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.

#### Solution

There are two charges,

q_{1} = 5 × 10^{−8} C

q_{2} = −3 × 10^{−8} C

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

r = Distance of point P from charge q_{1}

Let the electric potential (V) at point P be zero.

Potential at point P is the sum of potentials caused by charges q_{1} and q_{2} respectively.

∴ `"V" = q_1/(4piin_0"r") + "q"_2/(4piin_0("d" - "r"))` ........(i)

Where,

`in_0` = Permittivity of free space

For V = 0, equation (i) reduces to

`"q"_1/(4piin_0"r") = -"q"_2/(4piin_0("d" - "r"))`

`"q"_1/"r" = -"q"_2/("d" - "r")`

`(5xx10^-8)/"r" = -(-3 xx 10^-8)/(0.16 - "r")`

`0.16/"r" = 8/5`

∴ r = 0.1 m = 10 cm

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,

`"V" = "q"_1/(4piin_0"s") + "q"_2/(4 piin_0("s" -"d"))` ......(ii)

For V = 0, equation (ii) reduces to

`"q"_1/(4piin_0"s") = "q"_2/(4piin_0("s" - "d"))`

`"q"_1/"s" = -"q"_2/("s" - "d")`

`(5 xx 10^-8)/"s" = ((-3 xx 10^-8))/(("s" - 0.16))`

`1-0.16/"s" = 3/5`

`0.16/"s" = 2/5`

∴ s = 0.4 m = 40 m

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.