AP Physics 1 - Electrostatics

An excess charge of is put on an ideal neutral conducting sphere with radius . What is the Coulomb force this excess charge exerts on a point charge of that is from the surface of the sphere?

$k=8.99*10^9\frac{m^2N}{C^2}$

Explanation

Two principal realizations help with solving this problem, both derived from Gauss’ law for electricity:

The excess charge on an ideal conducting sphere is uniformly distributed over its surface
A uniform shell of charge acts, in terms of electric force, as if all the charge were contained in a point charge at the sphere’s center

With these realizations, an application of Coulomb’s law answers the question. If $q_{2}$ is the point charge outside the sphere, then the force $F_{q2}$ on $q_{2}$ is:

$F_{q2} = \frac{kq_{sphere}q_{2}}{r^{2}}$

In this equation, is Coulomb’s constant, $q_{sphere}$ is the excess charge on the spherical conductor, and is total distance in meters of $q_{2}$ from the center of the conducting sphere.

Using the given values in this equation, we can calculate the generated force:

$F_{q2}=\frac{(8.99*10^9\frac{m^2N}{C^2})(10*10^{-6}C)(5*10^{-6}C)}{(0.5m)^2}$

$F_{q2}=\frac{0.4495m^2N}{0.25m^2}=1.798N$

An excess charge of is put on an ideal neutral conducting sphere with radius . What is the Coulomb force this excess charge exerts on a point charge of that is from the surface of the sphere?

$k=8.99*10^9\frac{m^2N}{C^2}$

Explanation

Two principal realizations help with solving this problem, both derived from Gauss’ law for electricity:

The excess charge on an ideal conducting sphere is uniformly distributed over its surface
A uniform shell of charge acts, in terms of electric force, as if all the charge were contained in a point charge at the sphere’s center

With these realizations, an application of Coulomb’s law answers the question. If $q_{2}$ is the point charge outside the sphere, then the force $F_{q2}$ on $q_{2}$ is:

$F_{q2} = \frac{kq_{sphere}q_{2}}{r^{2}}$

In this equation, is Coulomb’s constant, $q_{sphere}$ is the excess charge on the spherical conductor, and is total distance in meters of $q_{2}$ from the center of the conducting sphere.

Using the given values in this equation, we can calculate the generated force:

$F_{q2}=\frac{(8.99*10^9\frac{m^2N}{C^2})(10*10^{-6}C)(5*10^{-6}C)}{(0.5m)^2}$

$F_{q2}=\frac{0.4495m^2N}{0.25m^2}=1.798N$

What is the magnitude of the electric force between two charged metals that are 3m apart, that have absolute value of the charges being 1C and 3C?

$k=9.0*10^9\frac{N\cdot m^2}{C^2}$

$\frac{10^9}{3}N$

Explanation

We are given all the necessary information to find the magnitude of the electric force by using Coulomb's law:

$\mid\mid F\mid\mid= \mid\frac{kq_1q_2}{r^2}\mid$

Where is Coulomb's constant given by $9.0*10^9\frac{N\cdot m^2}{C^2}$ , and are the respective charges, and is the distance between the charges. In our case:

$\mid\mid F\mid\mid= \mid\frac{9.0*10^9*1*3}{3^3}N\mid=10^9N$

What is the magnitude of the electric force between two charged metals that are 3m apart, that have absolute value of the charges being 1C and 3C?

$k=9.0*10^9\frac{N\cdot m^2}{C^2}$

$\frac{10^9}{3}N$

Explanation

We are given all the necessary information to find the magnitude of the electric force by using Coulomb's law:

$\mid\mid F\mid\mid= \mid\frac{kq_1q_2}{r^2}\mid$

Where is Coulomb's constant given by $9.0*10^9\frac{N\cdot m^2}{C^2}$ , and are the respective charges, and is the distance between the charges. In our case:

$\mid\mid F\mid\mid= \mid\frac{9.0*10^9*1*3}{3^3}N\mid=10^9N$

Determine the strength of a force of proton on another proton in the nucleus if they are $10^{-15}m$ apart.

$k=8.99*10^9\frac{N}{m^2C^2}$

$e=1.602*10^{-19}C$

$230*10^{-15}N$

$230*10^{15}N$

Explanation

Use Coulomb's law:

$F_c=\frac{k*Q*q}{r^2}$ , where is Coulomb's constant, are charges of the two points and is the distance between the charges.

$F_c=\frac{8.99*10^9*(1.602*10^{-19})^2}{(10^{-15})^2}=230N$

Determine the strength of a force of proton on another proton in the nucleus if they are $10^{-15}m$ apart.

$k=8.99*10^9\frac{N}{m^2C^2}$

$e=1.602*10^{-19}C$

$230*10^{-15}N$

$230*10^{15}N$

Explanation

Use Coulomb's law:

$F_c=\frac{k*Q*q}{r^2}$ , where is Coulomb's constant, are charges of the two points and is the distance between the charges.

$F_c=\frac{8.99*10^9*(1.602*10^{-19})^2}{(10^{-15})^2}=230N$

An electric field line is point from the left towards the right. Where will an electron move when placed in the field?

Towards the left

Towards the right

Upward

Downward

It will not move

Explanation

The electron will move towards the left because electric field lines always point towards the negative charge. The electron is negatively charged and will oppose the negative electric field on the right and move towards the positive end on the left.

Therefore the correct answer is that the electron will move to the left.

An electric field line is point from the left towards the right. Where will an electron move when placed in the field?

Towards the left

Towards the right

Upward

Downward

It will not move

Explanation

Therefore the correct answer is that the electron will move to the left.

Two protons are at a distance $r_{1}$ away from each other. There is a force $F_{1}$ acting on each proton due to the other. If the protons are moved so that they are now at a distance

$r_{2}=\frac{1}{2} r_{1}$ apart, what is the new force acting on each proton due to the other $\left (F_{2} \right )$ ?

$F_{2}=4 F_{1}$

$F_{2}=\frac{1}{2} F_{1}$

$F_{2}=2 F_{1}$

$F_{2}=\frac{1}{4} F_{1}$

Explanation

Coulomb's law shows that the force between two charged particles is inversely proportional to the square of the distance between the particles.

$F=k \frac{q_{1}q_{2}}{r^{2}}$

If the distance between the charges is reduced by $\frac{1}{2}$ , that means the $\frac{1}{2}$ is squared in the denominator and the will flip up to the top to give time the original force. More explicitly, if we plug in the given information the initial force will be:

$F_{1}=k \frac{q_{1}q_{2}}\left ( {r_{1}\right )^{2}}$

$F_{2}=k \frac{q_{1}q_{2}}\left ( {r_{2}\right )^{2}}=k \frac{q_{1}q_{2}}\left ( \frac{1}{2}{r_{1}\right )^{2}}$

$F_{2}=k \frac{q_{1}q_{2}}{ \frac{1}{4}\left (r_{1}\right )^{2}}=4 k \frac{q_{1}q_{2}}{\left (r_{1}\right )^{2}}$

$F_{2}=4 F_{1}$

If we have 2 charges, and , that are apart, what is the force exerted on by if we know that has a charge of $-13\mu C$ and has a charge of $-12\mu C$ ?