Question: A 10.0 g block with a charge of$+8.00\times {10}^{-5}C$is placed in an electric field $\overrightarrow{E}=\left(3000\hat{i}-600\hat{j}\right)N/C$. What are the (a) magnitude and (b) direction (relative to the positive direction of thexaxis) of the electrostatic force on the block? If the block is released from rest at the origin at time t = 0, what is its (c) x and (d)ycoordinates at t =3.00s?

Mass of the block, m = 10g

Charge in the electric field,$q=8\times {10}^{-5}C$

The electric field in vector notation,$\overrightarrow{E}=\left(3000\hat{i}-600\hat{j}\right)N/C$

Time of origin, t =0

Time of position, t = 3s

Using the basic concept of kinematics, Newtonian force, and the electric field related to the force, we can get the required quantities that include force, position and direction of the force.

Formulae:

The electrostatic force of a point charge,$\stackrel{\rightharpoonup }{F}=q\stackrel{\rightharpoonup }{E}$ (i)

The angle made by a vector with x-axis,$\theta ={\tan }^{-1}\left(\frac{y-component}{x-component}\right)$ (ii)

The second equation of kinematic motion, $x=v_0+\frac{1}{2}a{t}^2$(iii)

The force from the Newton’s second law, F =ma (iv)

The resultant magnitude of a vector, $\left|v\right|=\sqrt{v_x^2+v_y^2}$ (v)

The electrostatic force on a point charge can be given using the data in equation (i) as follows:

$$\begin{gathered} \overrightarrow{F}=\left(8.00\times {10}^{-5}C\right)\left(3.00\times \frac{{10}^{3}N}{C}\right)\hat{i}+\left(8.00\times {10}^{-5}C\right)\left(-600\frac{N}{C}\right)\hat{j} \\ =\left(0.24N\right)\hat{i}-\left(0.0480N\right)\overbrace{j} \end{gathered}$$

Now, the force has magnitude which can be given using equation (v) as:

$$\begin{gathered} F=\sqrt{{\left(0.240N\right)}^2+{\left(-0.0480N\right)}^2} \\ =0.245N \end{gathered}$$

Hence, the magnitude of the force is 0.245N

The angle the force F makes with the +x axis is given using equation (ii) as:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-0.0480N}{0.240N}\right) \\ =-11.3° \end{gathered}$$

Hence, the force is at an angle $-11.3°$that is $11.3°$measured in counter-clockwise from the +x axis.

With m = 0.0100 kg, the (x, y) coordinates at t = 3.00 s can be found by combining Newton’s second law with the kinematics equations. Thus, the x-coordinate is given using equation (iv) in equation (iii) as follows:

$$\begin{gathered} x=\frac{F_X{t}^2}{2m} \\ =\frac{\left(0.240N\right){\left(3.00s\right)}^2}{2\left(0.0100kg\right)} \\ =108m \end{gathered}$$

Hence, the value of the x-coordinate is 108m

Now, the y-coordinate is given using equation (iv) in equation (iii) as follows:
$$\begin{gathered} y=\frac{F_y{t}^2}{2m} \\ =\frac{\left(-0.0480\right){\left(3.00s\right)}^2}{2\left(0.0100kg\right)} \\ =-21.6m \end{gathered}$$

Hence, the value of the y-coordinate is -21.6m