Two charges 5 times 10-8- C and -3 times 10-8- C are located 0-16 m apart- At what point-s- on the line joining the two charges in the electric potential zero- Take the potential infinity to be zero-

Question

Two charges  5 times 10-8- C  and  -3 times 10-8- C  are located  0-16 m  apart- At what point-s- on the line joining the two charges in the electric potential zero- Take the potential infinity to be zero-

Toppr Admin · Accepted Answer

Using the formula for electrostatic potential-we get-Kq1x-Kq2-x-x2212-0-16-0-5x-x2212-3x-x2212-0-16-0-We get the value of x as-x-40cm-x-x2212-r-24cm

Two charges $5 \times 10^{- 8} C$ and $- 3 \times 10^{- 8} C$ are located $0.16 m$ apart. At what point(s) on the line joining the two charges in the electric potential zero? Take the potential at infinity to be zero.

Using the formula for electrostatic potential we get

$\frac{K q_{1}}{x} + \frac{K q_{2}}{(x - 0.16)} = 0$

$\frac{5}{x} - \frac{3}{x - 0.16} = 0$

We get the value of x as

$x = 40 c m$

$x - r = 24 c m$

Two charges 5×10−8C and −3×10−8C are located 0.16m apart. At what point(s) on the line joining the two charges in the electric potential zero? Take the potential at infinity to be zero.

Using the formula for electrostatic potential we get Kq1x+Kq2(x−0.16)=0 5x−3x−0.16=0 We get the value of x as x=40cm x−r=24cm

Two charges $5 \times 10^{- 8} C$ and $- 3 \times 10^{- 8} C$ are located $0.16 m$ apart. At what point(s) on the line joining the two charges in the electric potential zero? Take the potential at infinity to be zero.

Using the formula for electrostatic potential we get

$\frac{K q_{1}}{x} + \frac{K q_{2}}{(x - 0.16)} = 0$

$\frac{5}{x} - \frac{3}{x - 0.16} = 0$

We get the value of x as

$x = 40 c m$

$x - r = 24 c m$