# 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