Chapter 16, Problem 66GP

To determine

To Find: Net electric force on Q₁ .

Answer to Problem 66GP

  $0.45\mathrm{N}$

Towards right

Explanation of Solution

Given:

  ${\mathrm{Q}}_1=6.7\text{μC}$

  ${\mathrm{Q}}_2=1.8\text{μC}$

Electric field between plates,

  $\mathrm{E}=73\times {10}^3\text{N}/\text{C}$

Formula used:

Force on, ${\mathrm{Q}}_1$ are due to charge ${\mathrm{Q}}_2$ and electric field between plates . So, net force will be

  ${\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}={\displaystyle \sum _{}^{}\mathrm{F}=\frac{\mathrm{K}{\mathrm{Q}}_1{\mathrm{Q}}_2}{\mathrm{X}2}}-{\mathrm{Q}}_1\mathrm{E}$

Calculation:

The 2-point charges are ${\mathrm{Q}}_1=-9.7\mathrm{\mu }\text{C}$ and ${\mathrm{Q}}_2=1.8\mathrm{\mu }\text{C}$

Distance between the charge $\mathrm{r}=0.34\mathrm{m}$

Electric field between the plates $\mathrm{E}=73,000\text{N}/\text{C}$

The force on ${\mathrm{Q}}_1$ due to ${\mathrm{Q}}_2$ is ${\mathrm{F}}_1=\frac{\mathrm{k}{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}$ to the right

Or, ${\mathrm{F}}_1=\frac{\left(9\times {10}^9\right)\left(6.7\times {10}^{-6}\right)\left(1.8\times {10}^{-6}\right)}{{0.34}^2}\text{N}$

  $=0.94\mathrm{N}$ to the right

The force on ${\mathrm{Q}}_1$ due to the electric field is ${\mathrm{F}}_2={\mathrm{Q}}_1\mathrm{E}$ to the left.

Or, ${\mathrm{F}}_2=\left(6.7\times {10}^{-6}\right)(73000)\text{N}$

  $=0.49\text{N}$ to the left

Hence, the net electrostatic force on ${\mathrm{Q}}_1$ is $\mathrm{F}={\mathrm{F}}_1-{\mathrm{F}}_2$

  $=(0.94-0.49)\text{N}$

  $\mathrm{F}=0.45\text{N}$ to the right.

Conclusion:

Net force on Q is $0.45\mathrm{N}$ ( towards right)