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

Concept: Electric Potential — The potential due to a point charge is $V = \frac{kq}{r}$, and potentials add as scalars. We want the net potential to be zero.

Let the two charges be $q_1 = +5 \times 10^{-8}\ \text{C}$ and $q_2 = -3 \times 10^{-8}\ \text{C}$, separated by $d = 0.16\ \text{m}$. Place $q_1$ at $x = 0$ and $q_2$ at $x = d$.

Step 1: For a point on the line joining them, the potential is zero when:

Since $q_2$ is negative, $-\frac{q_2}{r_2}$ is positive, so the point must lie outside the segment between the charges, closer to the smaller magnitude charge.

Step 2: Let the point be at distance $x$ from $q_1$ on the side of $q_2$ (beyond $q_2$). Then $r_1 = d + x$, $r_2 = x$. The equation becomes:

Cancel $10^{-8}$ and solve:

So the point is $24\ \text{cm}$ from $q_2$ (or $40\ \text{cm}$ from $q_1$).

Step 3: Check if a point between the charges works. If between them, $r_1 = x$, $r_2 = d - x$, then:

This gives $r_1 = 10\ \text{cm}$, $r_2 = 6\ \text{cm}$. Check: $V = k\left(\frac{5\times10^{-8}}{0.1} + \frac{-3\times10^{-8}}{0.06}\right) = k(5\times10^{-7} - 5\times10^{-7}) = 0$. So this point also works.

The electric potential at a point due to a point charge is $V = kq/r$, where $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$ and $r$ is the distance from the charge. Potential is a scalar — it adds algebraically, not as a vector. That makes finding zero-potential points simpler than finding zero-field points: you just solve $V_1 + V_2 = 0$, with careful attention to signs.

Here, $q_1 = +5 \times 10^{-8}\ \text{C}$ and $q_2 = -3 \times 10^{-8}\ \text{C}$, separated by $d = 16\ \text{cm} = 0.16\ \text{m}$. We want points on the line joining them where the total potential is zero.

Because the charges have opposite signs, the potential can be zero in two distinct regions: between the charges (where one distance is small, the other large) and outside the smaller charge (where both distances are large but the signs differ). Let’s find both.

1. Set up the coordinate system. Place $q_1$ at $x = 0$ and $q_2$ at $x = 0.16\ \text{m}$. Let the point of interest be at distance $x$ from $q_1$, so its distance from $q_2$ is $|0.16 - x|$. The potential at that point is

Setting $V = 0$ and cancelling $k$ (nonzero) gives

2. Case 1: Point between the charges ($0 < x < 0.16$). Here $|0.16 - x| = 0.16 - x$ (positive). The equation becomes

Multiply through by $10^8$:

Cross-multiply: $5(0.16 - x) = 3x \ \Rightarrow\ 0.8 - 5x = 3x \ \Rightarrow\ 0.8 = 8x \ \Rightarrow\ x = 0.1\ \text{m} = 10\ \text{cm}.$

So one zero-potential point is $10\ \text{cm}$ from the positive charge, between the charges.

3. Case 2: Point outside the charges, beyond $q_2$ ($x > 0.16$). Here $|0.16 - x| = x - 0.16$. The equation is

Cross-multiply: $5(x - 0.16) = 3x \ \Rightarrow\ 5x - 0.8 = 3x \ \Rightarrow\ 2x = 0.8 \ \Rightarrow\ x = 0.4\ \text{m} = 40\ \text{cm}.$

So the second point is $40\ \text{cm}$ from the positive charge, on the side of the negative charge.

4. Case 3: Point outside, beyond $q_1$ ($x < 0$). Here $|0.16 - x| = 0.16 - x$ (since $0.16 - x > 0$). The equation becomes

But $x$ is negative, so $\frac{5}{x}$ is negative. The term $\frac{3}{0.16 - x}$ is positive. For the sum to be zero, the magnitudes must match, but solving gives $5(0.16 - x) = 3x \ \Rightarrow\ 0.8 - 5x = 3x \ \Rightarrow\ 0.8 = 8x \ \Rightarrow\ x = 0.1$, which is positive — a contradiction. So no solution exists on this side. (Intuitively, both terms would be negative if $x<0$, so they can’t sum to zero.)

Method: Finding Where the Potential Is Zero on the Line Joining Two Charges

This method applies whenever two point charges of opposite sign are given and you must find the point(s) on the line joining them where the net potential is zero.

Steps

Step 1: Set up a coordinate axis along the line joining the charges

Place one charge at the origin and the other at $x = d$ (the given separation). Let $x$ be the distance of the test point from the first charge.

Step 2: Write the zero-potential condition

Potential is a scalar, so the contributions simply add:

Since the charges have opposite sign, this ratio can be positive, so a real solution exists.

Step 3: Look for a solution between the charges

Between the charges, $r_1 + r_2 = d$. Substitute $r_2 = d - r_1$ and solve the resulting linear equation for $r_1$. This region always has exactly one solution when the charges have opposite sign, because the potential smoothly changes sign as you cross from one charge's side to the other.

Step 4: Look for a solution beyond the smaller-magnitude charge

Outside the segment, on the far side of the charge with the smaller magnitude, $r_1 - r_2 = d$ (or vice versa, depending on which side). Solve the same ratio equation with this new distance relationship. There is no solution on the far side of the larger-magnitude charge — check this by testing the sign of each term for that case.

Step 5: Report both distances, and verify by direct substitution

State each answer as a distance from a clearly-named reference charge, and as a sanity check substitute the numbers back into $q_1/r_1$ and $q_2/r_2$ to confirm they are equal in magnitude.