Problem

Two-point charges of $8 \, \mathrm{\mu C}$ and $-4 \, \mathrm{\mu C}$ are separated by a distance of $10 \, \mathrm{cm}$ in air. At what point on the line joining the two charges is the electric potential zero?

Data

• Charge $q_1 = 8 \, \mathrm{\mu C} = 8 \times 10^{-6} \, \mathrm{C}$
• Charge $q_2 = -4 \, \mathrm{\mu C} = -4 \times 10^{-6} \, \mathrm{C}$
• Separation: $r = 0.1 \, \mathrm{m}$

Prerequisite Concepts

• Electric Potential: The total potential $V$ at a point is the sum of potentials due to individual charges:

$$V = V_1 + V_2$$

• Zero potential condition: $V_1 = -V_2$.

Answer

Step 1: Zero Potential Condition

The potential at a distance $x$ from $q_1$ is:

$$V = k \frac{q_1}{x} + k \frac{q_2}{r - x} = 0$$

Step 2: Simplify

Eliminate $k$:

$$\frac{q_1}{x} = -\frac{q_2}{r - x}$$

Substitute charges:

$$\frac{8}{x} = \frac{4}{0.1 - x}$$

Step 3: Solve for $x$

Cross multiply:

$$8(0.1 - x) = 4x$$ $$0.8 - 8x = 4x$$ $$0.8 = 12x$$ $$x = \frac{0.8}{12} = 0.0667 \, \mathrm{m}$$

Step 4: Calculate Distance from $q_2$

$$r - x = 0.1 - 0.0667 = 0.0333 \, \mathrm{m}$$

Final Answer

• Distance from $q_1$: $0.0667 \, \mathrm{m}$ (or $6.67 \, \mathrm{cm}$)
• Distance from $q_2$: $0.0333 \, \mathrm{m}$ (or $3.33 \, \mathrm{cm}$).

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