Two charges


Two charges $5 \times 10^{-8} \mathrm{C}$ and $-3 \times 10^{-8} \mathrm{C}$ are located $16 \mathrm{~cm}$ 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.


There are two charges,

$q_{1}=5 \times 10^{-8} \mathrm{C}$

$q_{2}=-3 \times 10^{-8} \mathrm{C}$

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 q1

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

Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.

$\therefore V=\frac{q_{1}}{4 \pi \epsilon_{0} r}+\frac{q_{2}}{4 \pi \in_{0}(d-r)}$    ...(i)


$\epsilon_{0}=$ Permittivity of free space

For V = 0, equation (i) reduces to

$\frac{q_{1}}{4 \pi \in_{0} r}=-\frac{q_{2}}{4 \pi \in_{0}(d-r)}$


$\frac{5 \times 10^{-8}}{r}=-\frac{\left(-3 \times 10^{-8}\right)}{(0.16-r)}$



$\therefore r=0.1 \mathrm{~m}=10 \mathrm{~cm}$

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 from the negative charge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,

$V=\frac{q_{1}}{4 \pi \in_{0} s}+\frac{q_{2}}{4 \pi \in_{0}(s-d)}$  ...(ii)

For $V=0$, equation (ii) reduces to

$\frac{q_{1}}{4 \pi \epsilon_{0} s}=-\frac{q_{2}}{4 \pi \in_{0}(s-d)}$


$\frac{5 \times 10^{-8}}{s}=-\frac{\left(-3 \times 10^{-8}\right)}{(s-0.16)}$



$\therefore s=0.4 \mathrm{~m}=40 \mathrm{~cm}$

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

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