Two charges `5xx10^(-8)C and -3xx10^(-8)C` are located 16 cm apart. At what points on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.

Explanation

There are two charges,
`q_(1)=5xx10^(-8)C`
`q_(2)=-3xx10^(-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 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))/(4pi in_(0)r)+(q_(2))/(4pi in_(0)(d-r))......(i)`
Where,
`in_(0)`=Permittivity of of free space
for V=0, equation (i) reduces to
`(q_(1))/(4pi in_(0)r)=-(q_(2))/(4pi in_(0)(d-r))`
`(q_(1))/r=(-q_(2))/(d-r)`
`0.16/4 -1=3/5`
`0.16/r =8/5`
`:. r=0.1 m =10 cm `
Therefore, the potential is zero at distance of 10 cm from the positive charge between
the charges.
Supposite point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in figure in the following figure.

for this arragements , potential is given by
`V=(q_(1))/(4pi in_(0)s)+(q_(2))/(4pi in_(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)`
`(5xx10^(-8))/s=((-3xx10^(-8)))/((s-0.16))`
`1-0.16/s=3/5`
`0.16/s=2/5`
Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.