Boron I think look at the periodic table
Answer:
1/9 of that just outside the smaller sphere
Explanation:
The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:

where
k is the Coulomb's constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is

The second sphere has a radius which is 3 times that of the smaller sphere:

So, the electric field just outside the second sphere is

So, the correct answer is 1/9.
Answer:
<em><u>solution</u></em>
<em>3</em><em>0</em><em>8</em><em>=</em><em>2</em><em>0</em><em> </em><em>swings</em><em> </em>
<em> </em><em>?</em><em>:</em><em>:</em><em>:</em><em> </em><em>=</em><em>1</em>
<em>(</em><em> </em><em>3</em><em>0</em><em>8</em><em>×</em><em>1</em><em>)</em><em>÷</em><em>2</em><em>0</em>
<em>3</em><em>0</em><em>8</em><em>÷</em><em>2</em><em>0</em>
<em>1</em><em>5</em><em>4</em><em>÷</em><em>1</em><em>0</em>
<em>=</em><em>1</em><em>5</em><em>.</em><em>4</em>
<em>=</em>15.4
Is impossible say how many stars are in the sky haha