Questions: A charged particle ( q=7.5 × 10^-10 C) experiences a force of F=1.25 i-4.2 j N in an electric field. - Part (a) V Write an expression for the electric field vector E to which the charge is subject, in terms of the force F. E=F / q Part (b) Assume this field is generated by a point charge of Q=5 × 10^-9 C. How far away is this charge located? Give your answer in meters.

A charged particle ( q=7.5 × 10^-10 C) experiences a force of F=1.25 i-4.2 j N in an electric field.
- Part (a) V

Write an expression for the electric field vector E to which the charge is subject, in terms of the force F.
E=F / q
Part (b)
Assume this field is generated by a point charge of Q=5 × 10^-9 C. How far away is this charge located? Give your answer in meters.

Transcript text: A charged particle $\left(~ q=7.5 \times 10^{-10} \mathrm{C}\right)$ experiences a force of $\mathbf{F}=1.25 \mathbf{i}-4.2 \mathrm{j} \mathrm{N}$ in an electric field. - Part (a) $V$ Write an expression for the electric field vector $\mathbf{E}$ to which the charge is subject, in terms of the force $\mathbf{F}$. \[ E=F / q \] Part (b) Assume this field is generated by a point charge of $Q=5 \times 10^{-9} \mathrm{C}$. How far away is this charge located? Give your answer in meters.

Solution

Solution Steps

Step 1: Expression for the Electric Field Vector

To find the electric field vector $\mathbf{E}$ to which the charge $q$ is subject, we use the relationship between the electric field $\mathbf{E}$, the force $\mathbf{F}$, and the charge $q$:

\[ \mathbf{E} = \frac{\mathbf{F}}{q} \]

Given: \[ \mathbf{F} = 1.25 \mathbf{i} - 4.2 \mathbf{j} \, \text{N} \] \[ q = 7.5 \times 10^{-10} \, \text{C} \]

Substitute the values into the equation:

\[ \mathbf{E} = \frac{1.25 \mathbf{i} - 4.2 \mathbf{j}}{7.5 \times 10^{-10}} \]

Step 2: Calculate the Electric Field Vector

Perform the division for each component:

\[ E_x = \frac{1.25}{7.5 \times 10^{-10}} = 1.6667 \times 10^9 \, \text{N/C} \] \[ E_y = \frac{4.2}{7.5 \times 10^{-10}} = -5.6000 \times 10^9 \, \text{N/C} \]

Thus, the electric field vector is:

\[ \mathbf{E} = 1.6667 \times 10^9 \mathbf{i} - 5.6000 \times 10^9 \mathbf{j} \, \text{N/C} \]

$\boxed{\mathbf{E} = 1.6667 \times 10^9 \mathbf{i} - 5.6000 \times 10^9 \mathbf{j} \, \text{N/C}}$

Step 3: Distance from the Point Charge

To find the distance $d$ from the point charge $Q$ that generates this electric field, we use the formula for the electric field due to a point charge:

\[ E = \frac{kQ}{d^2} \]

Where:

$k = 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2$ (Coulomb's constant)
$Q = 5 \times 10^{-9} \, \text{C}$
$E$ is the magnitude of the electric field

First, find the magnitude of $\mathbf{E}$:

\[ |\mathbf{E}| = \sqrt{(1.6667 \times 10^9)^2 + (-5.6000 \times 10^9)^2} \]

Calculate the magnitude:

\[ |\mathbf{E}| = \sqrt{(1.6667 \times 10^9)^2 + (5.6000 \times 10^9)^2} = \sqrt{2.7778 \times 10^{18} + 3.1360 \times 10^{19}} = \sqrt{3.4138 \times 10^{19}} = 5.8445 \times 10^9 \, \text{N/C} \]

Now, solve for $d$:

\[ 5.8445 \times 10^9 = \frac{8.9875 \times 10^9 \times 5 \times 10^{-9}}{d^2} \]

\[ d^2 = \frac{8.9875 \times 10^9 \times 5 \times 10^{-9}}{5.8445 \times 10^9} \]

\[ d^2 = \frac{44.9375}{5.8445} = 7.6908 \]

\[ d = \sqrt{7.6908} = 2.7733 \, \text{m} \]

$\boxed{d = 2.7733 \, \text{m}}$