Focus: Electric Charge & Coulomb's Law
Two small spheres are brought near each other. If sphere A has a positive charge and sphere B has a negative charge, will they attract or repel each other? What if both spheres have a positive charge?
An object has a net negative charge due to an excess of $5 \times 10^{13}$ electrons. What is its total charge in Coulombs? (The elementary charge $e = 1.602 \times 10^{-19}$ C).
Two point charges exert a force $F$ on each other when they are a distance $d$ apart. What happens to the magnitude of this force if the distance between them is tripled to $3d$?
Why is a copper rod considered a good electrical conductor, while a glass rod is considered an insulator? Describe the difference in terms of electrons.
Two identical point charges are separated by 20 cm and repel each other with a force of 10 N. What is the magnitude of each charge? (Coulomb's constant $k = 8.99 \times 10^9$ N·m²/C²).
A point charge $q_1 = +2.0$ µC is located at the origin, and a second point charge $q_2 = -5.0$ µC is located on the x-axis at $x = 40$ cm. Calculate the magnitude and direction of the electrostatic force that $q_1$ exerts on $q_2$.
Three point charges are placed on the x-axis: $q_1 = +8$ µC is at the origin, $q_2 = -4$ µC is at $x = 20$ cm, and $q_3 = +18$ µC is at $x = 60$ cm. What is the net electrostatic force on charge $q_2$?
A charge $q_A = +1.0$ µC is at the origin, and a charge $q_B = +4.0$ µC is at $x = 90$ cm. At what position on the x-axis between the two charges should a third charge, $q_C$, be placed so that the net force on it is zero?
Calculate the ratio of the magnitude of the electrostatic force to the gravitational force between a proton and an electron. ($m_p = 1.67 \times 10^{-27}$ kg, $m_e = 9.11 \times 10^{-31}$ kg, $G = 6.67 \times 10^{-11}$ N·m²/kg²).
Two identical small spheres, each with a mass of 10 g, are hung from a point by two silk threads of length $L=50$ cm. They are given identical charges, and in equilibrium, they hang with an angle of $30^\circ$ between the threads. Find the magnitude of the charge on each sphere.
Three charges are arranged at the vertices of a right triangle. Charge $q_1 = -3.0$ nC is at the origin. Charge $q_2 = +5.0$ nC is on the y-axis at $y = 4.0$ m. Charge $q_3 = +2.0$ nC is on the x-axis at $x = 3.0$ m. Find the magnitude and direction of the net force on the charge at the origin ($q_1$).
Three point charges, each with magnitude $q = +5$ µC, are placed at the vertices of an equilateral triangle with side length $a = 30$ cm. What is the magnitude of the net electrostatic force on any one of the charges?
A total charge $Q$ is to be divided into two parts, $q_1$ and $q_2$, such that $q_1 + q_2 = Q$. What must be the ratio $q_1/Q$ for the electrostatic force between the two parts to be a maximum when they are separated by a fixed distance?
A small sphere of mass $m=2.0$ g and charge $q_1 = +5.0$ µC is placed on a frictionless horizontal table. A second sphere with charge $q_2$ is placed directly 15 cm above the first sphere. What charge must $q_2$ have for the first sphere to be lifted off the table?
Four identical point charges, each with charge $+q$, are placed at the corners of a square with side length $s$. Find the magnitude of the net electrostatic force on the charge at the top right corner.
Opposite charges attract, so the positive sphere A and negative sphere B will attract each other.
Like charges repel, so if both spheres are positive, they will repel each other.
The total charge $Q$ is the number of excess electrons $n$ times the elementary charge $e$. Since they are electrons, the charge is negative.
$Q = -n \cdot e = -(5 \times 10^{13}) \cdot (1.602 \times 10^{-19} \text{ C})$
$Q = -8.01 \times 10^{-6} \text{ C}$ or $-8.01$ µC.
The force $F$ is inversely proportional to the square of the distance: $F \propto \frac{1}{d^2}$.
If the new distance is $d' = 3d$, the new force $F'$ will be $F' \propto \frac{1}{(3d)^2} = \frac{1}{9d^2}$.
So, the new force will be $1/9$ of the original force: $F' = F/9$.
In a conductor like copper, the outermost electrons (valence electrons) are loosely bound to their atoms and are free to move throughout the material. These mobile electrons can easily carry an electric current.
In an insulator like glass, electrons are tightly bound to their atoms and cannot move freely. There are very few mobile charge carriers.
Using Coulomb's Law, $F = k \frac{|q_1 q_2|}{r^2}$. Since charges are identical, $q_1 = q_2 = q$.
$F = k \frac{q^2}{r^2}$. We rearrange to solve for $q$: $q = \sqrt{\frac{F r^2}{k}}$.
Convert distance to meters: $r = 20$ cm $= 0.20$ m.
$q = \sqrt{\frac{(10 \text{ N}) (0.20 \text{ m})^2}{8.99 \times 10^9 \text{ N·m²/C²}}} = \sqrt{4.449 \times 10^{-11} \text{ C²}} \approx 6.67 \times 10^{-6}$ C or 6.67 µC.
Use Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.
$F = (8.99 \times 10^9 \text{ N·m²/C²}) \frac{|(2.0 \times 10^{-6} \text{ C})(-5.0 \times 10^{-6} \text{ C})|}{(0.40 \text{ m})^2}$
$F = (8.99 \times 10^9) \frac{1.0 \times 10^{-11}}{0.16} \approx 0.562$ N.
Since $q_1$ is positive and $q_2$ is negative, the force is attractive. The force on $q_2$ is directed towards $q_1$, which is in the negative x-direction.
The net force on $q_2$ is the vector sum of the force from $q_1$ ($F_{12}$) and the force from $q_3$ ($F_{32}$).
Force from $q_1$: $F_{12} = k \frac{|q_1 q_2|}{r_{12}^2} = (8.99 \times 10^9) \frac{|(8 \times 10^{-6})(-4 \times 10^{-6})|}{(0.20)^2} = 7.19$ N. Since charges are opposite, this force is attractive, pulling $q_2$ to the left (negative direction).
Force from $q_3$: $F_{32} = k \frac{|q_3 q_2|}{r_{32}^2} = (8.99 \times 10^9) \frac{|(18 \times 10^{-6})(-4 \times 10^{-6})|}{(0.40)^2} = 4.05$ N. Since charges are opposite, this force is attractive, pulling $q_2$ to the right (positive direction).
Net Force: $F_{net} = -7.19 \text{ N} + 4.05 \text{ N} = -3.14$ N. The net force is 3.14 N in the negative x-direction.
For the net force on $q_C$ to be zero, the force from $q_A$ ($F_{AC}$) must be equal in magnitude and opposite in direction to the force from $q_B$ ($F_{BC}$). Let $x$ be the position of $q_C$.
$k \frac{|q_A q_C|}{x^2} = k \frac{|q_B q_C|}{(0.90 - x)^2}$. The terms $k$ and $q_C$ cancel out.
$\frac{1.0 \times 10^{-6}}{x^2} = \frac{4.0 \times 10^{-6}}{(0.90 - x)^2} \implies \frac{1}{x^2} = \frac{4}{(0.90 - x)^2}$.
Take the square root of both sides: $\frac{1}{x} = \frac{2}{0.90 - x} \implies 0.90 - x = 2x \implies 3x = 0.90$.
$x = 0.30$ m or 30 cm from the origin.
The electric force is $F_E = k \frac{e^2}{r^2}$ and the gravitational force is $F_G = G \frac{m_p m_e}{r^2}$. The distance $r$ is the same for both.
The ratio is $\frac{F_E}{F_G} = \frac{k e^2 / r^2}{G m_p m_e / r^2} = \frac{k e^2}{G m_p m_e}$.
$\frac{F_E}{F_G} = \frac{(8.99 \times 10^9)(1.602 \times 10^{-19})^2}{(6.67 \times 10^{-11})(1.67 \times 10^{-27})(9.11 \times 10^{-31})} \approx 2.27 \times 10^{39}$.
The electrostatic force is about $10^{39}$ times stronger than the gravitational force.
1. The distance between spheres is $r = 2L \sin(15^\circ) = 2(0.5)\sin(15^\circ) \approx 0.259$ m.
2. In equilibrium, the horizontal component of tension balances the electric force: $T \sin(15^\circ) = F_E$.
3. The vertical component of tension balances the weight: $T \cos(15^\circ) = mg$.
4. Divide the two equations: $\frac{T \sin(15^\circ)}{T \cos(15^\circ)} = \tan(15^\circ) = \frac{F_E}{mg}$.
5. Solve for $F_E$: $F_E = mg \tan(15^\circ) = (0.010 \text{ kg})(9.8 \text{ m/s²}) \tan(15^\circ) \approx 0.0263$ N.
6. Use Coulomb's law to find q: $F_E = k \frac{q^2}{r^2} \implies q = \sqrt{\frac{F_E r^2}{k}} = \sqrt{\frac{(0.0263)(0.259)^2}{8.99 \times 10^9}} \approx 4.43 \times 10^{-7}$ C or 0.443 µC.
1. Force from $q_2$ on $q_1$ ($F_{21}$) is attractive and points along the +y axis. $F_{21} = k \frac{|q_1 q_2|}{r_{21}^2} = (8.99 \times 10^9) \frac{|(-3\times 10^{-9})(5\times 10^{-9})|}{(4.0)^2} \approx 8.43 \times 10^{-9}$ N.
2. Force from $q_3$ on $q_1$ ($F_{31}$) is attractive and points along the +x axis. $F_{31} = k \frac{|q_1 q_3|}{r_{31}^2} = (8.99 \times 10^9) \frac{|(-3\times 10^{-9})(2\times 10^{-9})|}{(3.0)^2} \approx 5.99 \times 10^{-9}$ N.
3. The forces are perpendicular. Use the Pythagorean theorem to find the magnitude of the net force: $F_{net} = \sqrt{F_{21}^2 + F_{31}^2} = \sqrt{(8.43 \times 10^{-9})^2 + (5.99 \times 10^{-9})^2} \approx 1.03 \times 10^{-8}$ N.
4. The direction angle $\theta$ is $\tan^{-1}(\frac{F_{21}}{F_{31}}) = \tan^{-1}(\frac{8.43}{5.99}) \approx 54.6^\circ$ above the positive x-axis.
1. Consider the force on $q_1$. The force from $q_2$ ($F_{21}$) and from $q_3$ ($F_{31}$) are both repulsive and have the same magnitude: $F = k \frac{q^2}{a^2} = (8.99 \times 10^9) \frac{(5 \times 10^{-6})^2}{(0.30)^2} = 2.50$ N.
2. $F_{31}$ acts at an angle of $60^\circ$ from the line connecting $q_2$ and $q_1$. We use vector components. Let the x-axis be along the base. The force on the top charge has components:
$F_x = F_{21x} + F_{31x} = F \cos(60^\circ) - F \cos(60^\circ) = 0$.
$F_y = F_{21y} + F_{31y} = F \sin(60^\circ) + F \sin(60^\circ) = 2 F \sin(60^\circ)$.
3. $F_{net} = 2 (2.50 \text{ N}) \sin(60^\circ) = 5.0 \cdot \frac{\sqrt{3}}{2} \approx 4.33$ N.
1. Let $q_1 = q$. Then $q_2 = Q - q$. The force is $F = k \frac{q(Q-q)}{r^2} = \frac{k}{r^2}(qQ - q^2)$.
2. To find the maximum force, we take the derivative of $F$ with respect to $q$ and set it to zero: $\frac{dF}{dq} = 0$.
$\frac{dF}{dq} = \frac{k}{r^2}(Q - 2q)$. Setting this to zero: $Q - 2q = 0 \implies q = Q/2$.
3. This means $q_1 = Q/2$. The ratio is $q_1/Q = (Q/2)/Q = 1/2$. The force is maximum when the charge is split equally.
For the sphere to lift off, the upward electrostatic force $F_E$ must equal its downward weight $W$.
$F_E = W \implies k \frac{|q_1 q_2|}{r^2} = mg$.
The charges must be opposite to have an attractive (upward) force, so $q_2$ must be negative.
$|q_2| = \frac{mg r^2}{k |q_1|} = \frac{(0.002 \text{ kg})(9.8 \text{ m/s²})(0.15 \text{ m})^2}{(8.99 \times 10^9 \text{ N·m²/C²})(5.0 \times 10^{-6} \text{ C})} \approx 9.81 \times 10^{-9}$ C.
So, $q_2 = -9.81$ nC.
1. Consider the charge at the top right corner. It is repelled by three other charges. Let's call them Left, Bottom, and Diagonal.
2. Force from Left charge: $F_L = k \frac{q^2}{s^2}$ (in the +x direction).
3. Force from Bottom charge: $F_B = k \frac{q^2}{s^2}$ (in the +y direction).
4. Force from Diagonal charge: Distance is $s\sqrt{2}$. $F_D = k \frac{q^2}{(s\sqrt{2})^2} = k \frac{q^2}{2s^2}$. This force acts at $45^\circ$.
5. Sum the vector components:
$F_x = F_L + F_{Dx} = k\frac{q^2}{s^2} + F_D \cos(45^\circ) = k\frac{q^2}{s^2} + (k\frac{q^2}{2s^2})\frac{\sqrt{2}}{2} = k\frac{q^2}{s^2}(1 + \frac{\sqrt{2}}{4})$.
$F_y = F_B + F_{Dy} = k\frac{q^2}{s^2} + F_D \sin(45^\circ) = k\frac{q^2}{s^2} + (k\frac{q^2}{2s^2})\frac{\sqrt{2}}{2} = k\frac{q^2}{s^2}(1 + \frac{\sqrt{2}}{4})$.
6. $F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{2 \left[k\frac{q^2}{s^2}(1 + \frac{\sqrt{2}}{4})\right]^2} = \sqrt{2} \cdot k\frac{q^2}{s^2}(1 + \frac{\sqrt{2}}{4}) \approx 1.90 k\frac{q^2}{s^2}$.