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Two charges 3x10^(-8)C & -2x10^(-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 infinity to be 0. |
Answer» Solution : Let us take the ORIGIN O at the location of the positive charge. The line joining the two charges is taken to be the x-axis, the negative charge is taken to be on the right side of the origin (Fig.). Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no POSSIBILITY of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have. `(1)/(4pi epsi_0) [ (3 xx 10^(-8))/(x xx 10^(-2)) - (2 xx 10^(-8))/((15 - x) xx 10^(-2))] = 0` Where x is in cm. That is, ` 3/x - (2)/(15 - x) = 0` Which gives x = 9 cm If x lies on the extended line OA, the required condition is ` 3/x - (1)/(x - 15) = 0` Which gives x = 45 cm Thus , ELECTRIC potential is zero at 9cm and 45cm away from the positive charge on the side of the negative charge. NOTE that the formula for potential used in the culculation required choosing potential to be zero at infinity. |
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