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Expert-verified Found in: Page 713 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In the rectangle of Fig. 24-55, the sides have lengths 5.0 cm and 15 cm, q1= -5.0 mC, and q2= +2.0 mC. With V=0 at infinity, what is the electric potential at (a) corner A and (b) corner B? (c) How much work is required to move a charge q3= +3.0 mC from B to A along a diagonal of the rectangle? (d) Does this work increase or decrease the electric potential energy of the three-charge system? Is more, less, or the same work required if q3 is moved along a path that is (e) inside the rectangle but not on a diagonal and (f) outside the rectangle?

1. The electric potential at corner A is $60.0×{10}^{3}\mathrm{V}$.
2. The electric potential at corner B is $-7.8×{10}^{5}\mathrm{V}$.
3. The work required to move the third charge from B to A along a diagonal of the rectangle is $2.52\mathrm{J}$.
4. This work increases the electric potential energy of the three-charge system.
5. The same work is required inside the rectangle.
6. The same work is required outside the rectangle.
See the step by step solution

## Step 1: The given data

1. Length of the rectangle $\mathrm{L}=15\mathrm{cm}\mathrm{or}0.15\mathrm{m}$
2. Width of the rectangle $\mathrm{w}=5\mathrm{cm}\mathrm{or}0.05\mathrm{m}$
3. Charge at the top corner ${\mathrm{q}}_{\mathrm{t}}=-5.0×{10}^{-6}\mathrm{C}$
4. Charge at the bottom corner ${\mathrm{q}}_{2}=2.0×{10}^{-6}\mathrm{C}$
5. Charge at the bottom corner ${\mathrm{q}}_{3}=3.0×{10}^{-6}\mathrm{C}$
6. With V=0 at infinity.

## Step 2: Understanding the concept of energy and electric potential

Using the formula of the electric potential of a system, we can the desired values of the electric potentials at corners A and B. Again, this value will determine the potential energy at the corners and the difference value gives the work required to move the third particle. Again, as the work is conservative, it is independent of all the paths taken by the charges.

Formulae:

The electric potential at a point, $\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{r}}$ (i)

The potential energy of the system in terms of potential, U=qV (ii)

## Step 3: a) Calculation of the electric potential at corner A

The total electric potential at the corner is due to the charges present along the length and the width. So using the given data and equation (i), we get the potential as:

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{A}}& =& \mathrm{k}\left(\frac{{\mathrm{q}}_{1}}{\mathrm{L}}+\frac{{\mathrm{q}}_{2}}{\mathrm{W}}\right)\\ & =& \left(8.99×{10}^{9}{\mathrm{Nm}}^{2}/{\mathrm{C}}^{2}\right)\left(\frac{-5.0×{10}^{-6}\mathrm{C}}{0.15\mathrm{m}}+\frac{20×{10}^{-6}\mathrm{C}}{0.05\mathrm{m}}\right)\\ & =& 59.93×{10}^{3}\mathrm{V}\\ & \approx & 60.0×{10}^{3}\mathrm{V}\end{array}$

Hence, the value of the electric potential is $60.0×{10}^{3}\mathrm{V}$.

## Step 4: b) Calculation of the electric potential at corner B

Similarly, the electric potential at corner B is given using the given data in equation (i) as follows:

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{B}}& =& \mathrm{k}\left(\frac{{\mathrm{q}}_{1}}{\mathrm{W}}+\frac{{\mathrm{q}}_{2}}{\mathrm{L}}\right)\\ & =& \left(8.99×{10}^{8}{\mathrm{Nm}}^{2}/{\mathrm{C}}^{2}\right)\left(\frac{-5.0×{10}^{-6}\mathrm{C}}{0.05\mathrm{m}}+\frac{2.0×{10}^{-6}\mathrm{C}}{0.15\mathrm{m}}\right)\\ & =& -7.8×{10}^{5}\mathrm{V}\end{array}$

Hence, the value of the potential is data-custom-editor="chemistry" $-7.8×{10}^{5}\mathrm{V}$.

## Step 5: c) Calculation of the required work to move third particle from B to A

Work required moving a charge q from B to A be equal to difference in potential energy between point A and point B. So, using the above values, the work done is given using equation (ii) as follows:

$\begin{array}{rcl}\mathrm{W}& =& {\mathrm{U}}_{\mathrm{A}}-{\mathrm{U}}_{\mathrm{B}}\\ & =& {\mathrm{q}}_{3}\left({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}\right)\\ & =& \left(3.0×{10}^{-6}\mathrm{C}\right)\left(60.0×{10}^{3}\mathrm{V}-\left(-7.8×{10}^{5}\mathrm{V}\right)\right)\\ & =& 2.52\mathrm{J}\end{array}$

Hence, the value of the work is 2.52 J.

## Step 6: d) Calculation to check the effect of work on electric potential energy

Since the work done by the external agent is positive. So this work increases the electric potential energy of the three-charge system.

## Step 7: e) Calculation to check the behaviour of work inside the rectangle

The work done depends only on initial and final positions only. So the work done is independent of path. Hence the work done is same on all paths between two points.

## Step 8: f) Calculation to check the behaviour of work outside the rectangle

The work done depends only on initial and final positions only. So the work done is independent of path. Hence the work done is same on all paths between two points. ### Want to see more solutions like these? 