November 15, 2019

Exam 4

Physics 208 – Introductory Physics II
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

Exam 4
Electric Charge, Fields & Forces

§ Problem 1

A droplet of oil with a mass of m=1.5\;\mathrm{mg} and unknown charge Q is suspended in an electric field with a strength of E=2500\;\mathrm{N}/\mathrm{C} downward.

§ (a)

Create a free-body diagram showing the forces acting upon the oil drop. You may create your diagram on the figure shown or next to it. (10 points)

§ (b)

What is the value of the charge Q? (10 points)

\displaystyle|Q|E \displaystyle= \displaystyle mg
\displaystyle|Q| \displaystyle= \displaystyle\frac{mg}{E}
\displaystyle= \displaystyle\frac{\left(1.5\times 10^{{-3}}\;\mathrm{kg}\right)\left(9.8\;\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)}{2500\;\frac{\mathrm{N}}{\mathrm{C}}}
\displaystyle= \displaystyle 5.9\;\mu\mathrm{C}
\displaystyle Q \displaystyle= \displaystyle-5.9\;\mu\mathrm{C}

Note that the charge must be negative because the electric field and the force are in opposite directions for the oil drop to be suspended.

§ Problem 2

An unknown charge Q is placed 20.\;\mathrm{cm} from a 2.3\;\mu\mathrm{C} charge. In order to determine the value of Q a third charge with a value -0.15\;\mu\mathrm{C} is placed at the midpoint between the other two, and it is observed to experience a force of 0.50\;\mathrm{N} towards the 2.3\;\mu\mathrm{C} charge. Compute Q. (30 points)

There is definitely more than one way to proceed. I choose to consider the two forces acting upon the -0.15\;\mu\mathrm{C} charge taking into account that we know their sum, the net force or 0.50\;\mathrm{N} towards the 2.3\;\mu\mathrm{C} charge. First compute the force from that 2.3\;\mu\mathrm{C} charge which will be attractive and hence appear as shown on the diagram.

\displaystyle F_{{+2.3\;\mu\mathrm{C}}} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(2.3\times 10^{{-6}}\mathrm{C}\right)\left(0.15\times 10^{{-6}}\mathrm{C}\right)}{\left(0.10\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 0.3\underline{1}1\;\mathrm{N}

Notice that this is smaller than the net force also shown on the diagram, and hence the force of the unknown charge Q will need to be repulsive pushing the smallest charge towards the 2.3\;\mu\mathrm{C} charge, hence it has been drawn as shown. In other words,

\displaystyle F_{Q} \displaystyle= \displaystyle 0.50\;\mathrm{N}-0.3\underline{1}1\;\mathrm{N}
\displaystyle= \displaystyle 0.1\underline{9}0\;\mathrm{N}

Hence

\displaystyle 0.1\underline{9}0\;\mathrm{N} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(0.15\times 10^{{-6}}\mathrm{C}\right)\left|Q\right|}{\left(0.10\;\mathrm{m}\right)^{2}}
\displaystyle\left|Q\right| \displaystyle= \displaystyle 1.4\;\mu\mathrm{C}

Because the force is repulsive

\displaystyle Q \displaystyle= \displaystyle-1.4\;\mu\mathrm{C}

§ Problem 3

Three charges, Q_{1}=Q_{2}=Q_{3}=1.5\;\mu\mathrm{C}, are placed at three vertices of a square with side lengths of 25\;\mathrm{cm} placed and oriented as in the figure. Use the coordinate system provided.

§ (a)

What are the components E_{{1,x}} and E_{{1,y}} of the electric field due to Q_{1} evaluated at the fourth vertex at the bottom of the figure? Draw and label a vector representing \vec{E}_{1} on the figure.
(10 points)

\displaystyle E_{1} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(1.5\times 10^{{-6}}\mathrm{C}\right)}{\left(0.25\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 2.\underline{1}6\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle E_{{1,x}} \displaystyle= \displaystyle+\left(2.\underline{1}6\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}\right)\sin 45^{0}
\displaystyle= \displaystyle+1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle+1.5\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle E_{{1,y}} \displaystyle= \displaystyle-\left(2.\underline{1}6\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}\right)\cos 45^{0}
\displaystyle= \displaystyle-1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle-1.5\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}

Notice that, as drawn, the x component is opposite to the indicated angle and to the right while the y component is adjacent to the angle and downward.

§ (b)

What are the components E_{{2,x}} and E_{{2,y}} of the electric field due to Q_{2} evaluated at the fourth vertex at the bottom of the figure? Draw and label a vector representing \vec{E}_{2} on the figure.
(10 points)

The magnitudes are identical to the previous step, but the directions of the components must be reconsidered.

\displaystyle E_{{2,x}} \displaystyle= \displaystyle-1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle-1.5\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle E_{{2,y}} \displaystyle= \displaystyle-1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle-1.5\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}

§ (c)

What are the components E_{{3,x}} and E_{{3,y}} of the electric field due to Q_{3} evaluated at the fourth vertex at the bottom of the figure? Draw and label a vector representing \vec{E}_{3} on the figure.
(10 points)

Notice the increased separation between the opposite corners of the square as indicated on the diagram. Also notice that the direction is straight down, hence purely y component.

\displaystyle E_{3} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(1.5\times 10^{{-6}}\mathrm{C}\right)}{\left(\sqrt{2}\,(0.25\;\mathrm{m})\right)^{2}}
\displaystyle= \displaystyle 1.\underline{0}8\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle E_{{3,x}} \displaystyle= \displaystyle 0\;\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle E_{{3,y}} \displaystyle= \displaystyle-1.\underline{0}8\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle-1.1\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}

§ (d)

What are the components E_{{x}} and E_{{y}} of the resultant electric field due to all three charges evaluated at the fourth vertex at the bottom of the figure? Draw a vector representing \vec{E} on the figure. What is its magnitude, E, and its direction? Draw and label a vector representing \vec{E} on the figure.
(10 points)

\displaystyle E_{x} \displaystyle= \displaystyle E_{{1,x}}+E_{{2,x}}+E_{{2,x}}
\displaystyle= \displaystyle 1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}-1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}+0\,\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle 0\,\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle E_{y} \displaystyle= \displaystyle E_{{1,y}}+E_{{2,y}}+E_{{2,y}}
\displaystyle= \displaystyle-1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}-1.\underline{5}3\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}-1.\underline{0}8\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle-4.\underline{1}4\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle E \displaystyle= \displaystyle\sqrt{E_{x}^{2}+E_{y}^{2}}
\displaystyle= \displaystyle 4.\underline{1}4\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}
\displaystyle= \displaystyle 4.1\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}

downward, or in the -y direction. Could also state 90^{0} CW from the +x axis.

§ (e)

Suppose an electron is placed at the fourth vertex. What force, \vec{F}, magnitude and direction, would the electron experience?
(5 points)

The charge of the electron is

\displaystyle q \displaystyle= \displaystyle-e
\displaystyle= \displaystyle-1.602\times 10^{{-19}}\mathrm{C}

Because the charge is negative, the force will be opposite to the field, upward, or in the +y direction with a magnitude

\displaystyle F \displaystyle= \displaystyle eE
\displaystyle= \displaystyle\left(1.602\times 10^{{-19}}\mathrm{C}\right)\left(4.\underline{1}4\times 10^{5}\frac{\mathrm{N}}{\mathrm{C}}\right)
\displaystyle= \displaystyle 6.\underline{6}3\times 10^{{-14}}\mathrm{N}
\displaystyle= \displaystyle 6.6\times 10^{{-14}}\mathrm{N}

§ (f)

What acceleration, \vec{a}, magnitude and direction, would the electron experience?
(5 points)

\displaystyle a \displaystyle= \displaystyle\frac{F}{\mathrm{m}_{e}}
\displaystyle= \displaystyle\frac{6.\underline{6}3\times 10^{{-14}}\mathrm{N}}{9.11\times 10^{{-31}}\mathrm{kg}}
\displaystyle= \displaystyle 7.3\times 10^{{16}}\frac{\mathrm{m}}{\mathrm{s}^{2}}

upward, aligned with the force and opposite to the electric field.

 

Call Me Dr Rob logo