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Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 6m.
- a)17/6
- b)-17/6
- c)13/6
- d)-13/6
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Three charged cylindrical sheets are present in three spaces with &sig...
Answer: d
Explanation: The radius R = 6m encloses all the three Gaussian cylinders.
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units
Explanation: The radius R = 6m encloses all the three Gaussian cylinders.
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units
Most Upvoted Answer
Three charged cylindrical sheets are present in three spaces with &sig...
Different radii and charges. Let's call the radius of the first sheet r1, the second sheet r2, and the third sheet r3. Let's also call the charge of the first sheet Q1, the second sheet Q2, and the third sheet Q3.
To calculate the electric field at a point P outside the three sheets, we need to use the principle of superposition. This principle states that the electric field at a point due to multiple charges is the vector sum of the electric fields at that point due to each individual charge.
First, we need to calculate the electric field at point P due to the first sheet. We can use the formula for the electric field due to a charged cylindrical sheet, which is:
E1 = (1/2πε) * (Q1 / r1)
where ε is the permittivity of free space.
The electric field due to the second sheet can be calculated in the same way:
E2 = (1/2πε) * (Q2 / r2)
The electric field due to the third sheet can also be calculated in the same way:
E3 = (1/2πε) * (Q3 / r3)
Now, we can add the electric fields vectorially to get the total electric field at point P:
Etotal = E1 + E2 + E3
Note that if point P is inside one of the sheets, we need to use a different formula for the electric field due to a charged cylindrical sheet. In this case, we would only include the electric field due to the sheet that point P is inside of, and we would not include the electric fields due to the other sheets.
To calculate the electric field at a point P outside the three sheets, we need to use the principle of superposition. This principle states that the electric field at a point due to multiple charges is the vector sum of the electric fields at that point due to each individual charge.
First, we need to calculate the electric field at point P due to the first sheet. We can use the formula for the electric field due to a charged cylindrical sheet, which is:
E1 = (1/2πε) * (Q1 / r1)
where ε is the permittivity of free space.
The electric field due to the second sheet can be calculated in the same way:
E2 = (1/2πε) * (Q2 / r2)
The electric field due to the third sheet can also be calculated in the same way:
E3 = (1/2πε) * (Q3 / r3)
Now, we can add the electric fields vectorially to get the total electric field at point P:
Etotal = E1 + E2 + E3
Note that if point P is inside one of the sheets, we need to use a different formula for the electric field due to a charged cylindrical sheet. In this case, we would only include the electric field due to the sheet that point P is inside of, and we would not include the electric fields due to the other sheets.
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Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 6m.a)17/6b)-17/6c)13/6d)-13/6Correct answer is option 'D'. Can you explain this answer?
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