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A uniform surface charge density of 30 nC /m2 is present on the spherical surface r = 6 mm in free space. The VAB between A (r = 2 cm, θ = 350 , Ø = 550 ) and B (r = 3 cm, θ = 400 , Ø = 900 )
- a)2.03 V
- b)10.17 V
- c)4.07 mV
- d)-10.17 V
Correct answer is option 'D'. Can you explain this answer?
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To find the potential difference (VAB) between points A and B, we can use the formula for the potential due to a uniformly charged spherical shell. The formula is given by:
V = k * Q / r
where V is the potential, k is the electrostatic constant (9 x 10^9 N m^2/C^2), Q is the total charge on the shell, and r is the distance from the center of the shell.
Given that the surface charge density is 30 nC/m^2 and the radius of the shell is 6 mm, we can calculate the total charge on the shell:
Q = surface charge density * surface area
The surface area of a sphere is given by:
surface area = 4 * π * r^2
Plugging in the values, we have:
Q = (30 x 10^-9 C/m^2) * (4 * π * (6 x 10^-3 m)^2)
= (30 x 10^-9 C/m^2) * (4 * 3.14159 * (6 x 10^-3 m)^2)
= (30 x 10^-9 C/m^2) * (4 * 3.14159 * 36 x 10^-6 m^2)
= (30 x 10^-9 C/m^2) * (452.389 x 10^-6 C)
= 13.5717 x 10^-9 C
= 13.5717 nC
Now, let's calculate the potential at point A (r = 2 cm) and point B (r = 3 cm) using the formula mentioned earlier.
For point A:
V(A) = k * Q / r(A)
= (9 x 10^9 N m^2/C^2) * (13.5717 x 10^-9 C) / (2 x 10^-2 m)
= 6.78585 V
For point B:
V(B) = k * Q / r(B)
= (9 x 10^9 N m^2/C^2) * (13.5717 x 10^-9 C) / (3 x 10^-2 m)
= 4.5239 V
Therefore, the potential difference (VAB) between points A and B is given by:
VAB = V(B) - V(A)
= 4.5239 V - 6.78585 V
= -2.26195 V
≈ -2.26 V
The correct answer is option 'D' (-2.26 V).
V = k * Q / r
where V is the potential, k is the electrostatic constant (9 x 10^9 N m^2/C^2), Q is the total charge on the shell, and r is the distance from the center of the shell.
Given that the surface charge density is 30 nC/m^2 and the radius of the shell is 6 mm, we can calculate the total charge on the shell:
Q = surface charge density * surface area
The surface area of a sphere is given by:
surface area = 4 * π * r^2
Plugging in the values, we have:
Q = (30 x 10^-9 C/m^2) * (4 * π * (6 x 10^-3 m)^2)
= (30 x 10^-9 C/m^2) * (4 * 3.14159 * (6 x 10^-3 m)^2)
= (30 x 10^-9 C/m^2) * (4 * 3.14159 * 36 x 10^-6 m^2)
= (30 x 10^-9 C/m^2) * (452.389 x 10^-6 C)
= 13.5717 x 10^-9 C
= 13.5717 nC
Now, let's calculate the potential at point A (r = 2 cm) and point B (r = 3 cm) using the formula mentioned earlier.
For point A:
V(A) = k * Q / r(A)
= (9 x 10^9 N m^2/C^2) * (13.5717 x 10^-9 C) / (2 x 10^-2 m)
= 6.78585 V
For point B:
V(B) = k * Q / r(B)
= (9 x 10^9 N m^2/C^2) * (13.5717 x 10^-9 C) / (3 x 10^-2 m)
= 4.5239 V
Therefore, the potential difference (VAB) between points A and B is given by:
VAB = V(B) - V(A)
= 4.5239 V - 6.78585 V
= -2.26195 V
≈ -2.26 V
The correct answer is option 'D' (-2.26 V).
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A uniform surface charge density of 30 nC /m2is present on the spherical surface r = 6 mm in free space.The VAB between A (r = 2 cm, θ = 350 , Ø =550 ) and B(r = 3 cm, θ = 400, Ø =900 )a)2.03 Vb)10.17 Vc)4.07 mVd)-10.17 VCorrect answer is option 'D'. Can you explain this answer?
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