You are given charge +q at the origin. Consider a sphere s with centre (2,0,0)
Answers
Explanation:
Consider a spherical cap (a part of spherical surface) bounded by the ... (Does it depend on where on the spherical cap you look at?) 3. ... We will calculate the electric flux through
Answer:
The magnitude of the flux will be equal.
Explanation:
Question:
You are given a charge + Q at the origin O. Consider a sphere S with centre (√2,0, 0) of radius √2 m. Consider another sphere of radius √2 m centered at the origin.
Consider the spherical caps (i) PSQ (ii) PRO (iii) PWQ, with normals outward to the respective spheres, and (iv) the flat circle PTQ with normal along the x-axis.
(a) What is the sign of electric flux through each of the surface (i) - (iv)?
(b) What is the relation between the magnitudes of fluxes through surfaces (i) - (iv)?
Solution:
Given that:
The charge of + Q at the origin O.
we can consider a sphere S with centre (√2,0, 0) of radius √2 m. Consider another sphere of radius √2 m centered at the origin.
To find:
i) The sign of electric flux through each surface
ii) The relation between magnitude of fluxes through surfaces.
Solution:
i) The outwards of the normal will become positive and the flux is defined by E. infinite
The flux that passes through:
1) PSQ is negative.
d .E are antiparallel at 180 degree
2) PRQ is positive
The angle between d and E is called acute.
3) PWQ will become positive.
The angle will be acute.
4) PTQ will become positive.
The normal that is along the x axis will be an acute angle at x-axis for PTQ.
ii)The magnitude of the flux that will go through the PSQ and PTQ that are equal. The PTQ will be projected area of PSQ.
It is given by:
PTQ = PSQ cos θ.
Let us consider,
The flux through PTQ is,
Area of PTQ.E.
The flux through PSQ is,
Area of PSQ. E
From the figure given, we can conclude that all the magnitude of the flux is equal.
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