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.

Open in App

Solution

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

$x = 40 c m$

$x - r = 24 c m$

Suggest Corrections

Similar questions

Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Q. Two charges

3 \times 10^{- 8} C

and

- 2 \times 10^{- 8} C

are

15 c m

apart. At what point on the line joining the charges, theelectric potential is zero? Let the electric potential at infinity is zero.

Q. Two point charges

5 \times 10^{- 8} C

and

- 3 \times 10^{- 8} C

are located 16 cm apart. At what point on the line joining these charges the electric potential will be zero?

Q. Two charges

3 \times 10^{- 8} C

and

- 2 \times 10^{- 8}

are located

15 c m

apart. At what point on the line joining the two charges is the electric potential is zero?

Join BYJU'S Learning Program

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

$x = 40 c m$

$x - r = 24 c m$