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Chapter 3: Problem 3

Three point charges \(+Q\) are situated at three corners of a square of side \(a\). Find the E-field at the fourth corner.

Short Answer

Expert verified
Question: Determine the magnitude and direction of the net electric field at the fourth corner (point D) of a square with charges +Q at the other three corners, given the side length of the square is "a". Answer: The magnitude of the net electric field at point D can be found using the formula \(E_{net} = \sqrt{(E_{net_x})^2 + (E_{net_y})^2}\), where \(E_{net_x} = k\frac{Q}{a^2} + k\frac{Q}{2a^2}\cos(45^\circ)\) and \(E_{net_y} = -k\frac{Q}{2a^2}\cos(45^\circ) - k\frac{Q}{a^2}\). The direction of the net electric field can be found by calculating the angle: \(\theta = \tan^{-1}(\frac{E_{net_y}}{E_{net_x}})\).

Step by step solution

01

Diagram and labeling

First, draw a square and label the corners A, B, C, and D, with charges at corners A, B, C, and point D being the point where we need to find the electric field. The side length of the square is "a".
02

Calculating the electric field from each charge

We'll use Coulomb's law to find the electric field at point D due to each charge. Coulomb's law formula is: \(E = k\frac{Q}{r^2}\), where \(k\) is the electrostatic constant (\(8.99 \times 10^9 Nm^2/C^2\)), \(Q\) is the charge, and \(r\) is the distance between the charge and the point of interest (point D). The distances from charges A, B, and C to point D are: - For charge A: Distance \(r_A = a\) - For charge B: Distance \(r_B = a\sqrt{2}\) - For charge C: Distance \(r_C = a\) We can now calculate the electric fields due to each charge: - \(E_A = k\frac{Q}{r_A^2} = k\frac{Q}{a^2}\) - \(E_B = k\frac{Q}{r_B^2} = k\frac{Q}{2a^2}\) - \(E_C = k\frac{Q}{r_C^2} = k\frac{Q}{a^2}\)
03

Determining the direction of the electric field vectors

Now we need to find the direction of the electric field vectors of each charge at point D. - Electric field due to charge A points towards point D, along the x-axis; we designate this direction as \(+x\) direction. - Electric field due to charge B points towards point D, making an equal angle (\(45^\circ\)) with both the +x and -y axes. - Electric field due to charge C points towards point D, along the y-axis, in the negative y-direction that we designate as \(-y\) direction.
04

Calculating the components of the electric field vectors

To find the net electric field, we need to calculate the components of the electric field vectors for each charge. - \(E_A\) has only an x-component: \(E_{Ax} = E_A\), and \(E_{Ay} = 0\) - \(E_B\) has x and y components: \(E_{Bx} = E_B\cos(45^\circ)\), and \(E_{By} = -E_B\cos(45^\circ)\) (Note the negative sign is due to the -y direction) - \(E_C\) has only a y-component: \(E_{Cx} = 0\), and \(E_{Cy} = -E_C\) (Note the negative sign is due to the -y direction)
05

Summing up the components to get the net electric field

Now we can find the net electric field at point D by summing up the x and y components of each electric field vector. \(E_{net_x} = E_{Ax} + E_{Bx} = E_A + E_B\cos(45^\circ)\) \(E_{net_y} = E_{Ay} + E_{By} + E_{Cy} = -E_B\cos(45^\circ) - E_C\) Substitute the values from Step 2 to get: \(E_{net_x} = k\frac{Q}{a^2} + k\frac{Q}{2a^2}\cos(45^\circ)\) \(E_{net_y} = -k\frac{Q}{2a^2}\cos(45^\circ) - k\frac{Q}{a^2}\)
06

Calculate the magnitude and direction of the net electric field

Now, we can find the magnitude and direction of the net electric field at point D. The magnitude is found using the Pythagorean theorem: \(E_{net} = \sqrt{(E_{net_x})^2 + (E_{net_y})^2}\) and the direction is found by calculating the angle: \(\theta = \tan^{-1}(\frac{E_{net_y}}{E_{net_x}})\). After substituting the values and solving, we will obtain the magnitude and direction of the net electric field at point D.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is pivotal in understanding the interaction between stationary electric charges. It is expressed by the equation
\[ E = k\frac{Q}{r^2} \
where \(E\) represents the electric field, \(k\) is Coulomb's constant \(8.99 \times 10^9 Nm^2/C^2\), \(Q\) is the magnitude of the charge, and \(r\) is the distance between the charge and the point where the electric field is being calculated.

In our exercise, we applied Coulomb's Law to find the electric fields at point D due to the charges at the corners A, B, and C of a square. By executing this calculation for each charge separately, we can understand the contribution of each charge to the net electric field at the point of interest.
Electric Field Calculation
Electric field calculation at a point in space is determined by the strength and distance of nearby charges. As indicated by Coulomb's Law, the electric field emanates from charges and diminishes with the square of the distance from the charge. For our square configuration with charges at three corners, the distance varied, leading to different electric field magnitudes at the fourth corner from each charge.

For example, charge B is farther from point D than charges A and C, which results in a weaker contribution to the electric field at D. To accurately compute the electric field at point D, we calculate the vector components for each charge-induced field. This includes their magnitude, direction, and how they align with the coordinate axes, which significantly simplifies the determination of the net electric field.
Superposition Principle
The Superposition Principle is a cornerstone in the study of electric fields. It states that the net electric field created by multiple charges is the vector sum of the fields produced by each charge individually. This implies that electric fields add up linearly.

When we calculated the net electric field at point D, we used the superposition principle to add the components of the electric field vectors from charges A, B, and C. Each charge's contribution was calculated separately, considering both magnitude and direction, before being combined to produce the total electric field at point D. This principle is vital for predicting the behavior of complex charge distributions where direct calculation would otherwise be unfeasible.

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Most popular questions from this chapter

The potentials across the \(x\) and \(y\) plates of a cathode-ray oscilloscope produce deflections given, respectively, by \(A\) sin \(\omega t\) and \(A \cos \omega t\). Show that the trace is a circle of radius \(A\). What is the effect on the trace of applying at the same time a modulation to the accelerating potential so that it varies by \((1+\sin \omega t)\) times its original constant value?

Find the components of \(\mathbf{E}\) if (a) \(V=3 x y\) or (b) \(V=r^{2} \cos \theta\).

Point charges \(+Q\) and \(-Q\) are separated by a distance \(2 l\), and \(O\) is the point midway between them. Find the E-field at \(P\), a distance \(r\) from \(O\), (a) when OP is along the line joining the charges and \(r>I\) and (b) when OP is perpendicular to the line joining the charges. Show that when \(r D\) l both fields are nearly proportional to \(Q l / r^{3}\) - (The charges form a dipole and \(\mathrm{P}\) is in the end-on and broadside positions, respectively -see Sec. 3.9.).

The electron beam in a cathode-ray tube is deflected by plates \(2 \mathrm{~cm}\) long and \(5 \mathrm{~mm}\) apart. Estimate the linear deflection of the beam at a screen \(20 \mathrm{~cm}\) away from the plates when \(200 \mathrm{~V}\) is applied across them. The anode potential of the tube is \(500 \mathrm{~V}\).

A charge \(Q\) is distributed uniformly over the surface of a hemisphere of radius \(r\). What is the E-field at the centre of curvature?
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