Question

Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin, and the positive-axis is directed from the negative charge to the positive.

At what points along the x axis is the electric field zero.

Express your answer using two significant figures. If there is more than one answer, give your answers separated by commas (cm)

At what points along the x axis is the potential zero? Let V=0 at r=∞.

Express your answer using two significant figures. If there is more than one answer, give your answers separated by commas.(cm)

Answer 1

Answer 2

1. Electric Field Zero:

We need $E_{q_1}+E_{q_2}=0$. The only valid region is between the charges ($0<x<5$):

$$k\frac{\mathrm{\mid }{q}_1\mathrm{\mid }}{x^2}=k\frac{\mathrm{\mid }{q}_2\mathrm{\mid }}{(5-x{)}^2}$$

Cancel $k$ and plug in $q_1=-2.0\mu C$, $q_2=3.4\mu C$:

$$\frac{2.0}{x^2}=\frac{3.4}{(5-x{)}^2}$$

Take square roots:

$$\frac{\sqrt{2.0}}{x}=\frac{\sqrt{3.4}}{5-x}$$

Solve for $x$:

$$x=\frac{5\sqrt{2.0}}{\sqrt{2.0}+\sqrt{3.4}}=\frac{5\times 1.414}{1.414+1.843}=\frac{7.07}{3.257}\approx 2.17\text{cm}$$

Final Answer:
Electric field is zero at $\boxed{{\displaystyle 2.2\text{cm}}}$.

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2. Electric Potential Zero:

We need $V_{q_1}+V_{q_2}=0$. The only valid region is left of $q_1$ ($x<0$):

$$k\frac{q_1}{\mathrm{\mid }x\mathrm{\mid }}+k\frac{q_2}{5-x}=0$$

Cancel $k$ and plug in charges:

$$\frac{-2.0}{\mathrm{\mid }x\mathrm{\mid }}+\frac{3.4}{5-x}=0$$

Since $x<0$, $\mathrm{\mid }x\mathrm{\mid }=-x$:

$$\frac{3.4}{5-x}=\frac{2.0}{-x}$$

Cross-multiply and solve:

$$3.4(-x)=2.0(5-x)$$$$-3.4x=10-2.0x$$$$-1.4x=10⟹x=\frac{10}{-1.4}\approx -7.14\text{cm}$$

Final Answer:
Electric potential is zero at $\boxed{{\displaystyle -7.1\text{cm}}}$