Problem 93
Question
Four charges \(+2 q,-2 q,-3 q\), and \(+3 q\) are kept in the corners of a square of side \(a\). The total field at the centre \(O\) is, (A) Zero (B) \(\frac{2 \sqrt{2} q}{4 \pi \varepsilon_{0} a^{2}}\) (C) \(\frac{\sqrt{2} q}{4 \pi \varepsilon_{0} a^{2}}\) (D) \(\frac{10 \sqrt{2} q}{4 \pi \varepsilon_{0} a^{2}}\)
Step-by-Step Solution
Verified Answer
The net electric field at the center O of the square (considering directions) is given by \(E_{net} = \frac{\sqrt{2}q}{4 \pi \varepsilon_{0} a^{2}}\), which corresponds to option (C).
1Step 1: Name the Charges and Points
Let the charges be A (+2q), B (-2q), C (-3q), and D (+3q). Name the center of the square O.
2Step 2: Calculate the Distance from Each Charge to Point O
From the problem, we know that the side of the square is "a". Use the Pythagorean theorem to find the distance from each corner to point O (the center). Distance AO = BO = CO = DO = \(\sqrt{(\frac{a}{2})^{2}+(\frac{a}{2})^{2}} = \frac{a \sqrt{2}}{2}\).
3Step 3: Calculate the Electric Field at Point O Due to Each Charge
Use the electric field formula: \(E = \frac{kQ}{r^2}\), where k = \(\frac{1}{4 \pi \varepsilon_{0}}\). Calculate the electric field at point O due to each charge keeping the respective directions in mind.
4Step 4: Determine the Net Electric Field at Point O by Adding the Electric Fields Produced by the Charges
Add the electric fields produced by the charges: \(E_{net} = E_A + E_B + E_C + E_D\).
5Step 5: Compare the Calculated Net Electric Field to the Given Choices and Choose the Correct Answer
Comparing the calculated net electric field at point O to the given choices, choose the correct answer.
Key Concepts
Coulomb's LawCharge DistributionSuperposition Principle
Coulomb's Law
Coulomb's Law is essential in understanding electric fields. It describes the force between two point charges. This force is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula is given by:
The formula is given by:
- \( F = \frac{k |q_1 q_2|}{r^2} \)
- \( F \) is the force between the charges
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges
- \( r \) is the distance between the charges
- \( k \) is Coulomb's constant, \( \frac{1}{4 \pi \varepsilon_0} \)
- \( E = \frac{kQ}{r^2} \)
Charge Distribution
When dealing with problems involving multiple charges, understanding charge distribution is key. Here, charges are placed at specific positions, affecting the resulting electric field. In our problem, we have four charges at the corners of a square. The arrangement of these charges and their magnitudes determine the field outcome at the center.
Important points about charge distribution include:
Important points about charge distribution include:
- Symmetry of the configuration. This can sometimes simplify calculations, as symmetrical distributions might cancel out some effects.
- The magnitude and sign of each charge as they influence the direction and strength of the electric field produced.
- The exact placement of charges relative to each other and the point of interest, which in our case is the center of the square.
Superposition Principle
The superposition principle is a fundamental concept when dealing with multiple electric fields. It states that the total electric field created by multiple sources is the vector sum of the individual fields from each source. This principle simplifies complex problems by allowing us to calculate the effect of each charge individually and then add them up.
In our exercise:
- Calculate the electric field due to each charge using Coulomb's law.
- Consider the direction each field is acting. Some fields might have components that cancel each other out or add up depending on their directions.
- Add the vector fields to find the total electric field at the center of the square.
Other exercises in this chapter
Problem 91
Which one of the following statement is incorrect? (A) A moving charged particle produced electric and magnetic field both. (B) Equipotential surface is always
View solution Problem 92
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View solution Problem 95
Two spheres of radii \(r\) and \(R\) carry charges \(q\) and \(Q\), respectively. When they are connected by a wire, there will be no loss of energy of the syst
View solution Problem 96
If three moles of monatomic gas is mixed with 1 moles diatomic gas, the resultant value of \(\gamma\) for the mixture is (A) \(1.66\) (B) \(1.50\) (C) \(1.40\)
View solution