AP Physics C Electricity and Magnetism Quiz: Electric Charge And Electric Force

Electric charge is a conserved property of matter measured in coulombs (C); it comes in positive and negative types, and the total charge of an isolated system remains constant. Electric forces act at a distance: like charges repel, unlike charges attract, and the force magnitude depends on charge and separation. In space, two point charges $q_1=+4.0\times10^{-6}\ \text{C}$ and $q_2=+1.0\times10^{-6}\ \text{C}$ are fixed $0.30\ \text{m}$ apart. Coulomb's Law gives the magnitude: $F=k\frac{|q_1q_2|}{r^2}$. A numerical check: $F=(8.99\times10^{9})\frac{(4.0\times10^{-6})(1.0\times10^{-6})}{(0.30)^2}$. If a third charge is later introduced, the principle of superposition states the net force on any charge equals the vector sum of forces from each other charge calculated separately.

How does the force between charges change if the distance is doubled?

Electric charge (C) is a conserved property that can be positive or negative; like charges repel and unlike charges attract. Electric force is the interaction between charges, directed along the line joining them. Two point charges in space are fixed: $q_1=-5.0\times10^{-6}\ \text{C}$ and $q_2=+2.0\times10^{-6}\ \text{C}$ separated by $r=0.50\ \text{m}$. Coulomb's Law gives $F=k\frac{|q_1q_2|}{r^2}=(8.99\times10^{9})\frac{|(-5.0\times10^{-6})(2.0\times10^{-6})|}{(0.50)^2}$. If additional charges were introduced, superposition would require adding all forces vectorially to find the net force on either charge.

Calculate the force between two charges of $-5.0\ \mu\text{C}$ and $+2.0\ \mu\text{C}$, separated by $0.50\ \text{m}$.

Electric charge is a conserved property of matter measured in coulombs (C); it can be positive or negative, and interactions between charges produce electric forces. Electric force acts along the line joining charges and follows Coulomb's Law $F=k\frac{|q_1q_2|}{r^2}$. In space, two point charges $q_1=+6.0\times10^{-6}\ \text{C}$ and $q_2=-1.0\times10^{-6}\ \text{C}$ are separated by $r=0.30\ \text{m}$. Numerical setup: $F=(8.99\times10^{9})\frac{|(6.0\times10^{-6})(-1.0\times10^{-6})|}{(0.30)^2}$. If additional charges are introduced, the net force is found by superposition, adding the individual force vectors.

Calculate the force between two charges of $+6.0\ \mu\text{C}$ and $-1.0\ \mu\text{C}$, separated by $0.30\ \text{m}$.

Electric charge is a conserved property (C) that can be positive or negative; like charges repel and unlike charges attract. Electric force between point charges acts along the line joining them and obeys Coulomb's Law. Three point charges lie on a line: $q_1=+2.0\times10^{-6}\ \text{C}$ at $x=0$, $q_2=-1.0\times10^{-6}\ \text{C}$ at $x=0.30\ \text{m}$, and $q_3=+2.0\times10^{-6}\ \text{C}$ at $x=0.60\ \text{m}$. For $q_2$, the forces from $q_1$ and $q_3$ have magnitudes $F_{21}=k\frac{|q_2q_1|}{(0.30)^2},\quad F_{23}=k\frac{|q_2q_3|}{(0.30)^2}$ and superposition requires adding these vectors to find the net force.

Determine the net electric force on $q_2$ placed in an electric field.

Electric charge is measured in coulombs (C) and can be positive or negative; electric force between point charges follows Coulomb's Law and acts along the line between them. Three charges are on a line: $q_1=+4.0\,\mu\text{C}$ at $x=0$, $q_2=-4.0\,\mu\text{C}$ at $x=0.30\,\text{m}$, and $q_3=+1.0\,\mu\text{C}$ at $x=0.60\,\text{m}$. By superposition, the net force on $q_2$ is $\vec F_{21}+\vec F_{23}$. Calculate the net force on $q_2$ (take +x right).

Electric charge is a conserved property of matter (C) that can be positive or negative; forces between charges are electric forces that act along the line joining charges. In deep space, three point charges lie on the x-axis: $q_1=+1.0\times10^{-6}\ \text{C}$ at $x=0$, $q_2=+1.0\times10^{-6}\ \text{C}$ at $x=0.20\ \text{m}$, and $q_3=-2.0\times10^{-6}\ \text{C}$ at $x=0.40\ \text{m}$. Coulomb's Law gives each pairwise force magnitude, e.g. $F_{21}=k\frac{|q_2q_1|}{(0.20)^2},\quad F_{23}=k\frac{|q_2q_3|}{(0.20)^2}$ with $k=8.99\times10^{9}$. By superposition, the net force on $q_2$ is the vector sum of forces from $q_1$ and $q_3$.

Determine the net electric force on $q_2$ placed in an electric field.

Electric charge (C) comes in positive and negative; electric force is the push/pull between charges and depends on separation by Coulomb's Law $F\propto 1/r^2$. Two point charges $q_1=+1.0\,\mu\text{C}$ and $q_2=+3.0\,\mu\text{C}$ are initially separated by $r=0.20\,\text{m}$, producing force magnitude $F_0$ on each. If the distance is doubled to $0.40\,\text{m}$ while charges remain fixed, superposition would still apply if other charges existed, but here only the distance changes. How does the force magnitude change?

Electric charge (C) can be positive or negative; electric force is the interaction between charges and follows Coulomb's Law $F=k\dfrac{|q_1q_2|}{r^2}$. Two like point charges $q_1=+2.0\,\mu\text{C}$ and $q_2=+2.0\,\mu\text{C}$ are $0.30\,\text{m}$ apart on a line. A third point charge $q_3=+1.0\,\mu\text{C}$ is placed exactly midway between them. By superposition, the net force on $q_3$ equals the vector sum of the forces from $q_1$ and $q_2$. Explain the effect of introducing $q_3$ between the like charges by determining the net force on $q_3$.