1. Given data

• Charge at $A$: $q_A = +20\,\mu C$
• Charge at $B$: $q_B = -5\,\mu C$
• Separation: $AB = 5\,\text{cm}$

We need the position where a third charged particle experiences zero net electric force.

Since force on the third particle is $F = qE,$ this means we need the net electric field to be zero.

2. Choose regions to test

Place $A$ at $x=0$ and $B$ at $x=5\,\text{cm}$.

We check three regions:

1. Left of $A$
2. Between $A$ and $B$
3. Right of $B$

3. Check between the charges

Between $A(+20\mu C)$ and $B(-5\mu C)$:

• Field due to $+20\mu C$ is away from $A$, i.e. to the right.
• Field due to $-5\mu C$ is towards $B$, also to the right.

So both fields are in the same direction, hence they cannot cancel.

Therefore, no zero-field point lies between the charges.

So option C and D are incorrect.

4. Check left of $A$

Let the point be at distance $x$ cm to the left of $A$.

Then distances are:

• From $A$: $x$
• From $B$: $x+5$

For cancellation, magnitudes of fields must be equal: $\frac{k\cdot 20}{x^2} = \frac{k\cdot 5}{(x+5)^2}$

Cancel $k$: $\frac{20}{x^2} = \frac{5}{(x+5)^2}$ $4(x+5)^2 = x^2$ $2(x+5)=x$ which gives negative distance, impossible; or equivalently solving fully: $4(x^2+10x+25)=x^2$ $3x^2+40x+100=0$ This gives no physically suitable positive solution for the left side.

Hence, no valid point left of $A$.

So option A is incorrect.

5. Check right of $B$

Let the point be at distance $x$ cm to the right of $B$.

Then distances are:

• From $B$: $x$
• From $A$: $x+5$

For net field zero: $\frac{k\cdot 20}{(x+5)^2} = \frac{k\cdot 5}{x^2}$

Cancel $k$: $\frac{20}{(x+5)^2} = \frac{5}{x^2}$ $4x^2 = (x+5)^2$ $2x = x+5$ $x=5\,\text{cm}$

So the zero-field point is 5 cm to the right of the $-5\mu C$ charge.

6. Match with options

This corresponds to:

B: At 5 cm from $-5\,\mu C$ on the right side

7. Comparison with stored answer

Stored correct answer: B

My derived answer: B

So they agree.