Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of
Question:
Answers
N2 = ¼N1
Explanation:
First of all, let's define the terms;
N1 = number of electric field lines going through the sphere of radius R
N2 = number of electric field lines going through the sphere of radius 2R
Q = the charge enclosed at the centre of concentric spheres
ε_o = a constant known as "permittivity of the free space"
E1 = Electric field in the sphere of radius R.
E2 = Electric field in the sphere of radius 2R.
A1 = Area of sphere of radius R.
A2 = Area of sphere of radius 2R
Now, from Gauss's law, the electric flux through the sphere of radius R is given by;
Φ = Q/ε_o
We also know that;
Φ = EA
Thus;
E1 × A1 = Q/ε_o
E1 = Q/(ε_o × A1)
Where A1 = 4πR²
E1 = Q/(ε_o × 4πR²)
Similarly, for the sphere of radius 2R,we have;
E2 = Q/(ε_o × 4π(2R)²)
Factorizing out to get;
E2 = ¼Q/(ε_o × 4πR²)
Comparing E2 with E1, we arrive at;
E2 = ¼E1
Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;
N2 = ¼N1
Answer : no2. I’m pretty sure
Between p and q is the one
Do you possibly have a picture of the equation?
Start at 0 and move to point P. To do this, we simply move left 1/8 of a unit. Similarly, we move 1/8 of a unit to go from 0 to point Q
So the distance from P to 0 is the same as the distance from Q to 0.
Therefore, the absolute values are the same
Choice A) Jay is correct because each point is 1/8 a unit away from 0