Two-point charges +9q and -q are kept 10 cm apart. Find a point on the line joining the charges, where electric potential is zero.

Let the first charge be $Q_1 = +9q$ and the second charge be $Q_2 = -q$.
The distance between the charges is $d = 10$ cm.
The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{kQ}{r}$, where $k$ is Coulomb's constant.

We are looking for a point where the total electric potential is zero. Let this point be $P$.
The total potential at $P$ is the sum of the potentials due to $Q_1$ and $Q_2$:

where $r_1$ is the distance from $Q_1$ to $P$, and $r_2$ is the distance from $Q_2$ to $P$.
We want $V_P = 0$, so:

There are two regions on the line joining the charges where the potential can be zero:

• Region A: Between the charges. Let the point $P$ be at a distance $x$ from $Q_1 (+9q)$. Then $r_1 = x$. The distance from $Q_2 (-q)$ will be $r_2 = d - x = 10 - x$. Substitute these into the equation $9r_2 = r_1$:

  $$9(10 - x) = x$$ $$90 - 9x = x$$ $$90 = 10x$$ $$x = 9 \text{ cm}$$

  This means the point is 9 cm from $+9q$. This is a valid solution as $0 < 9 < 10$.

• Region B: Outside the charges, to the right of $-q$. Let the point $P$ be at a distance $x$ from $Q_1 (+9q)$. Then $r_1 = x$. The distance from $Q_2 (-q)$ will be $r_2 = x - d = x - 10$. Substitute these into the equation $9r_2 = r_1$:

  $$9(x - 10) = x$$ $$9x - 90 = x$$ $$8x = 90$$ $$x = \frac{90}{8} = \frac{45}{4} = 11.25 \text{ cm}$$

  This means the point is 11.25 cm from $+9q$. This is also a valid solution as $11.25 > 10$.

• Region C: Outside the charges, to the left of $+9q$. Let the point $P$ be at a distance $x'$ to the left of $Q_1 (+9q)$. Then $r_1 = x'$. The distance from $Q_2 (-q)$ will be $r_2 = d + x' = 10 + x'$. Substitute these into the equation $9r_2 = r_1$:

  $$9(10 + x') = x'$$ $$90 + 9x' = x'$$ $$90 = -8x'$$ $$x' = -\frac{90}{8} = -11.25 \text{ cm}$$

  Since distance cannot be negative, there is no solution in this region. This is expected because the zero potential point must be closer to the charge with the smaller magnitude, which is $-q$.

We found two points where the electric potential is zero:

1. 9 cm from $+9q$.
2. 11.25 cm from $+9q$ (which is 1.25 cm from $-q$).

Let's check the options:
(1) 1 cm from $+9q$ (Incorrect)
(2) 9 cm from $-q$. If this is between the charges, it means 1 cm from $+9q$. (Incorrect)
(3) 9 cm from $+9q$ (Correct, matches our first solution)
(4) 5 cm from $-q$. If this is between the charges, it means 5 cm from $+9q$. (Incorrect)