axis at distance 15 cm as shown in figure.  The potential at point P is zero. As seen logically the potential due to both charges cannot be zero for left side of O (x `LT` 0) and hence there are two possibilities for point P. (i) If point P is between O and A, then suppose P is from O at distance x cm and hence point A is at distance (15 - x) cm from P. Potential at P due to charge on point O is `V_(1) =(k(3xx10^(-8)))/(x xx 10^(-2))` and Potential at P due to charge on point A is `V_(2)=(k(-2xx10^(-8)))/((15-x)xx10^(-2))` but potential at P is `V=V_(1)+V_(2)` `O=(kxx3xx10^(-8))/(x xx10^(-2))-(kxx2xx10^(-8))/((15-x)xx10^(-2))` `O =(3)/(x)-(2)/(15-x)` `:. (2)/(15-x)=(3)/(x)` `:. 2X= 45 -3x` `:. 5x=45` `:. x=9` cm (ii) If neutral point obtain from right side of A them .  `:.` Now potential at P due to charge on point O is `V_(1)=(k(3xx10^(-8)))/(x xx10^(-2))` Potential at P due to charge on point A `V_(2) =(k(-2xx10^(-8)))/((x-15)xx10^(-2))=-(k(2xx10^(-8)))/((x-15)xx10^(-2))` Total potential at point P, `V= V_(1)+V_(2)` `:. O = (k(3xx10^(-8)))/(x xx10^(-2))-(k(2xx10^(-8)))/((x-15)xx10^(-2))` `:. (3)/(x)=(2)/(x-15)` `:. 3x-45=2x` `:. x= 45 cm` THUS, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge . Note : Here, CHOOSING a potential to be zero at infinity. |
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