Problem 34 - University Physics with Modern Physics

Question

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{cm} .\) (a) Calculate the electric fields \(\vec{\boldsymbol{E}}_{1}\) and \(\vec{\boldsymbol{E}}_{2}\) 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

\( \vec{E} = 6.48 \times 10^3 \hat{i} - 1.95 \times 10^4 \hat{j} \text{ N/C} \).

Step 1: Determine Distance from Charges to Point P

For charge \( q_1 \) located at the origin and \( q_2 \) at \( (3.00\, \text{cm}, 0) \), Point \( P \) is at \( (0, 4.00\, \text{cm}) \). The distances are calculated using the Pythagorean theorem. For \( q_1 \), the distance to \( P \) is \( d_1 = 4.00 \text{ cm} \). For \( q_2 \), \( d_2 = \sqrt{3.00^2 + 4.00^2} = 5.00 \text{ cm} \).

Step 2: Calculate Electric Field from Charge q1 at Point P

The electric field \( \vec{E}_1 \) due to charge \( q_1 \) is given by \[ \vec{E}_1 = \frac{k |q_1|}{d_1^2} \hat{j} \]. Plug in \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\), \( q_1 = -5.00 \times 10^{-9} \text{ C} \), and \( d_1 = 0.04 \text{ m}\). Thus, \( \vec{E}_1 = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-9}}{(0.04)^2} (-\hat{j}) = -2.81 \times 10^4 \hat{j} \text{ N/C} \).

Step 3: Calculate Electric Field from Charge q2 at Point P

The electric field \( \vec{E}_2 \) due to charge \( q_2 \) is found using \[ \vec{E}_2 = \frac{k q_2}{d_2^2} \times \frac{(3 \hat{i} + 4 \hat{j})}{5} \]. Substituting \( q_2 = 3.00 \times 10^{-9} \text{ C} \) and \( d_2 = 0.05 \text{ m} \) results in \( \vec{E}_2 = \frac{8.99 \times 10^9 \times 3.00 \times 10^{-9}}{(0.05)^2} \times \left(\frac{3 \hat{i} + 4 \hat{j}}{5}\right) = (6.48 \times 10^3 \hat{i} + 8.64 \times 10^3 \hat{j}) \text{ N/C} \).

Step 4: Calculate Resultant Electric Field at Point P

Add the electric fields \( \vec{E}_1 \) and \( \vec{E}_2 \) to find the resultant electric field \( \vec{E} = \vec{E}_1 + \vec{E}_2 \). Performing the vector addition: \( \vec{E} = (0 \hat{i} - 2.81 \times 10^4 \hat{j}) + (6.48 \times 10^3 \hat{i} + 8.64 \times 10^3 \hat{j}) = 6.48 \times 10^3 \hat{i} - 1.95 \times 10^4 \hat{j} \text{ N/C} \).

Key Concepts

Coulomb's Law

Coulomb's law is a fundamental principle used to calculate the electric field caused by point charges. It describes the force between two charges. The formula is \[ F = \frac{k |q_1 q_2|}{r^2} \] where

• \( F \) is the force between the charges
• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2\text{/C}^2) \)
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges
• \( r \) is the distance between the charges

This force is either attractive or repulsive, depending on the nature of the charges involved. If the charges are of opposite signs, the force is attractive; if the signs are the same, it is repulsive. When we're dealing with point charges, like in our exercise, we use Coulomb's law to help determine the electric field that a charge exerts at a certain location. This field is then used to figure out the forces on other charges placed in the vicinity.

Vector Addition

Vector addition is crucial for accurately combining multiple electric fields to find a resultant field. Electric fields, like vectors, have both magnitude and direction. In this exercise, we deal with combining the electric field vectors from two point charges to calculate the resultant field at a particular point.

• Each vector component is added separately: the \( i \)-components are summed together, and the \( j \)-components are summed separately.
• The results give us the total electric field as a new vector: \( \vec{E} = \vec{E}_1 + \vec{E}_2 \).

After computing the electric fields from each charge, we use vector addition to combine them. It's important to pay attention to signs and directions since they affect the outcome of the addition, as seen in the solution where the resultant field was calculated by adding \( 6.48 \times 10^3 \hat{i} \) and subtracting \( 1.95 \times 10^4 \hat{j} \). By knowing how to add vectors correctly, we ensure that our calculations respect both direction and magnitude.

Unit Vector Representation

Unit vector representation simplifies working with vectors by providing a clear indication of direction. A unit vector has a magnitude of one and indicates the direction of a vector. In electric field calculations, we often use the unit vectors \( \hat{i} \) and \( \hat{j} \) to represent directions along the x-axis and y-axis, respectively.

• A vector \( \vec{v} \) can be expressed as \( \vec{v} = v_x \hat{i} + v_y \hat{j} \), where \( v_x \) and \( v_y \) are the components of \( \vec{v} \) along the x and y axes.
• In this exercise, the electric field from each point charge was expressed in terms of these unit vectors, e.g., \( \vec{E}_1 = -2.81 \times 10^4 \hat{j} \).

Using unit vectors helps break down the problem into more manageable pieces and lets us focus on each directional component of forces or fields individually. This representation is especially helpful in multidimensional problems, where tracking direction is essential.

Electric Field Due to a Point Charge

An electric field due to a point charge illustrates how charges exert forces in space. The electric field from a point charge \( q \) at a distance \( r \) is given by the formula \[ E = \frac{k |q|}{r^2} \] where

• \( E \) is the electric field
• \( k \) is Coulomb's constant
• \( q \) is the charge creating the field
• \( r \) is the distance from the charge

The direction of this field depends on the charge's sign: it radiates outward from a positive charge and inward for a negative charge. In our exercise, each point charge creates an electric field at point \( P \). The calculations for these fields involve the charge's magnitude, the distance to \( P \), and appropriate unit vector directions. By calculating these separately for each charge, and then using vector addition, we find the total electric field at \( P \). Understanding this process helps elucidate how electric fields interact with charged objects in the environment, guiding numerous practical applications, from physics experiments to electronic device engineering.