Problem 44 - University Physics with Modern Physics

Question

Point charge \(q_1 = -\)5.00 nC is at the origin and point charge \(q_2 = +\)3.00 nC is on the \(x\)-axis at \(x = \)3.00 cm. Point \(P\) is on the \(y\)-axis at \(y = \)4.00 cm. (a) Calculate the electric fields \(\overrightarrow{E_1}\) and \(\overrightarrow{E_2}\) at point \(P\) due to the charges \(q_1\) and \(q_2\) . Express your results in terms of unit vectors (see Example 21.6). (b) Use the results of part (a) to obtain the resultant field at \(P\), expressed in unit vector form.

Step-by-Step Solution

Answer

\(\overrightarrow{E}_{\text{net}} = -6.48 \times 10^3\hat{i} + 1.95 \times 10^4\hat{j}\, \text{N/C}\).

Step 1: Identify coordinates of points and charges

The charge \(q_1 = -5.00\, \text{nC}\) is at the origin, \((x_1, y_1) = (0,0)\). The charge \(q_2 = +3.00\, \text{nC}\) is at \((x_2, y_2) = (3.00\, \text{cm}, 0)\). The point \(P\) is at \((x_P, y_P) = (0, 4.00\, \text{cm})\). Convert coordinates to meters if necessary. \(1\, \text{cm} = 0.01\, \text{m}\).

Step 2: Calculate distance between charges and point P

Using the Pythagorean theorem, calculate the distance \(r_1\) from \(q_1\) to \(P\) and \(r_2\) from \(q_2\) to \(P\). For \(q_1\): \[ r_1 = \sqrt{(0)^2 + (0.04)^2} = 0.04\, \text{m} \] For \(q_2\): \[ r_2 = \sqrt{(0.03)^2 + (0.04)^2} = 0.05\, \text{m} \].

Step 3: Calculate electric fields from each charge

Use the formula for electric field due to a point charge: \( E = \frac{k|q|}{r^2} \), where \(k = 8.99 \times 10^9\, \text{N m}^2/\text{C}^2\).For \(E_1\): \[ E_1 = \frac{k|q_1|}{r_1^2} = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-9}}{0.04^2} = 2.81 \times 10^4\, \text{N/C} \]The direction is toward the origin (positive \(y\)-axis) since \(q_1\) is negative: \[ \overrightarrow{E_1} = 2.81 \times 10^4 \hat{j}\, \text{N/C} \]For \(E_2\): \[ E_2 = \frac{k|q_2|}{r_2^2} = \frac{8.99 \times 10^9 \times 3.00 \times 10^{-9}}{0.05^2} = 1.08 \times 10^4\, \text{N/C} \]The \(x\) component is \(-\hat{i}\) and the \(y\) component is \(-\hat{j}\): \[ \overrightarrow{E_2} = 1.08 \times 10^4 (-0.6\hat{i} - 0.8\hat{j}) = -6.48 \times 10^3 \hat{i} - 8.64 \times 10^3 \hat{j}\, \text{N/C} \]

Step 4: Find resultant electric field at point P

Sum the electric fields at point \(P\) from both charges:\[ \overrightarrow{E}_{\text{net}} = \overrightarrow{E_1} + \overrightarrow{E_2} \]Substitute the values: \[ \overrightarrow{E}_{\text{net}} = 0\hat{i} + 2.81 \times 10^4\hat{j} - 6.48 \times 10^3\hat{i} - 8.64 \times 10^3\hat{j} \]Combine components: \[ \overrightarrow{E}_{\text{net}} = -6.48 \times 10^3\hat{i} + 1.95 \times 10^4\hat{j}\, \text{N/C} \]

Key Concepts

Point Charge

A point charge is a model used in physics to simplify problems involving electric fields and forces. This model considers a charged object as having all its charge concentrated at a single point in space. This simplification allows us to easily calculate the electric field and force interactions between charges without worrying about the charge's actual size or shape.
In our example, we have two point charges, \(q_1 = -5.00\, \text{nC}\) and \(q_2 = +3.00\, \text{nC}\), located at different positions on the coordinate system. These point charges produce electric fields in the surrounding space, influencing other charges nearby. The electric field \(E\) due to a point charge is given by:

• \(E = \frac{k|q|}{r^2}\)

where \(k\) is Coulomb's constant, \(q\) is the magnitude of the charge, and \(r\) is the distance from the charge to the point of interest. The direction of the field is away from positive charges and towards negative charges, reflecting a fundamental property of electric fields.

Superposition Principle

The superposition principle is a valuable concept when dealing with electric fields due to multiple charges. It states that the total electric field produced by multiple point charges is equal to the vector sum of the electric fields produced by each charge individually. This principle makes it easier to analyze complex fields by breaking them down into simpler, more manageable parts.
In the given problem, we apply the superposition principle by first calculating the electric fields \(\overrightarrow{E_1}\) and \(\overrightarrow{E_2}\) at point \(P\) due to each charge, \(q_1\) and \(q_2\), separately. Then, we add these fields together to find the resultant electric field \(\overrightarrow{E}_{\text{net}}\) at point \(P\).
Using vector addition, we compute:

• \(\overrightarrow{E}_{\text{net}} = \overrightarrow{E_1} + \overrightarrow{E_2}\)

This allows us to understand how the individual electric fields interact and combine to create a net effect at any point in space.

Unit Vector Notation

Unit vector notation is a convenient way to express vectors, such as electric fields, in terms of their direction and magnitude along coordinate axes. A unit vector has a magnitude of one and indicates direction, simplifying the representation of more complex vector quantities.
In our example exercise, each electric field component at point \(P\) is expressed in terms of unit vectors \(\hat{i}\) and \(\hat{j}\), which represent directions along the \(x\)- and \(y\)-axes, respectively. For instance, the electric field \(\overrightarrow{E_1}\) generated by the point charge \(q_1\) is directed entirely along the \(y\)-axis, described as:

• \(\overrightarrow{E_1} = 2.81 \times 10^4 \hat{j}\, \text{N/C}\)

Similarly, \(\overrightarrow{E_2}\) includes both \(\hat{i}\) and \(\hat{j}\) components due to its position relative to \(P\):

• \(\overrightarrow{E_2} = -6.48 \times 10^3 \hat{i} - 8.64 \times 10^3 \hat{j}\, \text{N/C}\)

Unit vector notation helps clarify the direction each component of the electric field is acting, making problem-solving more straightforward and visually intuitive.