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Six point charges (each of magnitude Q) are placed on the six vertices of a cube of side x, such that two adjacent vertices are vacant. Electrostatic field intensity at the center of the cube is?
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Six point charges (each of magnitude Q) are placed on the six vertices...
**Introduction**

To determine the electrostatic field intensity at the center of a cube, we need to consider the contributions from each of the six charges placed on the vertices. The field intensity at a point due to a single charge is given by Coulomb's law.

**Coulomb's Law**

Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:

F = k * (q1 * q2) / r^2

Where F is the electrostatic force, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

**Determining the Field Intensity**

To determine the field intensity at the center of the cube, we need to consider the contributions from each of the six charges placed on the vertices of the cube. Let's label the charges as Q1, Q2, Q3, Q4, Q5, and Q6.

The field intensity at the center of the cube due to each charge can be calculated using Coulomb's law. Since the charges are placed on the vertices of the cube, the distance between each charge and the center is half the length of a side of the cube.

Let's denote the side length of the cube as x. Then, the distance between the center and each charge is x/2.

Using Coulomb's law, the field intensity at the center of the cube due to each charge can be calculated as:

E1 = k * (Q1 / (x/2)^2)
E2 = k * (Q2 / (x/2)^2)
E3 = k * (Q3 / (x/2)^2)
E4 = k * (Q4 / (x/2)^2)
E5 = k * (Q5 / (x/2)^2)
E6 = k * (Q6 / (x/2)^2)

**Superposition Principle**

According to the superposition principle, the total field intensity at the center of the cube is the vector sum of the individual field intensities due to each charge. Since the charges are placed symmetrically on the vertices of the cube, the magnitudes of the field intensities due to opposite charges are equal.

Hence, the total field intensity at the center of the cube is given by:

E_total = E1 + E2 + E3 + E4 + E5 + E6

Substituting the expressions for E1, E2, E3, E4, E5, and E6, we get:

E_total = k * [(Q1 + Q2 + Q3 + Q4 + Q5 + Q6) / (x/2)^2]

Simplifying further, we have:

E_total = k * [(Q1 + Q2 + Q3 + Q4 + Q5 + Q6) / (x^2/4)]

Therefore, the electrostatic field intensity at the center of the cube is directly proportional to the sum of the magnitudes of the charges and inversely proportional to the square of the side length of the cube.
Community Answer
Six point charges (each of magnitude Q) are placed on the six vertices...
If all vertices are covered then intensity will be zero
now total intensity is equal to -(intensity by two adjacent)
(kq/(√3/4)^2)*2*cos$
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Six point charges (each of magnitude Q) are placed on the six vertices of a cube of side x, such that two adjacent vertices are vacant. Electrostatic field intensity at the center of the cube is?
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Six point charges (each of magnitude Q) are placed on the six vertices of a cube of side x, such that two adjacent vertices are vacant. Electrostatic field intensity at the center of the cube is? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Six point charges (each of magnitude Q) are placed on the six vertices of a cube of side x, such that two adjacent vertices are vacant. Electrostatic field intensity at the center of the cube is? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Six point charges (each of magnitude Q) are placed on the six vertices of a cube of side x, such that two adjacent vertices are vacant. Electrostatic field intensity at the center of the cube is?.
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