Chapter 17, Problem 24P

(a)

To determine

The potential $V_{ba}$ will be calculated.

Answer to Problem 24P

The answer is $8.6\times {10}^{3}V$ .

Explanation of Solution

Given data:

Charge $q=-3.8\mu C$

Distance $r_b=0.88m$

Distance $r_a=0.72m$

Formula used:

It is known that: $V=\frac{kq}{r}$

Where,

  $k$ is the constant factor.

  $V$ is the potential.

  $\text{q}$ is the charges.

  $\text{r}$ is the distance between two charges.

Calculation:

Apply the formula to calculate the potential.

  $V_b-V_a=\frac{kq}{r_b}-\frac{kq}{r_a}$

  $V_b-V_a=kq\left(\frac{1}{r_b}-\frac{1}{r_a}\right)$

  $V_b-V_a=\left(8.99\times {10}^{9}N\cdot m^2/C^2\right)\left(-3.8\times {10}^{-6}C\right)\left(\frac{1}{0.88m}-\frac{1}{0.72m}\right)$

  $V_b-V_a=8.6\times {10}^{3}V$

Conclusion: The potential difference due to point charge is $8.6\times {10}^{3}V$ .

(b)

To determine

  $\overrightarrow{E_b}-\overrightarrow{E_a}$ will be calculated.

Answer to Problem 24P

The answers are $7.9\times {10}^{4}V/m$ and $56°$ .

Explanation of Solution

Given data:

Charge $q=-3.8\mu C$

Distance $r_b=0.88m$

Distance $r_a=0.72m$

Formula used:

It is known that: $\overrightarrow{E}=\frac{k\left|q\right|}{r^2}$

Where,

  $\text{q}$ is the charges.

  $\text{r}$ is the distance between two charges.

  $k$ is the constant factor.

  $\overrightarrow{E}$ is the electric field.

Calculation:

Calculating the electric field in right and downward,

Apply the formula and substitute the data $\overrightarrow{E}=\frac{k\left|q\right|}{r^2}$

  ${\overrightarrow{E}}_b=\frac{k\left|q\right|}{r_b{}^2}right$

  $=\frac{\left(8.99\times {10}^{9}N\cdot m^2/C^2\right)\left(3.8\times {10}^{-6}C\right)}{{\left(0.88m\right)}^2}right$

  $=4.4114\times {10}^{4}V/m,right$

Electric field in downward direction,

  ${\overrightarrow{E}}_a=\frac{k\left|q\right|}{r_a{}^2}down$

  $=\frac{\left(8.99\times {10}^{9}N\cdot m^2/C^2\right)\left(3.8\times {10}^{-6}C\right)}{{\left(0.72m\right)}^2}down$

  $=6.5899\times {10}^{4}V/m,down$

The resultant of electric field,

  $\left|\overrightarrow{E_b}-\overrightarrow{E_a}\right|=\sqrt{{\text{E}}_a^2+E_b^2}$

  $\left|\overrightarrow{E_b}-\overrightarrow{E_a}\right|=\sqrt{{\left(4.4114\times {10}^{4}V/m\right)}^2+{\left(6.5899\times {10}^{4}V/m\right)}^2}$

  $\left|\overrightarrow{E_b}-\overrightarrow{E_a}\right|=7.9\times {10}^{4}V/m$

Calculating the angle,

  $\theta ={\tan }^{-1}\frac{E_a}{E_b}$

  $\theta ={\tan }^{-1}\frac{65899}{44114}$

  $\theta =56°$

Conclusion: The required magnitude is $7.9\times {10}^{4}V/m$ and direction is $56°$ .