Q49P - Fundamentals Of Physics

Question

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 the x axis) 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) y coordinates at t =3.00s?

Step-by-Step Solution

Answer

• a) The magnitude of the electrostatic force on the block is 0.245N
• b) The direction of the electrostatic force on the block is at an angle$-11.3°$ that is $11.3°$measured in counter-clockwise from the +x axis.
• c) The x-coordinate of the block at t=3s is 108m
• d) The x-coordinate of the block at t=3s is -21.6m

Step 1: The given data

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

Step 2: Understanding the concept of electric field

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)

 

Step 3: a) Calculation of the magnitude of the force

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

Step 4: b) Calculation of the direction of the force

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.

Step 5: c) Calculation of the x-coordinate of the position at t=3s

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

 

Step 6: d) Calculation of the y-coordinate of the position at t=3s

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