Question

A point charge of +7.45 pc (7.45 × 10-12
c.Is fixed at the origin. Another point charge of –4.28 pc is fixed on the y-axis, 2.79 mm from the origin. If you placed a proton at point p, which is on the x-axis, 9.83 mm from the origin, what would be the magnitude of the electric force on that proton?

Answer 1

Answer:$0.57\times10^{-16}N$

The electric force between two charges $q_1$ and $q_2$ separated by distance $r$can be defined as:

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

where, $k =\frac{1}{4\pi \epsilon_o}=9\times10^9Nm^2/C^2$

Charge of proton$q_2=1.6\times10^{-19}C$

Force due to charge at the origin on the proton:

$F_1=\frac{9\times10^9\times 7.45\times10^{-12}\times1.6\times10^{-19}}{(9.83\times10^{-3})^2}N=1.108\times10^{-16}N$

we need to calculate the distance between the charge on y-axis and the proton:

We will use the Pythagoras theorem,

$Hypotenuse=\sqrt{(perpendicular)^2+(base)^2}$

$r=\sqrt{(9.83\times10^{-3})^2+(2.79\times10^{-3})^2}=10.21\times10^{-3}m$

Force between these two charges:

$F_2=\frac{9\times10^9\times -4.28\times10^{-12}\times1.6\times10^{-19}}{(10.21\times10^{-3})^2}N=-0.576\times10^{-16}N$

The negative sign depicts that it is an attractive force because unlike charges attract.

Now we need to find the net force on the proton. we cannot simply add the two forces because force is a vector quantity. so, first we need to measure the angular separation between the forces.

$\theta= tan^{-1}\frac{2.79mm}{9.83mm}=tan^{-1}0.28=15.84^o$

Now, we will write the forces in vector form and do vector addition.

Net force along x-axis:

$F_x=F_1-F_2cos\theta=(1.108-0.576 cos15.84^o)\times10^{-16}N=0.55\times10^{-16}N$

Net force along y-axis:

$F_y=F_2sin\theta=0.576\times10^{-16}sin15.84^o=0.16\times10^{-16}N$

Net force on proton by the two charges:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(0.55\times10^{-16})^2+(0.16\times10^{-16})^2}=0.57\times10^{-16}N$

Answer 2

The magnitude of the net electric force on the proton is $\boxed{5.65 \times {10^{ - 17}}\,{\text{N}}}$ or $\boxed{0.565\times {10^{ - 16}}\,{\text{N}}}$.

Further explanation:

The net force on the proton placed at a point on x-axis is the resultant of the forces applied by the charge placed at origin and charge placed at the y-axis.

Given:

The positive charge which is fixed at the origin is ${q_1}=+ 7.45\,{\text{pC}}$.

The distance at which the negative charge fixed on the y-axis is $2.79\text{ mm}$.

The negative charge which is fixed on the y-axis is ${q_2}=- 4.28\,{\text{pC}}$.

The distance at which the proton is placed on x-axis is $9.83\text{ mm}$.

Concept:

The force between two charged particles is directly proportional to the product of both the charges and inversely proportional to the distance between them.

The force between the two charged particles is:

$F=\dfrac{k{q_A}{q_B}}{r^2}$                                                                 ...... (1)

In triangle ABC.

$\begin{aligned}\tan \theta&=\frac{{AB}}{{BC}} \\&=\frac{{2.79\,{\text{mm}}}}{{9.83\,{\text{mm}}}} \\ \end{aligned}$  

So,

$\begin{aligned}\theta&={\tan ^{ - 1}}\left( {\frac{{2.79}}{{9.83}}} \right) \\&=15.85^\circ\\ \end{aligned}$  

The value of the distance between charge placed on y-axis and charge placed on x-axis is:

$AC=\sqrt {{{\left( {AB} \right)}^2}+{{\left( {BC} \right)}^2}}$  

Substitute $2.79\text{ mm}$ for $AB$ and $9.83\text{ mm}$ for $AC$ in the above expression.

$\begin{aligned}AC&=\sqrt {{{\left( {2.79} \right)}^2} + {{\left( {9.83} \right)}^2}}\\&=10.22\,{\text{mm}} \\ \end{aligned}$

Let ${F_1}$ is the force exerted on proton of charge $q$ due to  the positive charge ${q_1}$ placed at the origin.

Substitute the values in equation (1).

$\begin{aligned}F_1&=\dfrac{{9\times10^9}\times{7.45\times10^{-12}}\times{1.6\times10^{-19}}}{(9.83\times10^{-3})^2}\\&=1.11\times10^{-16}\text{ N}\end{aligned}$

The direction of ${F_1}$ is away from the line joining BC as shown in the diagram.

Let ${F_2}$ is the force exerted on proton of charge $q$ due to negative charge $q_2$.

$\begin{aligned}F_2&=\dfrac{{9\times10^{9}}\times{4.28\times10^{-12}}{1.6\times10^{-19}}}{(10.22\times10^{-3})^2}\\ &=5.9\times10^{-17}\text{ N}\end{aligned}$

The direction of ${F_2}$ is directed towards the line joining AC as shown in the diagram.  

Resolving ${F_2}$ in the components along x and y direction.

The net force on proton in the x direction  is:

$\begin{aligned}{F_x}&={F_1}-{F_2}\cos \theta\\&=11.1 \times {10^{ - 17}}\,{\text{N}} - 5.9 \times {10^{ - 17}}\cos \left( {15.85} \right)\,{\text{N}} \\&=5.42\times10^{-17}\text{N}\\ \end{aligned}$  

The net force on proton in the y direction is:

$\begin{aligned}{F_y}&={F_2}\sin \theta\\&=5.9 \times {10^{ - 17}}\sin \left( {15.85} \right)\,{\text{N}} \\&=1.61\times10^{-17}\text{ N}\\ \end{aligned}$  

Magnitude of the net force is given as:

$\begin{aligned}{F_{{\text{net}}}}&=\sqrt {F_x^2 + F_y^2}\\&=\sqrt {{{\left( {5.42 \times {{10}^{ - 17}}\,{\text{N}}} \right)}^2} + {{\left( {1.61 \times {{10}^{ - 17}}\,{\text{N}}} \right)}^2}}\\&=5.65 \times {10^{ - 17}}\,{\text{N}} \\\end{aligned}$

Thus, the magnitude of the net electric force on the proton is $\boxed{5.65 \times {10^{ - 17}}\,{\text{N}}}$ or $\boxed{0.565\times {10^{ - 16}}\,{\text{N}}}$.

Learn more:

1.  Conservation of energy brainly.com/question/3943029

2.  The motion of a body under friction brainly.com/question/4033012

3. A ball falling under the acceleration due to gravity brainly.com/question/10934170

Answer Details:

Grade: College

Subject: Physics

Chapter: Electrostatics

Keywords:  

Point charge, +7.45 pc, +7.45times10^-12 c, fixed, origin, another, charge, -4.28 pc, -4.28times10^-12 c, y axis, 2.79 mm, proton, p, x axis, 9.83 mm, 2.79times10^-3 m, 9.83times10^-3 m, magnitude, electric force, 0.57times10^-16 N, 5.65times10^-17 N, electrostatic force.