Chapter 17, Problem 50QAP

To determine

(a)

The potential at point $a$which is $40.0\text{cm}$away from the origin on $x$-axis.

Answer to Problem 50QAP

The potential at point $a$which is $40.0\text{cm}$away from the origin on $x$-axis is $-2.37\times {10}^3\text{V}$.

Explanation of Solution

Given info:

Charge $q_1=+2.0\mu \text{C}$

Charge $q_2=-3.0\mu \text{C}$

Position of charge $q_1$is, $x=0$.

Position of charge $q_2$is, $y=40.0\text{cm}$.

Position of point $a$is, $x=40.0\text{cm}$.

Formula used:

Formula for the electric potential due to a point charge is given as,

  $V=k\frac{Q}{r}$

Calculation:

The distance of point $a$from charge $q_1$is, $r_1=40.0\text{cm}$.

The position of point $a$from charge $q_2$is, $r_2=40\sqrt{2}\text{cm}$.

The potential at point $a$due to both the charges can be calculated as,

  $\begin{array}{c}{V}_a=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}\\ V_a=9\times {10}^9\times \left[\frac{2.0\times {10}^{-6}}{0.40}+\frac{-3.0\times {10}^{-6}}{0.40\sqrt{2}}\right]\\ V_a=-2.37\times {10}^3\text{V}\end{array}$

Conclusion:

Thus, the potential at point $a$which is $40.0\text{cm}$away from the origin on $x$-axis is $-2.37\times {10}^3\text{V}$.

To determine

(b)

The potential difference $V_b-V_a$.

Answer to Problem 50QAP

The potential difference $V_b-V_a$is $-0.95\times {10}^3\text{V}$.

Explanation of Solution

Given info:

Charge $q_1=+2.0\mu \text{C}$

Charge $q_2=-3.0\mu \text{C}$

Position of charge $q_1$is, $x=0$.

Position of charge $q_2$is, $y=40.0\text{cm}$.

Position of point $a$is, $x=40.0\text{cm}$.

Position of point $b$is, $x=40.0\text{cm}$and $y=30.0\text{cm}$.

Formula used:

Formula for the electric potential due to a point charge is given as,

  $V=k\frac{Q}{r}$

Calculation:

The distance of point $b$from charge $q_1$is, $r_1=\sqrt{{0.40}^2+{0.30}^2}=0.50\text{m}$.

The position of point $a$from charge $q_2$is, $r_2=\sqrt{{0.40}^2+{0.10}^2}=0.41\text{m}$.

The potential at point $b$due to both the charges can be calculated as,

  $\begin{array}{c}{V}_b=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}\\ V_b=9\times {10}^9\times \left[\frac{2.0\times {10}^{-6}}{0.50}+\frac{-3.0\times {10}^{-6}}{0.41}\right]\\ V_b=-3.32\times {10}^3\text{V}\end{array}$

The potential difference between point $b$and point $a$can be calculated as,

  $\begin{array}{c}{V}_b-V_a=-3.32\times {10}^3\text{V}-\left(-2.37\times {10}^3\text{V}\right)\\ V_b-V_a=-0.95\times {10}^3\text{V}\end{array}$

Conclusion:

Thus, the potential difference $V_b-V_a$is $-0.95\times {10}^3\text{V}$.

To determine

(c)

The amount of work required to move an electron at rest at point $a$to rest at point $b$.

Answer to Problem 50QAP

The amount of work required to move an electron at rest at point $a$to rest at point $b$is $1.52\times {10}^{-16}\text{J}$.

Explanation of Solution

Given info:

Potential difference between point $b$and $a$is, $V_b-V_a=-0.95\times {10}^3\text{V}$.

Charge on the electron is, $q=-1.6\times {10}^{-19}\text{C}$.

Formula used:

Formula for the work required to move the electron from point $a$to point $b$is given as,

  $W=q\left(V_b-V_a\right)$

Calculation:

Substituting the given values in the above equation, we get

  $\begin{array}{c}W=\left(-1.6\times {10}^{-19}\text{C}\right)\left(-0.95\times {10}^3\text{V}\right)\\ W=1.52\times {10}^{-16}\text{J}\end{array}$

Conclusion:

Thus, the amount of work required to move an electron at rest at point $a$to rest at point $b$is $1.52\times {10}^{-16}\text{J}$.